MATH 237

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MATH 237 FINAL: SOS EXAM AID SESSION NIALL MACGILLIVRAY

December 14, 2010 This review package is comprehensive of the material that will be tested on your final examination. For those who attended the SOS review session, it should be apparent that this package starts with material covered from the very beginning of the course and ends with the material covered in December. It is my recommendation though that you focus your studies on the latter half of the midterm material. From past experience, upwards of 85% of the exam is post midterm material with a strong emphasis on double and triple integrals. Good Luck! REVIEW: Example 1: Let f : R2 → R where ( f (x, y) =

2y x2 +y 2

√x

: (x, y) 6= (0, 0)

0 . a) Determine the domain and range of f . b) Draw the level curves f = 1 and f = 0 c) Determine where f is continuous. d) Determine where f is differentiable.

: (x, y) = (0, 0)

2

SOLUTION. a) We know x2x+yy 2 is only undefined when the denominator is vanishing, that is when (x, y) is the origin. However, f itself is also defined at the origin, so it is clear that the domain is all of R2 . For the range, we note that f (0, 0) = 0. For any real number 2 k 6= 0, we check if it is possible that f = k. We can solve x2x+yy 2 = k to get the quadratic 0 = ky 2 − x2 y + kx2 , so by the quadratic formula, it follows that: 2

√

4

2

y = x ± x2k−4kx . Notice this has a real solution so long as x4 − 4kx2 ≥ 0. If we choose x2 ≥ 4k 2 , this is 2 satisfied, and so we can find (x, y) with x2x+yy 2 = k. Hence, we conclude the range of f is R. b) If f = 0, then we have x2 y = 0 or the origin. Thus the level curve of f = 0 are the lines x = 0 and y = 0. If f = 1, then we have x2 y = x2 + y 2 , or x2 (y − 1) = y 2 . Notice if 1

WATERLOO SOS E XAM -AID: MATH237 đ?‘Ś = 1, then đ?‘Ľ2 = đ?‘Ľ2 + 1, a contradiction, so đ?‘Ś ∕= 1. Then solving for đ?‘Ľ, we get đ?‘Ľ = Âąâˆš

âˆŁđ?‘ŚâˆŁ . đ?‘Śâˆ’1

Moreover, notice đ?‘Ľ2 + đ?‘Ś 2 ≼ 0 and đ?‘Ľ2 ≼ 0, so since đ?‘Ľ2 đ?‘Ś = đ?‘Ľ2 + đ?‘Ś 2 , we must have đ?‘Ś ≼ 0, so that đ?‘Ś . đ?‘Ľ = Âąâˆš đ?‘Śâˆ’1 Consider the case đ?‘Ľ ≼ 0, and the entire picture will come from symmetry. By 1-dimensional optimization, đ?‘Ľ is minimized when đ?‘Ś = 2, giving đ?‘Ľ = 2. As đ?‘Ś → 0, đ?‘Ľ → 0, and as đ?‘Ś → ∞, √ đ?‘Ľ → ∞, growing at a rate of approximately đ?‘Ś. We thus get the following picture:

(c) Recall the coordinate functions � and � are continuous, and sums and products of continuous functions are continuous, so that polynomials are continuous. Furthermore, quotients of contin2 � uous functions are continuous where the denominator is not 0, so that �2�+� 2 is continuous except possibly at (0, 0). To determine continuity at (0, 0), we must check if lim(�,�)→(0,0) � (�, �) = � (0, 0) = 0. Taking lines � = �� we get �2 �� �� = lim = 0. 2 2 (�,�)→(0,0) � + (��) (�,�)→(0,0) 1 + � lim

We then guess the limit is đ??ż = 0. Since đ?‘Ľ2 , đ?‘Ś 2 ≼ 0, âˆŁđ?‘Ľ2 âˆŁ ≤ âˆŁđ?‘Ľ2 + đ?‘Ś 2 âˆŁ. Then

đ?‘Ľ2 đ?‘Ś

âˆŁđ?‘Ľ2 + đ?‘Ś 2 âˆŁâˆŁđ?‘ŚâˆŁ

≤ − 0 = âˆŁđ?‘ŚâˆŁ.

đ?‘Ľ2 + đ?‘Ś 2

âˆŁđ?‘Ľ2 + đ?‘Ś 2 âˆŁ đ?‘Ľ2 đ?‘Ś = 0 by the Squeeze Theorem. Hence, đ?‘“ is also + đ?‘Ś2 (đ?‘Ľ,đ?‘Ś)→(0,0) continuous at (0, 0), so đ?‘“ is continuous on all of â„?2 .

Since

lim

âˆŁđ?‘ŚâˆŁ = 0 we get

lim

(�,�)→(0,0) �2

(d) Away from (0, 0), we have by our usual differentiation rules ∂đ?‘“ 2đ?‘Ľđ?‘Ś 2đ?‘Ľ3 đ?‘Ś = 2 − 2 , 2 ∂đ?‘Ľ đ?‘Ľ +đ?‘Ś (đ?‘Ľ + đ?‘Ś 2 )2 ∂đ?‘“ đ?‘Ľ2 2đ?‘Ľ2 đ?‘Ś 2 = 2 − 2 . 2 ∂đ?‘Ś đ?‘Ľ +đ?‘Ś (đ?‘Ľ + đ?‘Ś 2 )2 Since these are continuous away from (0, 0), đ?‘“ is differentiable away from (0, 0). Recall the definition of differentiability: đ?‘“ is differentiable if and only if there is a linear function đ??ż(đ?‘Ľ) = đ?‘“ (đ?‘Ž, đ?‘?) + đ?‘?(đ?‘Ľ − đ?‘Ž) + đ?‘‘(đ?‘Ś − đ?‘?) such that lim

�→�

âˆŁđ?‘…1,đ?‘Ž (đ?‘Ľ)âˆŁ = 0, âˆĽđ?‘Ľ − đ?‘ŽâˆĽ 2

WATERLOO SOS E XAM -AID: MATH237 where đ?‘…1,đ?‘Ž (đ?‘Ľ) = đ?‘“ (đ?‘Ľ) − đ??ż(đ?‘Ľ). By a theorem, it suffices to check the definition using the linear approximation, đ??ż(đ?‘Ž,đ?‘?) (đ?‘Ľ, đ?‘Ś) = ∂đ?‘“ đ?‘“ (đ?‘Ž, đ?‘?) + ∂đ?‘“ ∂đ?‘Ľ (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž) + ∂đ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ś − đ?‘?). For (0, 0), we must use the definition of partial derivatives: 0 đ?‘“ (0 + â„Ž, 0) − đ?‘“ (0, 0) ∂đ?‘“ 2 (0, 0) = lim = lim â„Ž = lim 0 = 0, ℎ→0 ℎ→0 â„Ž ℎ→0 ∂đ?‘Ľ â„Ž 0 ∂đ?‘“ đ?‘“ (0, 0 + â„Ž) − đ?‘“ (0, 0) 2 (0, 0) = lim = lim â„Ž = lim 0 = 0. ℎ→0 ℎ→0 â„Ž ℎ→0 ∂đ?‘Ś â„Ž

Since đ?‘“ (0, 0) = 0, the linear approximation is đ??ż(0,0) (đ?‘Ľ, đ?‘Ś) = 0, and the error is { 2 đ?‘Ľ đ?‘Ś if (đ?‘Ľ, đ?‘Ś) ∕= (0, 0), 2 2 đ?‘…1,(0,0) (đ?‘Ľ, đ?‘Ś) = đ?‘“ (đ?‘Ľ, đ?‘Ś) − đ??ż(0,0) (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ +đ?‘Ś 0 if (đ?‘Ľ, đ?‘Ś) = (0, 0), and the magnitude of displacement is âˆĽ(đ?‘Ľ, đ?‘Ś) − (0, 0)âˆĽ =

√ đ?‘Ľ2 + đ?‘Ś 2 .

Now along the line đ?‘Ś = đ?‘Ľ, we have 2

1 âˆŁ 2đ?‘Ľ đ?‘Ľ 2 âˆŁ âˆŁđ?‘…1,(0,0) (đ?‘Ľ, đ?‘Ľ)âˆŁ âˆŁđ?‘ĽâˆŁ 1 = √ đ?‘Ľ +đ?‘Ľ = √2 = √ ∕= 0, âˆĽ(đ?‘Ľ, đ?‘Ľ) − (0, 0)âˆĽ 2âˆŁđ?‘ĽâˆŁ 2 2 đ?‘Ľ2 + đ?‘Ľ2

so by definition, đ?‘“ is not differentiable at (0, 0).

E XAMPLE 2. Let đ?‘“ (đ?‘Ľ, đ?‘Ś) = tan đ?‘Ľ − cos đ?‘Ś. (a) Find the first and second partial derivatives of đ?‘“ . đ?œ‹ (b) Find the linear approximation of đ?‘“ at ( 3đ?œ‹ 4 , 2 ). đ?œ‹ (c) Find the second degree Taylor polynomial of đ?‘“ at ( 3đ?œ‹ 4 , 2 ). đ?œ‹ 2 2 (d) Show that the error in the linear approximation is at most 2[(đ?‘Ľâˆ’ 3đ?œ‹ 4 ) + (đ?‘Ś − 2 ) ] if đ?œ‹ and 0 ≤ đ?‘Ś ≤ 2 .

S OLUTION.

3đ?œ‹ 4

≤đ?‘Ľâ‰¤đ?œ‹

(a) Differentiating, đ?‘“đ?‘Ľ = sec2 đ?‘Ľ, đ?‘“đ?‘Ľđ?‘Ľ = 2 sec2 đ?‘Ľ tan đ?‘Ľ,

đ?‘“đ?‘Ś = sin đ?‘Ś. đ?‘“đ?‘Ľđ?‘Ś = 0,

đ?‘“đ?‘Śđ?‘Ś = cos đ?‘Ś.

(b) We have 3đ?œ‹ đ?œ‹ 3đ?œ‹ đ?œ‹ đ?œ‹ (đ?‘Ľ, đ?‘Ś) = đ?‘“ ( đ??ż( 3đ?œ‹ 4 , 2 ) + đ?‘“đ?‘Ľ ( 4 , 2 )(đ?‘Ľ − 4 ,2) đ?œ‹ = tan( 3đ?œ‹ 4 ) − cos( 2 ) +

3đ?œ‹ 3đ?œ‹ đ?œ‹ đ?œ‹ 4 ) + đ?‘“đ?‘Ś ( 4 , 2 )(đ?‘Ś − 2 ) 3đ?œ‹ đ?œ‹ đ?œ‹ sec2 ( 3đ?œ‹ 4 )(đ?‘Ľ − 4 ) + sin( 2 )(đ?‘Ś − 2 )

3

= −1 + 2(đ?‘Ľ −

3đ?œ‹ 4 )

+ (đ?‘Ś − đ?œ‹2 ).

WATERLOO SOS E XAM -AID: MATH237 (c) By (a), ( ∇đ?‘“

3đ?œ‹ đ?œ‹ , 4 2

)

( = (2, 1),

đ??ťđ?‘“

Then (

3đ?œ‹ đ?‘ƒ2,đ?‘Ž (đ?‘Ľ, đ?‘Ś) = −1 + 2 đ?‘Ľ − 4

)

3đ?œ‹ đ?œ‹ , 4 2

)

[ =

−4 0

0 0

] .

[ ( )2 ] 3đ?œ‹ đ?œ‹) 1 + −4 đ?‘Ľ − . + đ?‘Śâˆ’ 2 2 4 (

đ?œ‹ (d) For 3đ?œ‹ 4 ≤ đ?‘Ľ ≤ đ?œ‹ and 0 ≤ đ?‘Ś ≤ 2 , đ?‘“ has continuous second partial derivatives, so we can apply đ?œ‹ Taylor’s theorem to get that there exists a point đ?‘? on the line segment joining ( 3đ?œ‹ 4 , 2 ) and đ?‘Ľ such that đ?œ‹ (đ?‘Ľ, đ?‘Ś) + đ?‘… đ?œ‹ (đ?‘Ľ, đ?‘Ś), đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ??ż( 3đ?œ‹ 1,( 3đ?œ‹ 4 ,2) 4 ,2)

where

[ ]

( ( )2 )

1 (đ?œ‹ ) )2

3đ?œ‹ 3đ?œ‹ ( đ?œ‹

đ?œ‹ (đ?‘Ľ, đ?‘Ś)âˆŁ =

âˆŁđ?‘…1,( 3đ?œ‹ đ?‘“đ?‘Ľđ?‘Ľ (đ?‘?) đ?‘Ľ − + 2đ?‘“đ?‘Ľđ?‘Ś (đ?‘?) đ?‘Ľ − − 2 + đ?‘“đ?‘Śđ?‘Ś (đ?‘?) −2

, 4 ,2)

2 4 4 2 2 đ?œ‹ (đ?‘Ľ, đ?‘Ś)âˆŁ. Since the theorem does so that the error in the linear approximation is given by âˆŁđ?‘…1,( 3đ?œ‹ 4 ,2) not give us the value of đ?‘?, we will simply find an upper bound for this error. On 3đ?œ‹ 4 ≤ đ?‘Ľ ≤ đ?œ‹, we have that cos đ?‘Ľ is decreasing and non-positive so that sec2 đ?‘Ľ is decreasing and non-negative. (Indeed, if đ?‘Ľ ≤ đ?‘Ś, then cos đ?‘Ľ ≼ cos đ?‘Ś. Since cos đ?‘Ľ ≤ 0, cos2 đ?‘Ľ ≤ cos đ?‘Ľ cos đ?‘Ś, and since cos đ?‘Ś ≤ 0, cos đ?‘Ľ cos đ?‘Ś ≤ cos2 đ?‘Ś. Then cos2 đ?‘Ľ ≤ cos2 đ?‘Ś, so that cos2 đ?‘Ľ is increasing. Reciprocating reverses this inequality, so that sec2 đ?‘Ľ is decreasing. Clearly squaring forces it to be non-negative.) As 2 well, on 3đ?œ‹ 4 ≤ đ?‘Ľ ≤ đ?œ‹ we have that tan đ?‘Ľ is increasing and non-positive. Thus, sec đ?‘Ľ tan đ?‘Ľ is 2 2 increasing and non-positive. (For đ?‘Ľ ≤ đ?‘Ś, we have sec đ?‘Ľ ≼ sec đ?‘Ś and tan đ?‘Ľ ≤ tan đ?‘Ś. Then since tan đ?‘Ľ ≤ 0, sec2 đ?‘Ľ tan đ?‘Ľ ≤ sec2 đ?‘Ś tan đ?‘Ľ, and since sec2 đ?‘Ś ≼ 0, sec2 đ?‘Ś tan đ?‘Ľ ≤ sec2 đ?‘Ś tan đ?‘Ś. Clearly it is 3đ?œ‹ non-positive.) In particular, âˆŁ2 sec2 đ?‘Ľ tan đ?‘ŚâˆŁ takes a maximum on 3đ?œ‹ 4 ≤ đ?‘Ľ ≤ đ?œ‹ at the end point 4 , so that

3đ?œ‹

2 3đ?œ‹

âˆŁđ?‘“đ?‘Ľđ?‘Ľ âˆŁ ≤ 2 sec tan = âˆŁ2(2)(−1)âˆŁ = 4. 4 4

We have đ?‘“đ?‘Ľđ?‘Ś = 0, and finally âˆŁđ?‘“đ?‘Śđ?‘Ś âˆŁ = âˆŁ cos đ?‘ŚâˆŁ ≤ 1 everywhere. Then by triangle inequality, [ ]

( ( )2 ) )

(đ?œ‹ )2

1 3đ?œ‹ 3đ?œ‹ ( đ?œ‹

đ?œ‹ (đ?‘Ľ, đ?‘Ś)âˆŁ ≤ âˆŁđ?‘…1,( 3đ?œ‹ âˆŁđ?‘“đ?‘Ľđ?‘Ľ (đ?‘?)âˆŁ đ?‘Ľ − + 2âˆŁđ?‘“đ?‘Ľđ?‘Ś (đ?‘?)âˆŁ đ?‘Ľ − − 2 + âˆŁđ?‘“đ?‘Śđ?‘Ś (đ?‘?)âˆŁ −2 4 ,2) 2 4 4 2 2 [ ( [ ( ] ] )2 ( )2 )2 (đ?œ‹ )2 1 3đ?œ‹ đ?œ‹ 3đ?œ‹ 1 ≤ 4 đ?‘Ľâˆ’ + −2 ≤ 4 đ?‘Ľâˆ’ +4 −2 2 4 2 2 4 2 [( ] )2 ( đ?œ‹ )2 3đ?œ‹ + đ?‘Śâˆ’ , =2 đ?‘Ľâˆ’ 4 2 as required.

E XAMPLE 3. Define � : �2 → � by � (�, �) = �3 + �2 � + � 2 . (a) Find the directional derivative of � at the point (2, 1) in the direction (5, 12). (b) Find the largest rate of change of � at the point (2, 1), and the direction in which it occurs.

4

WATERLOO SOS E XAM -AID: MATH237 S OLUTION. (a) Since đ?‘“ is differentiable at (2, 1), a theorem tells us that đ??ˇđ?‘˘Ë† đ?‘“ (đ?‘Ž) = ∇đ?‘“ (đ?‘Ž) â‹… đ?‘˘ ˆ. We must first normalize: ( ) (5, 12) 5 12 đ?‘˘ ˆ= = , . âˆĽ(5, 12)âˆĽ 13 13 We have ∇đ?‘“ = (3đ?‘Ľ2 + 2đ?‘Ľđ?‘Ś, đ?‘Ľ2 + 2đ?‘Ś), so that đ??ˇđ?‘˘Ë† đ?‘“ ((2, 1)) = ∇đ?‘“ (2, 1) â‹… đ?‘˘ ˆ ) ( 5 12 , = (3(2)2 + 2(2)(1), (2)2 + 2(1)) â‹… 13 13 ( ) 5 12 80 72 152 = (16, 6) â‹… , = + = . 13 13 13 13 13 (b) Notice đ?‘“ is differentiable at (2, 1) and ∇đ?‘“ (2, 1) = (3đ?‘Ľ2 + 2đ?‘Ľđ?‘Ś, đ?‘Ľ2 + 2đ?‘Ś)âˆŁ(đ?‘Ľ,đ?‘Ś)=(2,1) = (16, 6) ∕= (0, 0). Then by a theorem, the largest rate of change of đ?‘“ is âˆĽâˆ‡đ?‘“ (2, 1)âˆĽ = âˆĽ(16, 6)âˆĽ =

√

256 + 36 =

√

√ 292 = 2 73,

and the direction in which it occurs is ∇đ?‘“ (2, 1) = (16, 6).

E XAMPLE 4. Suppose � = � (�, �), with � =

√

đ?‘ 2 + đ?‘Ą2 , đ?‘Ś = đ?‘’đ?‘ đ?‘Ą , and đ?‘“ is twice differentiable.

(a) Find the chain rule for đ?‘˘đ?‘ . (b) Find the chain rule for đ?‘˘đ?‘ đ?‘Ą . S OLUTION.

(a) The dependence diagram is � �

đ?‘Ľ đ?‘

đ?‘

đ?‘Ą

đ?‘Ą

Then by our algorithm, đ?‘˘đ?‘ (đ?‘ , đ?‘Ą) = đ?‘“đ?‘Ľ (đ?‘Ľ, đ?‘Ś)đ?‘Ľđ?‘ (đ?‘ , đ?‘Ą) + đ?‘“đ?‘Ś (đ?‘Ľ, đ?‘Ś)đ?‘Śđ?‘ (đ?‘ , đ?‘Ą) = đ?‘“đ?‘Ľ (đ?‘Ľ, đ?‘Ś) √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Ś (đ?‘Ľ, đ?‘Ś)đ?‘Ąđ?‘’đ?‘ đ?‘Ą . (b) By (a), đ?‘˘đ?‘ is dependent on đ?‘Ľ, đ?‘Ś, đ?‘ , đ?‘Ą, where đ?‘Ľ, đ?‘Ś are each dependent on đ?‘ , đ?‘Ą. We get the dependence diagram

5

WATERLOO SOS E XAM -AID: MATH237 đ?‘˘đ?‘ đ?‘Ś

đ?‘Ľ đ?‘

đ?‘Ą

đ?‘

đ?‘

đ?‘Ą

đ?‘Ą

Then by our algorithm, đ?‘˘đ?‘ đ?‘Ą = We then calculate

∂đ?‘˘đ?‘ ∂đ?‘Ľ

∂đ?‘˘đ?‘ ∂đ?‘Ľ ∂đ?‘˘đ?‘ ∂đ?‘Ś ∂đ?‘˘đ?‘ + + . ∂đ?‘Ľ ∂đ?‘Ą ∂đ?‘Ś ∂đ?‘Ą ∂đ?‘Ą

in the usual way. Since đ?‘ and đ?‘Ą are fixed, ∂đ?‘“đ?‘Ľ √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Ś đ?‘Ąđ?‘’đ?‘ đ?‘Ą ∂đ?‘˘đ?‘ = ∂đ?‘Ľ ∂đ?‘Ľ ∂đ?‘“đ?‘Ľ đ?‘ ∂đ?‘“đ?‘Ś đ?‘ đ?‘Ą √ = + đ?‘Ąđ?‘’ 2 2 ∂đ?‘Ľ đ?‘ + đ?‘Ą ∂đ?‘Ľ = đ?‘“đ?‘Ľđ?‘Ľ √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Śđ?‘Ľ đ?‘Ąđ?‘’đ?‘ đ?‘Ą .

Similarly, ∂đ?‘˘đ?‘ = đ?‘“đ?‘Ľđ?‘Ś √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Śđ?‘Ś đ?‘Ąđ?‘’đ?‘ đ?‘Ą . ∂đ?‘Ś Finally, holding đ?‘Ľ, đ?‘Ś, đ?‘ fixed gives ( ) ∂đ?‘˘đ?‘ = đ?‘“đ?‘Ľ − √ 2đ?‘ đ?‘Ą 2 3 + đ?‘“đ?‘Ś (đ?‘’đ?‘ đ?‘Ą + đ?‘ đ?‘Ąđ?‘’đ?‘ đ?‘Ą ). đ?‘ +đ?‘Ą ∂đ?‘Ą Then ) ( ) ( ( đ?‘˘đ?‘ đ?‘Ą = đ?‘“đ?‘Ľđ?‘Ľ √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Śđ?‘Ľ đ?‘Ąđ?‘’đ?‘ đ?‘Ą √đ?‘ 2đ?‘Ą+đ?‘Ą2 + đ?‘“đ?‘Ľđ?‘Ś √đ?‘ 2đ?‘ +đ?‘Ą2 + đ?‘“đ?‘Śđ?‘Ś đ?‘Ąđ?‘’đ?‘ đ?‘Ą đ?‘ đ?‘’đ?‘ đ?‘Ą + đ?‘“đ?‘Ľ − √

2

đ?‘ đ?‘Ą 3 đ?‘ 2 +đ?‘Ą2

)

+ đ?‘“đ?‘Ś (đ?‘’đ?‘ đ?‘Ą + đ?‘ đ?‘Ąđ?‘’đ?‘ đ?‘Ą ).

Generalizations of Taylor Polynomials/Taylor’s Theorem (8.3)

∂ ∂ and đ??ˇ2 = ∂đ?‘Ś . We may formally multiply to Recall the operator notation for the partial derivative: đ??ˇ1 = ∂đ?‘Ľ get [(đ?‘Ľ − đ?‘Ž)đ??ˇ1 + (đ?‘Ś − đ?‘?)đ??ˇ2 ]2 = (đ?‘Ľ − đ?‘Ž)2 đ??ˇ12 + 2(đ?‘Ľ − đ?‘Ž)(đ?‘Ś − đ?‘?)đ??ˇ1 đ??ˇ2 + (đ?‘Ś − đ?‘?)2 đ??ˇ22 ,

where đ??ˇ12 = đ??ˇ1 đ??ˇ1 means to apply đ??ˇ1 twice (take the second partial derivative with respect to the first variable). In this notation, we can define the đ?‘˜th degree Taylor polynomial. D EFINITION 2.1. Define đ?‘ƒ1,đ?‘Ž (đ?‘Ľ) = đ?‘“ (đ?‘Ž) + [(đ?‘Ľ − đ?‘Ž)đ??ˇ1 + (đ?‘Ś − đ?‘?)đ??ˇ2 ]đ?‘“ (đ?‘Ž). For an integer đ?‘˜ ≼ 2, the kth degree Taylor polynomial is given by đ?‘ƒđ?‘˜,đ?‘Ž (đ?‘Ľ) = đ?‘ƒđ?‘˜âˆ’1,đ?‘Ž +

1 [(đ?‘Ľ − đ?‘Ž)đ??ˇ1 + (đ?‘Ś − đ?‘?)đ??ˇ2 ]đ?‘˜ đ?‘“ (đ?‘Ž), đ?‘˜!

where the expression [(đ?‘Ľ − đ?‘Ž)đ??ˇ1 + (đ?‘Ś − đ?‘?)đ??ˇ2 ]đ?‘˜ may be formally expanded using the binomial theorem. 6

WATERLOO SOS E XAM -AID: MATH237 T HEOREM 2.2. Let đ?‘“ : â„?2 → â„?, đ?‘“ ∈ đ??ś đ?‘˜+1 at every point in the line segment joining đ?‘Ž and đ?‘Ľ. Then there exists a point đ?‘? on the line segment joining đ?‘Ž and đ?‘Ľ such that đ?‘“ (đ?‘Ľ) = đ?‘ƒđ?‘˜,đ?‘Ž (đ?‘Ľ) + đ?‘…đ?‘˜,đ?‘Ž (đ?‘Ľ), where đ?‘…đ?‘˜,đ?‘Ž (đ?‘Ľ) =

[ ]đ?‘˜+1 1 ∂ ∂ đ?‘“ (đ?‘?). (đ?‘Ľ − đ?‘Ž) + (đ?‘Ś − đ?‘?) (đ?‘˜ + 1)! ∂đ?‘Ľ ∂đ?‘Ś

C OROLLARY 2.3. If đ?‘“ ∈ đ??ś đ?‘˜ in some neighborhood of đ?‘Ž, then âˆŁđ?‘“ (đ?‘Ľ) − đ?‘ƒđ?‘˜,đ?‘Ž (đ?‘Ľ)âˆŁ = 0. đ?‘Ľâ†’đ?‘Ž âˆĽđ?‘Ľ − đ?‘ŽâˆĽđ?‘˜ lim

C OROLLARY 2.4. If đ?‘“ ∈ đ??ś đ?‘˜+1 in some closed neighborhood đ?‘ (đ?‘Ž) of đ?‘Ž, then there exists a constant đ?‘€ > 0 such that âˆŁđ?‘“ (đ?‘Ľ) − đ?‘ƒđ?‘˜,đ?‘Ž (đ?‘Ľ)âˆŁ ≤ đ?‘€ âˆĽđ?‘Ľ − đ?‘ŽâˆĽđ?‘˜+1 , for all đ?‘Ľ ∈ đ?‘ (đ?‘Ž). We can further generalize this by looking at đ?‘“ : â„?đ?‘› → â„?, and using the operator notation, to get [(đ?‘Ľ1 − đ?‘Ž1 )đ??ˇ1 + â‹… â‹… â‹… + (đ?‘Ľđ?‘› − đ?‘Žđ?‘› )đ??ˇđ?‘› ] = [(đ?‘Ľ − đ?‘Ž) â‹… ∇], where ∇ = (đ??ˇ1 , . . . , đ??ˇđ?‘› ), to replace the differential operator [(đ?‘Ľ − đ?‘Ž)đ??ˇ1 + (đ?‘Ś − đ?‘?)đ??ˇ2 ].

đ?œ‹ E XERCISE 1. Let đ?‘“ (đ?‘Ľ, đ?‘Ś) = tan đ?‘Ľ − sin đ?‘Ľ cos đ?‘Ś. Find the third degree Taylor polynomial of đ?‘“ at ( 3đ?œ‹ 4 , 2 ).

3

Local Extrema and Critical Points (9.1)

D EFINITION 3.1. Suppose đ?‘“ : â„?2 → â„?. A point (đ?‘Ž, đ?‘?) is a: (a) local minimum point of đ?‘“ if đ?‘“ (đ?‘Ľ, đ?‘Ś) ≼ đ?‘“ (đ?‘Ž, đ?‘?) for all (đ?‘Ľ, đ?‘Ś) in some neighborhood of (đ?‘Ž, đ?‘?). (b) local maximum point of đ?‘“ if đ?‘“ (đ?‘Ľ, đ?‘Ś) ≤ đ?‘“ (đ?‘Ž, đ?‘?) for all (đ?‘Ľ, đ?‘Ś) in some neighborhood of (đ?‘Ž, đ?‘?). (c) critical point of đ?‘“ if đ?‘“đ?‘Ľ (đ?‘Ž, đ?‘?) = 0 = đ?‘“đ?‘Ś (đ?‘Ž, đ?‘?), or at least one of đ?‘“đ?‘Ľ or đ?‘“đ?‘Ś does not exist at (đ?‘Ž, đ?‘?). If (đ?‘Ž, đ?‘?) is a local minimum or local maximum point of đ?‘“ , we may call (đ?‘Ž, đ?‘?) a local extremum point of đ?‘“ . T HEOREM 3.2. Let đ?‘“ : â„?2 → â„?. If (đ?‘Ž, đ?‘?) is a local extremum point of đ?‘“ , then (đ?‘Ž, đ?‘?) is a critical point of đ?‘“ . 7

WATERLOO SOS E XAM -AID: MATH237 Notice this doesn’t mean that every critical point is either a local minimum or a local maximum point. It is possible for the partials to vanish (or not exist) at a point, yet the function is neither the smallest nor the largest value in a neighborhood around the point. D EFINITION 3.3. A critical point (đ?‘Ž, đ?‘?) of đ?‘“ : â„?2 → â„? is a saddle point of đ?‘“ if in every neighborhood of (đ?‘Ž, đ?‘?) there exist points (đ?‘Ľ1 , đ?‘Ś1 ) and (đ?‘Ľ2 , đ?‘Ś2 ) such that đ?‘“ (đ?‘Ľ1 , đ?‘Ś1 ) > đ?‘“ (đ?‘Ž, đ?‘?),

đ?‘“ (đ?‘Ľ2 , đ?‘Ś2 ) < đ?‘“ (đ?‘Ž, đ?‘?).

E XAMPLE 5. Find the critical points of the following functions and determine if they are local minimum points or local maximum points. đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 + đ?‘Ś 2 ,

đ?‘”(đ?‘Ľ, đ?‘Ś) = −đ?‘Ľ2 − đ?‘Ś 2 ,

â„Ž(đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 − đ?‘Ś 2 .

S OLUTION. We have đ?‘“đ?‘Ľ = 2đ?‘Ľ and đ?‘“đ?‘Ś = 2đ?‘Ś, so (0, 0) is the only critical point of đ?‘“ , and đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 + đ?‘Ś 2 ≼ 0 = đ?‘“ (0, 0), so (0, 0) is a local minimum point of đ?‘“ . We have đ?‘”đ?‘Ľ = −2đ?‘Ľ and đ?‘”đ?‘Ś = −2đ?‘Ś, so (0, 0) is the only critical point of đ?‘”, and đ?‘”(đ?‘Ľ, đ?‘Ś) = −đ?‘Ľ2 − đ?‘Ś 2 ≤ 0 = đ?‘”(0, 0), so (0, 0) is a local maximum point of đ?‘”. We have â„Žđ?‘Ľ = 2đ?‘Ľ and â„Žđ?‘Ś = −2đ?‘Ś, so (0, 0) is the only critical point of â„Ž, and for đ?‘Ľ ∕= 0, đ?‘Ś ∕= 0, â„Ž(đ?‘Ľ, 0) = đ?‘Ľ2 > 0 = â„Ž(0, 0),

â„Ž(0, đ?‘Ś) = −đ?‘Ś 2 < 0 = â„Ž(0, 0)

so (0, 0) is neither a local minimum point nor a local maximum point of đ?‘“ . In general, it is not easy to classify a critical point, but we will have a test for this later.

E XAMPLE 6. Define đ?‘“ : â„?2 → â„? by đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ3 + đ?‘Ś 3 + 3đ?‘Ľ2 − 3đ?‘Ś 2 − 2. Find the critical points of đ?‘“ . S OLUTION. Notice đ?‘“đ?‘Ľ and đ?‘“đ?‘Ś exist everywhere. We then solve (1)

0 = đ?‘“đ?‘Ľ = 3đ?‘Ľ2 + 6đ?‘Ľ,

(2)

0 = đ?‘“đ?‘Ś = 3đ?‘Ś 2 − 6đ?‘Ś.

(2) yields đ?‘Ľ = 0 or đ?‘Ľ = −2, while (1) yields đ?‘Ś = 0 or đ?‘Ś = 2. Thus, the critical points are (0, 0),

(0, 2),

(−2, 0),

E XERCISE 2. Find the critical points of the following functions. (a) đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ3 + 6đ?‘Ľ2 − 12đ?‘Ľđ?‘Ś + 9đ?‘Ľ + 3đ?‘Ś 2 . (b) đ?‘”(đ?‘Ľ, đ?‘Ś) = âˆŁđ?‘Ľ2 − đ?‘ĽâˆŁ + đ?‘Ś 2 .

8

(−2, 2).

WATERLOO SOS E XAM -AID: MATH237

4

Second Derivative Test (9.2)

Recall the second degree Taylor polynomial gives us an approximation of a function đ?‘“ : â„?2 → â„? at the point (đ?‘Ž, đ?‘?): đ?‘“ (đ?‘Ľ, đ?‘Ś) ≈ đ?‘ƒ2,(đ?‘Ž,đ?‘?) (đ?‘Ľ, đ?‘Ś) =đ?‘“ ((đ?‘Ž, đ?‘?)) + đ?‘“đ?‘Ľ (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž) + đ?‘“đ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ś − đ?‘?) 1 + [đ?‘“đ?‘Ľđ?‘Ľ (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)2 + 2đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)(đ?‘Ś − đ?‘?) + đ?‘“đ?‘Śđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ś − đ?‘?)2 ], 2 for (đ?‘Ľ, đ?‘Ś) sufficiently close to (đ?‘Ž, đ?‘?). If (đ?‘Ž, đ?‘?) is a critical point of đ?‘“ such that đ?‘“đ?‘Ľ (đ?‘Ž, đ?‘?) = 0 = đ?‘“đ?‘Ś (đ?‘Ž, đ?‘?), then this yields đ?‘“ (đ?‘Ľ, đ?‘Ś) − đ?‘“ (đ?‘Ž, đ?‘?) ≈

1 [đ?‘“đ?‘Ľđ?‘Ľ (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)2 + 2đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)(đ?‘Ś − đ?‘?) + đ?‘“đ?‘Śđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ś − đ?‘?)2 ], 2

for (đ?‘Ľ, đ?‘Ś) sufficiently close to (đ?‘Ž, đ?‘?). Then the sign on the right will determine the sign of đ?‘“ (đ?‘Ľ, đ?‘Ś)−đ?‘“ (đ?‘Ž, đ?‘?), and so will determine whether (đ?‘Ž, đ?‘?) is a local minimum, a local maximum, or a saddle point. This expression on the right is known as a quadratic form; we will examine properties of these objects in order to develop a test for classifying critical points. D EFINITION 4.1. A function đ?‘„ : â„?2 → â„? of the form đ?‘„(đ?‘˘, đ?‘Ł) = đ?‘Ž11 đ?‘˘2 + 2đ?‘Ž12 đ?‘˘đ?‘Ł + đ?‘Ž22 đ?‘Ł 2 , where đ?‘Ž11 , đ?‘Ž12 , đ?‘Ž22 are real constants, is called a quadratic form on â„?2 . Observe we may write the action of đ?‘„ on (đ?‘˘, đ?‘Ł) in matrix form: [ ][ ] đ?‘Ž11 đ?‘Ž12 đ?‘˘ đ?‘„(đ?‘˘, đ?‘Ł) = [ đ?‘˘ đ?‘Ł ] , đ?‘Ž12 đ?‘Ž22 đ?‘Ł so a quadratic form on â„?2 is determined by a 2 Ă— 2 symmetric matrix. D EFINITION 4.2. đ?‘„ is said to be (a) positive definite if đ?‘„(đ?‘˘, đ?‘Ł) > 0 for all (đ?‘˘, đ?‘Ł) ∕= (0, 0). (b) negative definite if đ?‘„(đ?‘˘, đ?‘Ł) < 0 for all (đ?‘˘, đ?‘Ł) ∕= (0, 0). (c) indefinite if đ?‘„(đ?‘˘, đ?‘Ł) < 0 for some (đ?‘˘, đ?‘Ł) ∕= (0, 0) and đ?‘„(đ?‘˘, đ?‘Ł) > 0 for some other (đ?‘˘, đ?‘Ł) ∕= (0, 0). (d) semidefinite if đ?‘„ does not satisfy any of (a)-(c). In general, it can be difficult to classify a quadratic form. Fortunately, we have the following theorem: T HEOREM 4.3. Let đ?‘„(đ?‘˘, đ?‘Ł) = đ?‘Ž11 đ?‘˘2 + 2đ?‘Ž12 đ?‘˘đ?‘Ł + đ?‘Ž22 đ?‘Ł 2 and let đ??ˇ = đ?‘Ž11 đ?‘Ž22 − đ?‘Ž212 . (Notice đ??ˇ is the determinant of the associated matrix form of đ?‘„.) Then đ?‘„ is (a) positive definite if and only if đ??ˇ > 0 and đ?‘Ž11 > 0. (b) negative definite if and only if đ??ˇ > 0 and đ?‘Ž11 < 0.

9

WATERLOO SOS E XAM -AID: MATH237 (c) indefinite if and only if đ??ˇ < 0. (d) semidefinite if and only if đ??ˇ = 0. Let us return to our equation đ?‘“ (đ?‘Ľ, đ?‘Ś) − đ?‘“ (đ?‘Ž, đ?‘?) ≈

1 [đ?‘“đ?‘Ľđ?‘Ľ (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)2 + 2đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ľ − đ?‘Ž)(đ?‘Ś − đ?‘?) + đ?‘“đ?‘Śđ?‘Ś (đ?‘Ž, đ?‘?)(đ?‘Ś − đ?‘?)2 ]. 2

Using đ?‘˘ = đ?‘Ľ − đ?‘Ž, đ?‘Ł = đ?‘Ś − đ?‘?, đ?‘“ (đ?‘Ľ, đ?‘Ś) − đ?‘“ (đ?‘Ž, đ?‘?) ≈

1 [đ?‘“đ?‘Ľđ?‘Ľ (đ?‘Ž, đ?‘?)đ?‘˘2 + 2đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?)đ?‘˘đ?‘Ł + đ?‘“đ?‘Śđ?‘Ś (đ?‘Ž, đ?‘?)đ?‘Ł 2 ]. 2

Notice the matrix form of the quadratic form on the right is the Hessian matrix of đ?‘“ at (đ?‘Ž, đ?‘?), [ ] đ?‘“đ?‘Ľđ?‘Ľ (đ?‘Ž, đ?‘?) đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?) đ??ťđ?‘“ (đ?‘Ž, đ?‘?) = . đ?‘“đ?‘Ľđ?‘Ś (đ?‘Ž, đ?‘?) đ?‘“đ?‘Śđ?‘Ś (đ?‘Ž, đ?‘?) Aside: Recall the Hessian is not symmetric, đ?‘“đ?‘Śđ?‘Ľ is the lower left entry. Why is it that we have đ?‘“đ?‘Ľđ?‘Ś instead? For the second degree Taylor polynomial, we assume đ?‘“ ∈ đ??ś 2 , so that đ?‘“đ?‘Ľđ?‘Ś = đ?‘“đ?‘Śđ?‘Ľ . It is thus plausible that if đ??ťđ?‘“ (đ?‘Ž, đ?‘?) is positive definite, then đ?‘“ (đ?‘Ľ, đ?‘Ś) − đ?‘“ (đ?‘Ž, đ?‘?) > 0 for (đ?‘˘, đ?‘Ł) ∕= (0, 0), i.e. for (đ?‘Ľ, đ?‘Ś) ∕= (đ?‘Ž, đ?‘?) ((đ?‘Ľ, đ?‘Ś) sufficiently close to (đ?‘Ž, đ?‘?)). That is, if đ??ťđ?‘“ (đ?‘Ž, đ?‘?) is positive definite, it seems as though (đ?‘Ž, đ?‘?) is a local minimum point of đ?‘“ . Similar arguments will follow for when đ??ťđ?‘“ (đ?‘Ž, đ?‘?) is negative definite or indefinite, but this is not a proof due to the approximations. Rather, it is a useful way of gaining insight and a method of remembering the consequences of the next theorem. T HEOREM 4.4 [S ECOND PARTIAL D ERIVATIVES T EST ]. Let đ?‘“ : â„?2 → â„? be such that đ?‘“ ∈ đ??ś 2 in some neighborhood of đ?‘Ž and that đ?‘“đ?‘Ľ (đ?‘Ž) = 0 = đ?‘“đ?‘Ś (đ?‘Ž). (a) If đ??ťđ?‘“ (đ?‘Ž) is positive definite, then đ?‘Ž is a local minimum point of đ?‘“ . (b) If đ??ťđ?‘“ (đ?‘Ž) is negative definite, then đ?‘Ž is a local maximum point of đ?‘“ . (c) If đ??ťđ?‘“ (đ?‘Ž) is indefinite, then đ?‘Ž is a saddle point of đ?‘“ . Theorems 4.3 and 4.4 can be used together to classify (most) critical points.

E XAMPLE 7. Define đ?‘“ : â„?2 → â„? by đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ3 + đ?‘Ś 3 + 3đ?‘Ľ2 − 3đ?‘Ś 2 − 2. Find and classify the critical points of đ?‘“. S OLUTION. Recall in Example 6 we showed the critical points are (0, 0),

(0, 2),

(−2, 0),

(−2, 2).

The second partial derivatives are given by đ?‘“đ?‘Ľđ?‘Ľ = 6đ?‘Ľ + 6,

đ?‘“đ?‘Ľđ?‘Ś = 0,

so the Hessian matrix of đ?‘“ at (đ?‘Ž, đ?‘?) is given by [ ] 6đ?‘Ž + 6 0 . 0 6đ?‘? − 6 10

đ?‘“đ?‘Śđ?‘Ś = 6đ?‘Ś − 6,

WATERLOO SOS E XAM -AID: MATH237 ] . det đ??ťđ?‘“ (0, 0) = −36 < 0, so by the Second

[

6 0

0 −6

6 0

0 6

] . det đ??ťđ?‘“ (0, 2) = 36 > 0 and 6 > 0, so by the Second

At (0, 0), the Hessian matrix is đ??ťđ?‘“ (0, 0) = Derivative Test, (0, 0) is a saddle point. [ At (0, 2), the Hessian matrix is đ??ťđ?‘“ (0, 2) = Derivative Test, (0, 2) is a local minimum point. [

] −6 0 . det đ??ťđ?‘“ (−2, 0) = 36 > 0 and −6 < 0, so by 0 −6 the Second Derivative Test, (−2, 0) is a local maximum point. [ ] −6 0 At (−2, 2), the Hessian matrix is đ??ťđ?‘“ (−2, 2) = . det đ??ťđ?‘“ (−2, 2) = −36 < 0, so by the Second 0 6 Derivative Test, (−2, 2) is a saddle point. At (−2, 0), the Hessian matrix is đ??ťđ?‘“ (−2, 0) =

E XERCISE 3. Find and classify the critical points of the following functions. (a) đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ3 − 9đ?‘Ľ2 + 24đ?‘Ľ − 24đ?‘Ľđ?‘Ś 2 . (b) đ?‘”(đ?‘Ľ, đ?‘Ś) = −2đ?‘Ľ3 − 3đ?‘Ś 4 + 6đ?‘Ľđ?‘Ś 2 . (c) â„Ž(đ?‘Ľ, đ?‘Ś) = đ?‘Ľ4 + đ?‘Ś 4 − 4đ?‘Ľđ?‘Ś + 1. You may notice that the Second Derivative Test is inconclusive if the Hessian matrix (quadratic form) is semidefinite, i.e. when the determinant is 0. In this case, the critical point might be a local minimum, a local maximum, or a saddle point. You may see this when examing the functions đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ4 + đ?‘Ś 4 , đ?‘”(đ?‘Ľ, đ?‘Ś) = −đ?‘Ľ4 − đ?‘Ś 4 , â„Ž(đ?‘Ľ, đ?‘Ś) = đ?‘Ľ4 − đ?‘Ś 4 . It is left as an exercise to see why (0, 0) is a local minimum point of đ?‘“ , a local maximum point of đ?‘”, and a saddle point of â„Ž. D EFINITION 4.5. If đ??ťđ?‘“ (đ?‘Ž) is semidefinite, so that the Second Derivative Test gives no conclusion, we say that the critical point đ?‘Ž is degenerate. E XAMPLE 8. Define đ?‘“ : â„?2 → â„? by đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ3 đ?‘Ś 2 + đ?‘Ľ2 đ?‘Ś. Show that (0, 0) is a degenerate critical point of đ?‘“ and classify it. S OLUTION. Differentiating, đ?‘“đ?‘Ľ = 3đ?‘Ľ2 đ?‘Ś 2 + 2đ?‘Ľđ?‘Ś, đ?‘“đ?‘Ľđ?‘Ľ = 6đ?‘Ľđ?‘Ś 2 + 2đ?‘Ś,

đ?‘“đ?‘Ś = 2đ?‘Ľ3 đ?‘Ś + đ?‘Ľ2 .

đ?‘“đ?‘Ľđ?‘Ś = 6đ?‘Ľ2 đ?‘Ś,

đ?‘“đ?‘Śđ?‘Ś = 2đ?‘Ľ3 . [

0 Then ∇đ?‘“ (0, 0) = (0, 0) so that (0, 0) is indeed a critical point of đ?‘“ , and đ??ťđ?‘“ (0, 0) = 0 det đ??ťđ?‘“ (0, 0) = 0 and (0, 0) is thus a degenerate point. To classify it, notice for đ?‘Ľ > 0 we have đ?‘“ (đ?‘Ľ, đ?‘Ľ) = đ?‘Ľ5 + đ?‘Ľ3 > 0, while for đ?‘Ľ < 0 we have đ?‘“ (đ?‘Ľ, đ?‘Ľ) = đ?‘Ľ5 + đ?‘Ľ3 < 0. Hence, (0, 0) is a saddle point. 11

0 0

] , so that

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 4. Let đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 đ?‘Ś 2 đ?‘’−(đ?‘Ľ

2

+đ?‘Ś 2 )

. Show that (0, 0) is a degenerate critical point of đ?‘“ and classify it.

We can of course generalize the notions of local minimum points, local maximum points and critical points to functions of đ?‘› variables in the obvious way. The Hessian matrix of đ?‘“ : â„?đ?‘› → â„? at đ?‘Ž is the đ?‘› Ă— đ?‘› symmetric matrix given by [ 2 ] ∂ đ?‘“ (đ?‘Ž) , đ??ťđ?‘“ (đ?‘Ž) = ∂đ?‘Ľđ?‘– ∂đ?‘Ľđ?‘— where đ?‘–, đ?‘— = 1, . . . , đ?‘›. We can classify the Hessian matrix as positive definite, negative definite, indefinite or semidefinite by considering the associated quadratic form in â„?đ?‘› , đ?‘„(đ?‘˘) =

đ?‘› ∑

∂2đ?‘“ (đ?‘Ž)đ?‘˘đ?‘– đ?‘˘đ?‘— , ∂đ?‘Ľđ?‘– ∂đ?‘Ľđ?‘— đ?‘–,đ?‘—=1

and the Second Derivative Test now holds in â„?đ?‘› . To classify the Hessian matrix, we may use the determinant, as in the case đ?‘› = 2, which requires more algebra than should be necessary for this course.

5

Extreme Values (10.1-10.2)

D EFINITION 5.1. Given a function đ?‘“ : â„?2 → â„? and a set đ?‘† ⊂ â„?2 , (a) A point (đ?‘Ž, đ?‘?) ∈ đ?‘† is a(n) (absolute) minimum point of đ?‘“ on đ?‘† if đ?‘“ (đ?‘Ľ, đ?‘Ś) ≼ đ?‘“ (đ?‘Ž, đ?‘?) for all (đ?‘Ľ, đ?‘Ś) ∈ đ?‘†. The value đ?‘“ (đ?‘Ž, đ?‘?) is the (absolute) minimum value of đ?‘“ on đ?‘†. (b) A point (đ?‘Ž, đ?‘?) ∈ đ?‘† is a(n) (absolute) maximum point of đ?‘“ on đ?‘† if đ?‘“ (đ?‘Ľ, đ?‘Ś) ≤ đ?‘“ (đ?‘Ž, đ?‘?) for all (đ?‘Ľ, đ?‘Ś) ∈ đ?‘†. The value đ?‘“ (đ?‘Ž, đ?‘?) is the (absolute) maximum value of đ?‘“ on đ?‘†. If (đ?‘Ž, đ?‘?) is a minimum or maximum point of đ?‘“ , we may call (đ?‘Ž, đ?‘?) an extremum point of đ?‘“ , and đ?‘“ (đ?‘Ž, đ?‘?) an extreme value of đ?‘“ . Recall for 1 variable, the Extreme Value Theorem tells us that continuous functions will attain their minimum and maximum on any finite (bounded) closed interval. We have a generalization to higher dimensions which requires some new definitions. D EFINITION 5.2. A set đ?‘† ⊂ â„?2 is bounded if and only if it is contained in some neighborhood of the origin. D EFINITION 5.3. Given a set đ?‘† ⊂ â„?2 , a point đ?‘? is a boundary point of đ?‘† if and only if every neighborhood of đ?‘? contains at least one point in đ?‘† and one point not in đ?‘†. The set of all boundary points of đ?‘† is denoted đ??ľ(đ?‘†) and is called the boundary of đ?‘†. A set đ?‘† ⊂ â„?2 is closed if and only if đ?‘† contains its boundary. T HEOREM 5.4. Suppose đ?‘“ : â„?2 → â„? is continuous on a closed and bounded set đ?‘† ⊂ â„?2 . Then there exist đ?‘?1 , đ?‘?2 ∈ đ?‘† such that for all đ?‘Ľ ∈ đ?‘†, đ?‘“ (đ?‘?1 ) ≤ đ?‘“ (đ?‘Ľ) ≤ đ?‘“ (đ?‘?2 ). 12

WATERLOO SOS E XAM -AID: MATH237 Recall that in 1 variable, the extreme (maximum/minimum) values of a differentiable function occur at a critical point or one of the endpoints. This concept generalizes to higher dimensions: Let đ?‘† ⊂ â„?2 be closed and bounded, with boundary đ??ľ(đ?‘†) and suppose đ?‘“ : â„?2 → â„? is differentiable on đ?‘†. Then the maximum (minimum) value of đ?‘“ on đ?‘† occur either at a critical point of đ?‘“ in đ?‘†, or at a point in đ??ľ(đ?‘†). That is, we have the following algorithm for finding the maximum (minimum) point of đ?‘“ on đ?‘†: (1) Find all critical points of đ?‘“ in đ?‘†. (2) Find the maximum (minimum) points of đ?‘“ on the boundary đ??ľ(đ?‘†). (3) Evaluate đ?‘“ at the points found in (1) and (2). (4) Take the point corresponding to the largest (smallest) value found in (3). Notice we may apply this alogirthm inductively in step (2) to further reduce our work.

E XAMPLE 9. Find the (absolute) maximum and minimum of đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 + 2đ?‘Ľđ?‘Ś + 3đ?‘Ś 2 in the region bounded by the triangle with vertices (1, 2), (1, −1), (4, −1). S OLUTION. We have ∇đ?‘“ = (2đ?‘Ľ + 2đ?‘Ś, 6đ?‘Ś + 2đ?‘Ľ). Taking the difference of 2đ?‘Ľ + 2đ?‘Ś = 0 and 6đ?‘Ś + 2đ?‘Ľ = 0, we get 4đ?‘Ś = 0, or đ?‘Ś = 0. Then 2đ?‘Ľ + 2đ?‘Ś = 0 gives đ?‘Ľ = 0 as well, so the only critical point is (0, 0), which lies outside of the triangle. We then look for maximum and minimum points of đ?‘“ on the boundary of the triangle, which is composed of three line segments. Notice each of these line segments are closed and bounded with đ?‘“ differentiable on them, so we can apply the algorithm again for each segment. On đ??ż1 , the line joining (1, 2) to (1, −1), the đ?‘Ľ-coordinate is identically 1. We get đ?‘”1 (đ?‘Ś) := đ?‘“ (1, đ?‘Ś) = 12 + 2(1)đ?‘Ś + 3đ?‘Ś 2 = 1 + 2đ?‘Ś + 3đ?‘Ś 2 , so that đ?‘”1′ (đ?‘Ś) = 6đ?‘Ś + 2. This gives đ?‘Ś = − 31 , so we have a critical point (1, − 31 ) which lies on this line. On đ??ż2 , the line joining (1, −1) to (4, −1), the đ?‘Ś-coordinate is identically −1. We get đ?‘”2 (đ?‘Ľ) := đ?‘“ (đ?‘Ľ, −1) = đ?‘Ľ2 + 2đ?‘Ľ(−1) + 3(−1)2 = đ?‘Ľ2 − 2đ?‘Ľ + 3, so that đ?‘”2′ (đ?‘Ľ) = 2đ?‘Ľ − 2. This gives đ?‘Ľ = 1, so we have a critical point (1, −1) which lies on this line. On đ??ż2 , the line joining (4, −1) to (1, 2), points satisfy đ?‘Ś = 3 − đ?‘Ľ. We get đ?‘”3 (đ?‘Ľ) := đ?‘“ (đ?‘Ľ, 3 − đ?‘Ľ) = đ?‘Ľ2 + 2đ?‘Ľ(3 − đ?‘Ľ) + 3(3 − đ?‘Ľ)2 = đ?‘Ľ2 + 6đ?‘Ľ − 2đ?‘Ľ2 + 27 − 18đ?‘Ľ + 3đ?‘Ľ2 = 2đ?‘Ľ2 − 12đ?‘Ľ + 27, so that đ?‘”3′ (đ?‘Ľ) = 4đ?‘Ľ − 12. This gives đ?‘Ľ = 3, so we have a critical point (3, 0) which lies on this line. By the algorithm, we also have to check the boundaries, đ??ľ(đ??ż1 ), đ??ľ(đ??ż2 ), đ??ľ(đ??ż3 ), which are the three points (1, 2), (1, −1), (4, −1). Thus, for step (3) in the algorithm we check the value of đ?‘“ at the points (1, − 31 ),

(1, −1),

(3, 0),

(1, 2),

(4, −1).

We have, by evaluating, đ?‘“ (1, − 31 ) = 23 ,

đ?‘“ (1, −1) = 2,

đ?‘“ (3, 0) = 9,

đ?‘“ (1, 2) = 17,

đ?‘“ (4, −1) = 11.

Then the minimum is 32 , attained at (1, − 13 ), and the maximum is 17, attained at (1, 2).

E XERCISE 5. Find the (absolute) maximum and minimum of đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ľ2 + đ?‘Ś 2 − đ?‘Ľ − đ?‘Ś + 1 in the region đ?‘† = {(đ?‘Ľ, đ?‘Ś) : đ?‘Ľ2 + đ?‘Ś 2 ≤ 1}.

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WATERLOO SOS E XAM -AID: MATH237

6

Optimization: Lagrange Multipliers (10.3)

You may have noticed that when finding extrema on the boundary (step (2) of our algorithm), we parametrized the boundary to solve for a new function. This can be difficult to do sometimes, so we will develop another algorithm to find extrema of đ?‘“ on a curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜, without the need to find a parametrization. We wish to find the maximum (minimum) value of a differentiable function đ?‘“ subject to the constraint đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜, or geometrically, on the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜. If đ?‘“ (đ?‘Ľ, đ?‘Ś) has a local maximum (minimum) at (đ?‘Ž, đ?‘?) relative to nearby points on the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜ and ∇đ?‘”(đ?‘Ž, đ?‘?) ∕= (0, 0), then using the Implicit Function Theorem and Chain Rule will gives us a constant đ?œ† such that ∇đ?‘“ (đ?‘Ž, đ?‘?) = đ?œ†âˆ‡đ?‘”(đ?‘Ž, đ?‘?). Thus, if we include the exceptional case of ∇đ?‘”(đ?‘Ž, đ?‘?) = (0, 0), we get a method of finding extrema of đ?‘“ on the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜. (1) Find (đ?‘Ž, đ?‘?) satisfying ∇đ?‘“ (đ?‘Ž, đ?‘?) = đ?œ†âˆ‡đ?‘”(đ?‘Ž, đ?‘?) for some constant đ?œ† (with đ?‘”(đ?‘Ž, đ?‘?) = đ?‘˜). (2) Find (đ?‘Ž, đ?‘?) satisfying ∇đ?‘”(đ?‘Ž, đ?‘?) = (0, 0). (3) Find end points (đ?‘Ž, đ?‘?) of the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜. (4) Take the point corresponding to the largest (smallest) value found in (1)-(3). R EMARK 6.1. (a) Notice to find points (đ?‘Ž, đ?‘?) in (1), we have to solve a system of 3 equations in 3 unknowns: đ?‘“đ?‘Ľ (đ?‘Ľ, đ?‘Ś) = đ?œ†đ?‘”đ?‘Ľ (đ?‘Ľ, đ?‘Ś) đ?‘“đ?‘Ś (đ?‘Ľ, đ?‘Ś) = đ?œ†đ?‘”đ?‘Ś (đ?‘Ľ, đ?‘Ś) đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜ for đ?‘Ľ and đ?‘Ś. The constant đ?œ† is the Lagrange multiplier is not required and should be eliminated. (b) If the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜ is unbounded, then we can still use this algorithm, but instead of the “end pointsâ€? in (3), we find limâˆĽ(đ?‘Ľ,đ?‘Ś)âˆĽâ†’âˆž đ?‘“ (đ?‘Ľ, đ?‘Ś) for (đ?‘Ľ, đ?‘Ś) satisfying đ?‘”(đ?‘Ľ, đ?‘Ś) = đ?‘˜.

E XAMPLE 10. Find the maximum and minimum of đ?‘“ (đ?‘Ľ, đ?‘Ś) = đ?‘Ś on the dumbbell curve defined by đ?‘Ś 2 + đ?‘Ľ6 − đ?‘Ľ4 = 0. S OLUTION. We begin by looking for points (đ?‘Ž, đ?‘?) that satisfy ∇đ?‘“ (đ?‘Ž, đ?‘?) = đ?œ†âˆ‡đ?‘”(đ?‘Ž, đ?‘?). We have ∇đ?‘“ = (0, 1),

∇đ?‘” = (6đ?‘Ľ5 − 4đ?‘Ľ3 , 2đ?‘Ś) = (2đ?‘Ľ3 (3đ?‘Ľ2 − 2), 2đ?‘Ś).

We need to solve 0 = đ?œ†2đ?‘Ľ3 (3đ?‘Ľ2 − 2) 1 = đ?œ†2đ?‘Ś 0 = đ?‘Ś 2 + đ?‘Ľ6 − đ?‘Ľ4 √ The second equation gives đ?œ† ∕= 0, so the first equation gives đ?‘Ľ = 0 or đ?‘Ľ = Âą 23 . If đ?‘Ľ = 0, then the third √ equation gives đ?‘Ś = 0, which contradicts the second equation. If đ?‘Ľ = Âą 23 , then the third equation gives ( √ )6 ( √ )4 ( )3 ( )2 2 2 2 2 8 4 4 đ?‘Ś =− Âą + Âą =− + =− + = , 3 3 3 3 27 9 27 2

14

WATERLOO SOS E XAM -AID: MATH237 so that đ?‘Ś =

2 . Âą 3√ 3

This gives the four points (Âą

√

2 2 √ 3 , Âą 3 3 ).

√ Next, we solve ∇đ?‘” = (0, 0). We have 2đ?‘Ľ3 (3đ?‘Ľ2 − 2) = 0 and 2đ?‘Ś = 0, which implies đ?‘Ľ = 0 or đ?‘Ľ = Âą 23 , √ and đ?‘Ś = 0. (Âą 23 , 0) does not lie on the curve đ?‘”(đ?‘Ľ, đ?‘Ś) = 0, so we are left with (0, 0). Since đ?‘”(đ?‘Ľ, đ?‘Ś) is a closed curve, there are no end points to check. Evaluating đ?‘“ at all the necessary points gives ( √ ) 2 2 2 đ?‘“ Âą ,Âą √ =Âą √ , đ?‘“ (0, 0) = 0. 3 3 3 3 3 Thus, the maximum value is

2 √ 3 3 √

at either of the two points (Âą

at either of the two points (Âą

√

2 2 √ 3, 3 3)

2 and the minimum value is − 3√ 3

2 2 √ 3 , − 3 3 ).

The method of Lagrange multipliers can be generalized to functions of đ?‘› variables đ?‘“ : â„?đ?‘› → â„? with đ?‘&#x; constraints of the form đ?‘”1 (đ?‘Ľ) = đ?‘˜1 ,

đ?‘”2 (đ?‘Ľ) = đ?‘˜2 ,

...,

đ?‘”đ?‘&#x; (đ?‘Ľ) = đ?‘˜đ?‘&#x; ,

where đ?‘Ľ = (đ?‘Ľ1 , đ?‘Ľ2 , . . . , đ?‘Ľđ?‘› ). The first step of the algorithm then becomes solving for points đ?‘Ž that satisfy ∇đ?‘“ (đ?‘Ž) = đ?œ†1 ∇đ?‘”1 (đ?‘Ž) + â‹… â‹… â‹… + đ?œ†đ?‘&#x; ∇đ?‘”đ?‘&#x; (đ?‘Ž),

đ?‘”đ?‘– (đ?‘Ž) = đ?‘˜đ?‘– ,

for 1 ≤ đ?‘– ≤ đ?‘&#x;. Although it is straightforward to extend the concept to đ?‘› variables, having more than one constraint can be tricky, so in this course you will only be considering one constraint, đ?‘› dimensional Lagrange multiplier questions.

E XAMPLE 11. Find the minimum distance from the origin (0, 0, 0) to the ellipsoid đ?‘Ľ2 + 2đ?‘Ś 2 + (đ?‘§ − 2)2 = 1. √ S OLUTION. To minimize distance to the origin, we are looking to minimize đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 such that (đ?‘Ľ, đ?‘Ś, đ?‘§) lies on the ellipse. This is equivalent to minimizing đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) = đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 , subject to the constraint đ?‘”(đ?‘Ľ, đ?‘Ś, đ?‘§) = đ?‘Ľ2 + 2đ?‘Ś 2 + (đ?‘§ − 2)2 = 1. We begin by looking for points đ?‘Ž that satisfy ∇đ?‘“ (đ?‘Ž) = đ?œ†âˆ‡đ?‘”(đ?‘Ž). We have ∇đ?‘“ = (2đ?‘Ľ, 2đ?‘Ś, 2đ?‘§),

∇đ?‘” = (2đ?‘Ľ, 4đ?‘Ś, 2(đ?‘§ − 2)).

We need to solve (3)

2đ?‘Ľ = đ?œ†2đ?‘Ľ

(4)

2đ?‘Ś = đ?œ†4đ?‘Ś

(5)

2đ?‘§ = đ?œ†2(đ?‘§ − 2)

(6)

1 = đ?‘Ľ2 + 2đ?‘Ś 2 + (đ?‘§ − 2)2

(3) gives đ?œ† = 1 or đ?‘Ľ = 0, while (4) gives đ?œ† = 21 or đ?‘Ś = 0. If both đ?‘Ľ = 0 = đ?‘Ś, then (6) gives (đ?‘§ − 2)2 = 1, so that đ?‘§ = 1 or đ?‘§ = 3. This gives two possibilities, (0, 0, 1) and (0, 0, 3). If đ?‘Ľ = 0 and đ?‘Ś ∕= 0, then (4) gives đ?œ† = 21 . Into (5), we have 2đ?‘§ = đ?‘§ − 2, so that đ?‘§ = −2. Into (6), we have 2 0 + 2đ?‘Ś 2 + (−2 − 2)2 = 2đ?‘Ś 2 + 16 = 1, which has no (real) solution for đ?‘Ś. If đ?‘Ľ ∕= 0, then (3) gives đ?œ† = 1. Into (5), we have đ?‘§ = đ?‘§ − 2, which has no solution for đ?‘§. 15

WATERLOO SOS E XAM -AID: MATH237 Next, we solve ∇đ?‘” = (0, 0, 0). We have 2đ?‘Ľ = 0, 4đ?‘Ś = 0, and 2(đ?‘§ − 2) = 0, which implies đ?‘Ľ = 0, đ?‘Ś = 0, and đ?‘§ = 2. However, đ?‘”(0, 0, 2) = 0 ∕= 1. Since đ?‘”(đ?‘Ľ, đ?‘Ś, đ?‘§) is a closed surface, there are no boundary points to check. Evaluating đ?‘“ at all the necessary points gives đ?‘“ (0, 0, 1) = 1,

đ?‘“ (0, 0, 3) = 9.

Thus, the minimum distance from the origin to the given ellipsoid is 1, and is at the point (0, 0, 1).

E XERCISE 6. Find the largest volume a rectangular box can have if its surface area is fixed at 10 m2 .

2

E XERCISE 7. Suppose a mountain’s shape is defined by the graph of the function đ?‘“ (đ?‘Ľ, đ?‘Ś) = 3 − đ?‘Ľ2 − đ?‘Ś4 , so that đ?‘“ (đ?‘Ľ, đ?‘Ś) is the altitude of the mountain at map coordinate (đ?‘Ľ, đ?‘Ś), in kilometers. There is a trail marked out around the mountain side, defined by the circle đ?‘Ľ2 + đ?‘Ś 2 = 1. If you hike along this trail, what are the highest and lowest altitudes you will reach?

7

Polar Coordinates (11.1)

D EFINITION 7.1. Consider â„?2 . We begin by defining our frame of reference: choose a point đ?‘‚ in the plane and call it the pole(or origin). From đ?‘‚ we construct a ray called the polar axis. Generally, đ?‘‚ corresponds to the origin in Cartesian coordinates, (0, 0), and the polar axis is the positive đ?‘Ľ-axis. For any point đ?‘ƒ in the plane, we can represent its position by the ordered pair (đ?‘&#x;, đ?œƒ) where đ?‘&#x; ≼ 0 is the length of the line đ?‘‚đ?‘ƒ and đ?œƒ is the angle between the polar axis and đ?‘‚đ?‘ƒ (positive if measured in the counterclockwise direction). (đ?‘&#x;, đ?œƒ) are called the polar coordinates of đ?‘ƒ . We will represent đ?‘‚ by (0, đ?œƒ) for any đ?œƒ. Polar coordinates are useful when there is a symmetry about the pole đ?‘‚. R EMARK 7.2. In Cartesian coordinates, every point has a unique representation (đ?‘Ľ, đ?‘Ś). In polar coordinates, a point đ?‘ƒ can have infinitely many representations: (đ?‘&#x;, đ?œƒ) = (đ?‘&#x;, đ?œƒ + 2đ?œ‹đ?‘˜), for đ?‘˜ ∈ ℤ. However, we usually are not concerned with every solution, and may place a restriction such as 0 ≤ đ?œƒ < 2đ?œ‹, or −đ?œ‹ < đ?œƒ ≤ đ?œ‹. Placing đ?‘‚ at the origin of the Cartesian plane and the polar axis along the positive đ?‘Ľ-axis, we can get a relationship between the coordinates of a point (đ?‘Ľ, đ?‘Ś) = (đ?‘&#x;, đ?œƒ) in the two coordinate systems: √ đ?‘Ľ = đ?‘&#x; cos đ?œƒ, đ?‘&#x; = đ?‘Ľ2 + đ?‘Ś 2 , đ?‘Ś đ?‘Ś = đ?‘&#x; sin đ?œƒ, tan đ?œƒ = . đ?‘Ľ

E XERCISE 8. Convert the following points in Cartesian coordinates into polar coordinates. (a) (0, 1) √ √ (b) (− 2, 2)

16

WATERLOO SOS E XAM -AID: MATH237 E XERCISE 9. Convert the following points in polar coordinates into Cartesian coordinates. (a) (3, đ?œ‹6 ) (b) (2, − 5đ?œ‹ 4 ) We can take conversion of points one step further, and convert equations of curves between the two systems.

E XAMPLE 12. Convert the equation of the curve (đ?‘Ľ2 + đ?‘Ś 2 )2 = đ?‘Ľ2 − đ?‘Ś 2 to polar coordinates. S OLUTION. Since đ?‘Ľ = đ?‘&#x; cos đ?œƒ and đ?‘Ś = đ?‘&#x; sin đ?œƒ, we get (đ?‘Ľ2 + đ?‘Ś 2 )2 = đ?‘Ľ2 − đ?‘Ś 2 đ?‘&#x;4 = đ?‘&#x;2 cos2 đ?œƒ − đ?‘&#x;2 sin2 đ?œƒ đ?‘&#x;4 = đ?‘&#x;2 cos 2đ?œƒ. We can simplify this equation to đ?‘&#x;2 = cos 2đ?œƒ since the pole is still included in the graph by taking đ?œƒ = 5đ?œ‹ Since đ?‘&#x; ≼ 0, we have the restriction cos 2đ?œƒ ≼ 0, so that − đ?œ‹4 ≤ đ?œƒ ≤ đ?œ‹4 or 3đ?œ‹ 4 ≤đ?œƒ ≤ 4 .

đ?œ‹ 4.

E XAMPLE 13. Convert the equation of the curve đ?‘&#x; = 1 − cos đ?œƒ to Cartesian coordinates. S OLUTION. Since đ?‘&#x;2 = đ?‘Ľ2 + đ?‘Ś 2 and đ?‘Ľ = đ?‘&#x; cos đ?œƒ, multiplying our equation by đ?‘&#x; yields đ?‘&#x; = 1 − cos đ?œƒ đ?‘&#x;2 = đ?‘&#x; − đ?‘&#x; cos đ?œƒ √ đ?‘Ľ2 + đ?‘Ś 2 = đ?‘Ľ2 + đ?‘Ś 2 − đ?‘Ľ (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘Ľ)2 = đ?‘Ľ2 + đ?‘Ś 2 . Using sectors of a circle, we can produce a formula for calculating area in polar coordinates in the same way we find area in Cartesian coordinates: The area of the region bounded by đ?œƒ = đ?‘Ž, đ?œƒ = đ?‘?, and đ?‘&#x; = đ?‘“ (đ?œƒ) is given by âˆŤ đ?‘? 1 đ??´= [đ?‘“ (đ?œƒ)]2 đ?‘‘đ?œƒ. đ?‘Ž 2

E XERCISE 10. Find the area inside the “eight curveâ€?, defined by đ?‘&#x;2 = 3 sec4 đ?œƒ cos(2đ?œƒ). We can find the area between two curves as well, by finding the points of intersection, then splitting the desired region into easily integratable regions, taking care to subtract the area where it is necessary.

17

WATERLOO SOS E XAM -AID: MATH237

8

Cylindrical Coordinates (11.2)

D EFINITION 8.1. One way to extend polar coordinates into three dimensions is to introduce another axis, called the axis of symmetry, through the pole perpendicular to the plane. We can thus represent any point in the space by the cylindrical coordinates (đ?‘&#x;, đ?œƒ, đ?‘§), where đ?‘&#x; and đ?œƒ are as in polar coordinates and đ?‘§ is the vertical position relative to the plane. We have the restrictions đ?‘&#x; ≼ 0, and possibly a restriction on đ?œƒ, such as 0 ≤ đ?œƒ < 2đ?œ‹, or −đ?œ‹ < đ?œƒ ≤ đ?œ‹. Cylindrical coordinates are useful when there is a symmetry about an axis. Placing đ?‘‚ at the origin of the Cartesian plane, the polar axis along the positive đ?‘Ľ-axis, and the axis of symmetry along the đ?‘§-axis, we can get a relationship between the coordinates of a point (đ?‘Ľ, đ?‘Ś, đ?‘§) = (đ?‘&#x;, đ?œƒ, đ?‘§) in the two coordinate systems: √ đ?‘Ľ = đ?‘&#x; cos đ?œƒ, đ?‘&#x; = đ?‘Ľ2 + đ?‘Ś 2 , đ?‘Ś đ?‘Ś = đ?‘&#x; sin đ?œƒ, tan đ?œƒ = , đ?‘Ľ đ?‘§ = đ?‘§, đ?‘§ = đ?‘§. Keep in mind this definition can be modified by interchanging the roles of any of the variables. For example, we may come across a situation in which it makes more sense to write đ?‘Ľ = đ?‘&#x; cos đ?œƒ, đ?‘Ś = đ?‘Ś, and đ?‘§ = đ?‘&#x; sin đ?œƒ.

E XERCISE 11. Convert the following points in Cartesian coordinates into cylindrical coordinates. √ (a) (1, − 3, 2) (b) (−1, 1, −1)

E XERCISE 12. Convert the following points in cylindrical coordinates into Cartesian coordinates. (a) (2, đ?œ‹2 , 1) (b) (3, − đ?œ‹3 , 2) We can take conversion of points one step further, and convert equations of surfaces between the two systems, just as in the case for polar coordinates.

E XAMPLE 14. Convert the equation of the curve (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 = 2đ?‘§(đ?‘Ľ2 + đ?‘Ś 2 ) to cylindrical coordinates. S OLUTION. Since đ?‘Ľ2 + đ?‘Ś 2 = đ?‘&#x;2 , this easily becomes (đ?‘&#x;2 + đ?‘§ 2 )2 = 2đ?‘§đ?‘&#x;2 . The bounds on đ?‘&#x; and đ?‘§ will be discussed in a later example.

E XAMPLE 15. Convert the equation of the curve đ?‘§ = sin 2đ?œƒ to Cartesian coordinates. S OLUTION. We have đ?‘§ = 2 sin đ?œƒ cos đ?œƒ. If đ?‘&#x; > 0, this becomes đ?‘§ = 2đ?‘&#x; sin đ?œƒđ?‘&#x; cos đ?œƒ/đ?‘&#x;2 . Since đ?‘Ľ = đ?‘&#x; cos đ?œƒ, đ?‘Ś = đ?‘&#x; sin đ?œƒ, and đ?‘&#x;2 = đ?‘Ľ2 + đ?‘Ś 2 , this becomes đ?‘§ = đ?‘Ľ22đ?‘Ľđ?‘Ś +đ?‘Ś 2 . We must also include the point (0, 0, 0), since we omitted it earlier. 18

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 13. Convert the equation of the curve � 2 = 2�� to cylindrical coordinates.

E XERCISE 14. Convert the equation of the curve đ?‘§đ?‘&#x; = tan đ?œƒ(sin đ?œƒ + cos đ?œƒ) to Cartesian coordinates.

9

Spherical Coordinates(11.3)

D EFINITION 9.1. Another way to extend the polar coordinates to three dimensions is to begin by constructing another axis đ?‘§ perpendicular to the polar plane, then any point đ?‘ƒ in the space is represented by the coordinates (đ?œŒ, đ?œƒ, đ?œ™), where đ?œŒ ≼ 0 is the length of the line đ?‘‚đ?‘ƒ , đ?œƒ is the same angle as in polar and cylindrical coordinates, and đ?œ™ is the angle between the positive đ?‘§-axis and the line đ?‘‚đ?‘ƒ . Then 0 ≤ đ?œ™ ≤ đ?œ‹, as đ?œƒ will determine orientation about the đ?‘§-axis. Spherical coordinates are useful when there is a symmetry about the origin. Placing đ?‘‚ at the origin of the Cartesian plane, the polar axis along the positive đ?‘Ľ-axis, and the axis of symmetry along the đ?‘§-axis, we can get a relationship between the coordinates of a point (đ?‘Ľ, đ?‘Ś, đ?‘§) = (đ?œŒ, đ?œƒ, đ?œ™) in the two coordinate systems: √ đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ, đ?œŒ = đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 , đ?‘Ś đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ, tan đ?œƒ = , đ?‘Ľ đ?‘§ đ?‘§ = đ?œŒ cos đ?œ™, cos đ?œ™ = √ . 2 đ?‘Ľ + đ?‘Ś2 + đ?‘§2 Keep in mind this definition can be modified by interchanging the roles of any of the variables. For example, we may come across a situation in which it makes more sense to write đ?‘Ľ = đ?œŒ cos đ?œ™, đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ, and đ?‘§ = đ?œŒ sin đ?œ™ cos đ?œƒ.

E XERCISE 15. Convert the following points in Cartesian coordinates into spherical coordinates. √ (a) (2, −2 6, 2) (b) (−1, 1, −1)

E XERCISE 16. Convert the following points in spherical coordinates into Cartesian coordinates. đ?œ‹ (a) (2, 5đ?œ‹ 4 , 2)

(b) (4, đ?œ‹6 , − đ?œ‹3 ) We can take conversion of points one step further, and convert equations of surfaces between the two systems, just as in the case for polar coordinates.

19

WATERLOO SOS E XAM -AID: MATH237 E XAMPLE 16. Convert the equation of the curve (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 = 2đ?‘§(đ?‘Ľ2 + đ?‘Ś 2 ) to spherical coordinates. S OLUTION. Since đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ and đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ, we have đ?‘Ľ2 + đ?‘Ś 2 = đ?œŒ2 sin2 đ?œ™. Since đ?œŒ2 = đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 and đ?‘§ = đ?œŒ cos đ?œ™, (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 = 2đ?‘§(đ?‘Ľ2 + đ?‘Ś 2 ) đ?œŒ4 = 2đ?œŒ cos đ?œ™(đ?œŒ2 sin2 đ?œ™) đ?œŒ4 = đ?œŒ3 sin 2đ?œ™ sin đ?œ™. We can simplify this equation to đ?œŒ = sin 2đ?œ™ sin đ?œ™ since the pole is still included in the graph by taking đ?œƒ = 0. Since đ?œŒ ≼ 0, we have the restriction sin 2đ?œ™ sin đ?œ™ ≼ 0, so that 0 ≤ đ?œ™ ≤ đ?œ‹2 (recall 0 ≤ đ?œ™ ≤ đ?œ‹). Finally, 0 ≤ đ?œŒ ≤ sin 2đ?œ™ sin đ?œ™.

E XAMPLE 17. Convert the equation of the curve sin2 đ?œ™ cos 2đ?œƒ = đ?œŒ cos đ?œ™ to Cartesian coordinates. S OLUTION. Since đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ, đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ and đ?‘§ = đ?œŒ cos đ?œ™, sin2 đ?œ™ cos 2đ?œƒ = đ?œŒ cos đ?œ™ đ?œŒ2 sin2 đ?œ™(cos2 đ?œƒ − sin2 đ?œƒ) = đ?œŒ cos đ?œ™đ?œŒ2 đ?œŒ2 sin2 đ?œ™ cos2 đ?œƒ − đ?œŒ2 sin2 đ?œ™ sin2 đ?œƒ = đ?œŒ cos đ?œ™đ?œŒ2 đ?‘Ľ2 − đ?‘Ś 2 = đ?‘§(đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 ).

E XERCISE 17. Convert the equation of the curve � 2 = 2�� to spherical coordinates.

E XERCISE 18. Convert the equation of the curve đ?œŒ3 = tan2 đ?œ™ to Cartesian coordinates.

10

Linear Approximation of �2 → �2 Mappings (12.1-12.2)

D EFINITION 10.1. Prior to now, we have examined functions which map from �� to �. We now consider a generalization of this, by looking at functions whose domain is in �2 and whose range is in �2 . For scalar functions �, � : �2 → �, the pair of equations � = � (�, �) � = �(�, �) 2

associates with each point (đ?‘Ľ, đ?‘Ś) ∈ â„? (in the domains of đ?‘“ and đ?‘”) a unique point (đ?‘˘, đ?‘Ł) ∈ â„?2 , and thus defines a vector-valued function đ??š : â„?2 → â„?2 . Writing đ?‘Ľ = (đ?‘Ľ, đ?‘Ś) and đ?‘˘, the equations can be written as a vector equation đ?‘˘ = đ??š (đ?‘Ľ), where đ??š (đ?‘Ľ) = (đ?‘“ (đ?‘Ľ), đ?‘”(đ?‘Ľ)) ∈ â„?2 . We call đ??š : â„?2 → â„?2 a mapping of â„?2 into â„?2 . The scalar functions đ?‘“ and đ?‘” are called the component functions of the mapping. If a mapping acts on any subset đ?‘† ⊂ â„?2 in its domain, it will determine a set đ??š (đ?‘†) ⊂ â„?2 in its range, called the image of đ?‘† under đ??š . 20

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 19. Find the image of the rectangle đ?‘… = {(đ?‘&#x;, đ?œƒ) : 1 ≤ đ?‘&#x; ≤ 3, đ?œ‹6 ≤ đ?œƒ ≤ đ?œ‹} under the mapping đ??š : â„?2 → â„?2 defined by (đ?‘Ľ, đ?‘Ś) = đ??š (đ?‘&#x;, đ?œƒ) = (đ?‘&#x; cos đ?œƒ, đ?‘&#x; sin đ?œƒ). Consider a mapping đ??š : â„?2 → â„?2 defined by đ?‘˘ = đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘Ł = đ?‘”(đ?‘Ľ, đ?‘Ś). Assume đ??š has continuous partial derivatives, in the sense that the component functions đ?‘“ and đ?‘” have continuous partial derivatives. We are interested in approximating đ??š (đ?‘“ (đ?‘Ž, đ?‘?), đ?‘”(đ?‘Ž, đ?‘?)) for points near (đ?‘Ž, đ?‘?); we do this by using the linear approximation formula for đ?‘“ (đ?‘Ľ, đ?‘Ś) and đ?‘”(đ?‘Ľ, đ?‘Ś) separately: Δđ?‘˘ ≈ đ?‘“đ?‘Ľ (đ?‘Ž, đ?‘?)Δđ?‘Ľ + đ?‘“đ?‘Ś (đ?‘Ž, đ?‘?)Δđ?‘Ś, Δđ?‘Ł ≈ đ?‘”đ?‘Ľ (đ?‘Ž, đ?‘?)Δđ?‘Ľ + đ?‘”đ?‘Ś (đ?‘Ž, đ?‘?)Δđ?‘Ś, for Δđ?‘Ľ and Δđ?‘Ś sufficiently small. In matrix form, this reads [ ] [ ][ ] Δđ?‘˘ đ?‘“đ?‘Ľ (đ?‘Ž, đ?‘?) đ?‘“đ?‘Ś (đ?‘Ž, đ?‘?) Δđ?‘Ľ , ≈ Δđ?‘Ł đ?‘”đ?‘Ľ (đ?‘Ž, đ?‘?) đ?‘”đ?‘Ś (đ?‘Ž, đ?‘?) Δđ?‘Ś for Δđ?‘Ľ and Δđ?‘Ś sufficiently small. This looks like our usual form of the linear approximation, except we have the 2 Ă— 2 matrix in place of the derivative. With this as motivation, we define: D EFINITION 10.2. The derivative matrix of a mapping đ??š : â„?2 → â„?2 with component functions đ?‘“, đ?‘”, i.e. đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘“ (đ?‘Ľ, đ?‘Ś), đ?‘”(đ?‘Ľ, đ?‘Ś)), is denoted đ??ˇđ??š and defined by [ ] đ?‘“đ?‘Ľ đ?‘“đ?‘Ś đ??ˇđ??š = . đ?‘”đ?‘Ľ đ?‘”đ?‘Ś If we introduce the vector notation [ Δđ?‘˘ =

Δ� Δ�

]

[ ,

Δ� =

Δ� Δ�

] ,

then the linear approximation formula for mappings becomes Δđ?‘˘ ≈ đ??ˇđ??š (đ?‘Ž)Δđ?‘Ľ, for Δđ?‘Ľ sufficiently small. Using linear approximation to estimate the image of a point under a mapping is the same kind of work we have already looked at when using linear approximation for scalar functions, but applied multiple times.

E XAMPLE 18. Define a mapping đ??š : â„?2 → â„?2 by √ √ (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = ( đ?‘Ľ − đ?‘Ś, 7 đ?‘Ľ + đ?‘Ś). √ √ Use the linear approximation to approximate đ??š (125, 5) = ( 120, 7 130). 21

WATERLOO SOS E XAM -AID: MATH237 S OLUTION. The derivative matrix of đ??š is [ đ??ˇđ??š (đ?‘Ľ, đ?‘Ś) =

√1 2 đ?‘Ľâˆ’đ?‘Ś √1 7 7 đ?‘Ľ+đ?‘Ś 6

1 − 2√đ?‘Ľâˆ’đ?‘Ś 7

1 √ 7 đ?‘Ľ+đ?‘Ś 6

] .

Choose đ?‘Ž = (124.5, 3.5). The image is đ??š (125.5, 4.5) = (11, 2). Then we have the approximation ][ [ ] [ ] [ ] [ 1 ] 1 − Δđ?‘˘ 0.5 Δđ?‘Ľ −0.04545 2(11) 2(11) ≈ đ??ˇđ??š (124.5, 3.5) = ≈ . 1 1 Δđ?‘Ł 1.5 Δđ?‘Ś 0.01042 7(26 ) 7(26 ) Thus đ??š (125, 5) ≈ (11, 2) + (−0.04545, 0.01042) = (10.95455, 2.01042). The calculator value is (10.95445, 2.00443). One can generalize the notion of a vector-valued mapping to a mapping đ??š : â„?đ?‘› → â„?đ?‘š defined by a set of đ?‘š component functions, đ?‘˘1 = đ?‘“1 (đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› ) .. . đ?‘˘đ?‘š = đ?‘“đ?‘š (đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› ), or in vector notation, đ?‘˘ = đ??š (đ?‘Ľ) = (đ?‘“1 (đ?‘Ľ), . . . , đ?‘“đ?‘š (đ?‘Ľ)). We assume that đ??š has continuous partial derivatives, and define the derivative matrix of đ??š to be the đ?‘š Ă— đ?‘› matrix ⎥ ∂đ?‘“1 ∂đ?‘“1 ⎤ â‹… â‹… â‹… ∂đ?‘Ľ ∂đ?‘Ľ1 đ?‘› ⎢ .. ⎼ . đ??ˇđ??š (đ?‘Ľ) = ⎣ ... . ⎌ ∂đ?‘“đ?‘š ∂đ?‘“đ?‘š â‹… â‹… â‹… ∂đ?‘Ľđ?‘› ∂đ?‘Ľ1 The linear approximation generalizes in the expected way: Δđ?‘˘ ≈ đ??ˇđ??š (đ?‘Ž)Δđ?‘Ľ, where

⎥

⎥

⎤ Δđ?‘˘1 ⎢ ⎼ Δđ?‘˘ = ⎣ ... ⎌ ∈ â„?đ?‘š , Δđ?‘˘đ?‘š

⎤ Δđ?‘Ľ1 ⎢ ⎼ Δđ?‘Ľ = ⎣ ... ⎌ ∈ â„?đ?‘› . Δđ?‘Ľđ?‘›

E XERCISE 20. Use the linear approximation to simultaneously approximate √

and

1 2 2 (4.04)

+ 0.98.

22

√

(0.98)3 + 4(1.99),

√ 3

(1.99)2 + 4.04,

WATERLOO SOS E XAM -AID: MATH237

11

Composite Mappings and Chain Rule (12.3)

Suppose we have two mappings, đ??š : â„?2 → â„?2 and đ??ş : â„?2 → â„?2 , defined by đ??š (đ?‘˘, đ?‘Ł) = (đ?‘?(đ?‘˘, đ?‘Ł), đ?‘ž(đ?‘˘, đ?‘Ł)),

đ??ş(đ?‘Ľ, đ?‘Ś) = (đ?‘˘(đ?‘Ľ, đ?‘Ś), đ?‘Ł(đ?‘Ľ, đ?‘Ś)).

Then we can define a composite mapping đ??š ∘ đ??ş by (đ??š ∘ đ??ş)(đ?‘Ľ, đ?‘Ś) = (đ?‘?(đ?‘˘(đ?‘Ľ, đ?‘Ś), đ?‘Ł(đ?‘Ľ, đ?‘Ś)), đ?‘ž(đ?‘˘(đ?‘Ľ, đ?‘Ś), đ?‘Ł(đ?‘Ľ, đ?‘Ś))), which maps from the domain of đ??ş into the range of đ??š (provided the range of đ??ş is in the domain of đ??š ). Similar to the scalr function case, we are interested in the derivative matrix đ??ˇ(đ??š ∘ đ??ş) and how it is related to đ??ˇđ??š and đ??ˇđ??ş. T HEOREM 11.1 [C HAIN R ULE (M ATRIX F ORM )]. Let đ??š : â„?2 → â„?2 and đ??ş : â„?2 → â„?2 be mappings. Suppose đ??ş has continuous partial derivatives at đ?‘Ľ and đ??š has continuous partial derivatives at đ?‘˘ = đ??ş(đ?‘Ľ). Then the composite mapping đ??š ∘ đ??ş has continuous partial derivatives at đ?‘Ľ and moreover, đ??ˇ(đ??š ∘ đ??ş)(đ?‘Ľ) = đ??ˇđ??š (đ?‘˘)đ??ˇđ??ş(đ?‘Ľ).

E XAMPLE 19. Define the mappings (đ?‘?, đ?‘ž) = đ??š (đ?‘˘, đ?‘Ł) = ((đ?‘˘+đ?‘Ł)2 , (đ?‘˘âˆ’đ?‘Ł)2 ) and (đ?‘˘, đ?‘Ł) = đ??ş(đ?‘Ľ, đ?‘Ś) = (đ?‘Ľ cos đ?‘Ś, đ?‘Ľ sin đ?‘Ś). Form the composite mapping đ??š ∘ đ??ş. Find the derivative matrices đ??ˇđ??ş, đ??ˇđ??š and đ??ˇ(đ??š ∘ đ??ş) and verify the Chain Rule for matrices. S OLUTION. The composite mapping is given by (đ?‘?, đ?‘ž) = (đ??š ∘ đ??ş)(đ?‘Ľ, đ?‘Ś) = đ??š (đ?‘Ľ cos đ?‘Ś, đ?‘Ľ sin đ?‘Ś) = (đ?‘Ľ2 (cos2 đ?‘Ś + 2 cos đ?‘Ś sin đ?‘Ś + sin2 đ?‘Ś), đ?‘Ľ2 (cos2 đ?‘Ś − 2 cos đ?‘Ś sin đ?‘Ś + sin2 đ?‘Ś)) = (đ?‘Ľ2 (1 + sin 2đ?‘Ś), đ?‘Ľ2 (1 − sin 2đ?‘Ś)). The derivative matrices are [ đ??ˇđ??š (đ?‘˘) = [ đ??ˇ(đ??š ∘ đ??ş)(đ?‘Ľ) =

2(đ?‘˘ + đ?‘Ł) 2(đ?‘˘ + đ?‘Ł) 2(đ?‘˘ − đ?‘Ł) −2(đ?‘˘ − đ?‘Ł)

]

[ ,

2đ?‘Ľ(1 + sin 2đ?‘Ś) 2đ?‘Ľ2 cos 2đ?‘Ś 2đ?‘Ľ(1 − sin 2đ?‘Ś) −2đ?‘Ľ2 cos 2đ?‘Ś

đ??ˇđ??ş(đ?‘Ľ) =

cos đ?‘Ś sin đ?‘Ś

−đ?‘Ľ sin đ?‘Ś đ?‘Ľ cos đ?‘Ś

] .

The matrix product gives (using đ?‘˘ = đ?‘Ľ cos đ?‘Ś, đ?‘Ł = đ?‘Ľ sin đ?‘Ś) [ ][ ] 2(đ?‘˘ + đ?‘Ł) 2(đ?‘˘ + đ?‘Ł) cos đ?‘Ś −đ?‘Ľ sin đ?‘Ś đ??ˇđ??š (đ?‘˘)đ??ˇđ??ş(đ?‘Ľ) = 2(đ?‘˘ − đ?‘Ł) −2(đ?‘˘ − đ?‘Ł) sin đ?‘Ś đ?‘Ľ cos đ?‘Ś [ ] 2(đ?‘˘ + đ?‘Ł)(cos đ?‘Ś + sin đ?‘Ś) 2đ?‘Ľ(đ?‘˘ + đ?‘Ł)(cos đ?‘Ś − sin đ?‘Ś) = 2(đ?‘˘ − đ?‘Ł)(cos đ?‘Ś − sin đ?‘Ś) −2đ?‘Ľ(đ?‘˘ − đ?‘Ł)(cos đ?‘Ś + sin đ?‘Ś) [ ] 2đ?‘Ľ(cos đ?‘Ś + sin đ?‘Ś)2 2đ?‘Ľ2 (cos2 đ?‘Ś − sin2 đ?‘Ś) = 2đ?‘Ľ(cos đ?‘Ś − sin đ?‘Ś)2 −2đ?‘Ľ2 (cos2 đ?‘Ś − sin2 đ?‘Ś) [ ] 2đ?‘Ľ(1 + sin 2đ?‘Ś) 2đ?‘Ľ2 cos 2đ?‘Ś = 2đ?‘Ľ(1 − sin 2đ?‘Ś) −2đ?‘Ľ2 cos 2đ?‘Ś = đ??ˇ(đ??š ∘ đ??ş)(đ?‘Ľ), where we have used the double angle identities in the last equality. 23

] ,

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 21. Define the mappings (đ?‘?, đ?‘ž) = đ??š (đ?‘˘, đ?‘Ł) = (đ?‘˘2 + đ?‘Ł 2 , đ?‘˘2 − đ?‘Ł 2 ) and (đ?‘˘, đ?‘Ł) = đ??ş(đ?‘Ľ, đ?‘Ś) = (đ?‘’đ?‘Ľâˆ’đ?‘Ś , đ?‘’đ?‘Ľ+đ?‘Ś ). Use the linear approximation to approximate the image of (đ?‘˘, đ?‘Ł) = (0.97, 1.01) under đ??š ∘ đ??ş.

12

Jacobian and Inverse Mapping Theorem (13.1-13.2)

D EFINITION 12.1. A mapping đ??š : â„?2 → â„?2 is one-to-one on a subset đ??ˇđ?‘Ľđ?‘Ś ⊂ â„?2 if and only if ⇒

đ??š (đ?‘Ž) = đ??š (đ?‘?)

for all đ?‘Ž, đ?‘? ∈ đ??ˇđ?‘Ľđ?‘Ś .

đ?‘Ž = đ?‘?,

D EFINITION 12.2. Suppose that a mapping đ??š : â„?2 → â„?2 is one-to-one on đ??ˇđ?‘Ľđ?‘Ś and that the image of đ??ˇđ?‘Ľđ?‘Ś under đ??š is đ??ˇđ?‘˘đ?‘Ł ⊂ â„?2 . Then đ??š has an inverse mapping đ??š −1 mapping đ??ˇđ?‘˘đ?‘Ł onto đ??ˇđ?‘Ľđ?‘Ś such that (đ?‘Ľ, đ?‘Ś) = đ??š −1 (đ?‘˘, đ?‘Ł)

if and only if

đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘˘, đ?‘Ł).

We are interested in the relationship between the derivative matrix đ??ˇđ??š and the invertibility of đ??š . T HEOREM 12.3. Let đ??š : â„?2 → â„?2 be a mapping that maps đ??ˇđ?‘Ľđ?‘Ś onto đ??ˇđ?‘˘đ?‘Ł . Suppose đ??š has continuous partial derivatives at đ?‘Ľ ∈ đ??ˇđ?‘Ľđ?‘Ś and has an inverse mapping đ??š −1 of đ??š which has continuous partial derivatives at đ?‘˘ = đ??š (đ?‘Ľ) ∈ đ??ˇđ?‘˘đ?‘Ł . Then đ??ˇđ??š −1 (đ?‘˘)đ??ˇđ??š (đ?‘Ľ) = đ??ź.

E XERCISE 22. Define the mapping (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘Ľ(đ?‘Ľ2 + đ?‘Ľ + 2đ?‘Ś) + đ?‘Ś 2 , đ?‘Ľ + đ?‘Ś). Find the inverse mapping đ??š −1 . Find the derivative matrices đ??ˇđ??š , đ??ˇđ??š −1 and verify that đ??ˇđ??š −1 (đ?‘˘) is the matrix inverse of đ??ˇđ??š (đ?‘Ľ).

D EFINITION 12.4. Suppose đ??š : â„?2 → â„?2 is a mapping defined by (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘“ (đ?‘Ľ, đ?‘Ś), đ?‘”(đ?‘Ľ, đ?‘Ś)). The Jacobian of đ??š , denoted ∂(đ?‘˘,đ?‘Ł) ∂(đ?‘Ľ,đ?‘Ś) , is defined by [ ] ∂đ?‘˘ ∂đ?‘˘ ∂(đ?‘˘, đ?‘Ł) ∂đ?‘Ľ ∂đ?‘Ś = det[đ??ˇđ??š (đ?‘Ľ)] = det ∂đ?‘Ł ∂đ?‘Ł . ∂(đ?‘Ľ, đ?‘Ś) ∂đ?‘Ľ ∂đ?‘Ś

E XAMPLE 20. Calculate the Jacobian

∂(đ?‘Ľ,đ?‘Ś) ∂(đ?‘&#x;,đ?œƒ)

of the mapping defined by

đ?‘Ľ = đ?‘&#x; cos đ?œƒ,

đ?‘Ś = đ?‘&#x; sin đ?œƒ.

S OLUTION. ∂(đ?‘Ľ, đ?‘Ś) = det ∂(đ?‘&#x;, đ?œƒ)

[

∂đ?‘Ľ ∂đ?‘&#x; ∂đ?‘Ś ∂đ?‘&#x;

∂đ?‘Ľ ∂đ?œƒ ∂đ?‘Ś ∂đ?œƒ

]

[ = det

cos đ?œƒ sin đ?œƒ

−đ?‘&#x; sin đ?œƒ đ?‘&#x; cos đ?œƒ

]

= đ?‘&#x; cos2 đ?œƒ + đ?‘&#x; sin2 đ?œƒ = đ?‘&#x;.

Notice this mapping is the change of Cartesian coordinates to polar coordinates. 24

WATERLOO SOS E XAM -AID: MATH237

C OROLLARY 12.5. Let đ??š : â„?2 → â„?2 be a mapping that maps đ??ˇđ?‘Ľđ?‘Ś onto đ??ˇđ?‘˘đ?‘Ł , defined by (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘“ (đ?‘Ľ, đ?‘Ś), đ?‘”(đ?‘Ľ, đ?‘Ś)). Suppose đ??š has continuous partial derivatives at đ?‘Ľ ∈ đ??ˇđ?‘Ľđ?‘Ś and has an inverse mapping đ??š −1 of đ??š which has continuous partial derivatives at đ?‘˘ = đ??š (đ?‘Ľ) ∈ đ??ˇđ?‘˘đ?‘Ł . Then the Jacobian of đ??š is non-zero, i.e. ∂(đ?‘˘, đ?‘Ł) ∕= 0, ∂(đ?‘Ľ, đ?‘Ś)

on đ??ˇđ?‘Ľđ?‘Ś .

Moreover, the Jacobian of the inverse mapping satisfies ∂(đ?‘Ľ, đ?‘Ś) = ∂(đ?‘˘, đ?‘Ł)

1 ∂(đ?‘˘,đ?‘Ł) ∂(đ?‘Ľ,đ?‘Ś)

.

E XERCISE 23. Verify the Jacobian of the inverse polar mapping satisfies this theorem. Notice the converse of this corollary is not true, i.e. a non-zero Jacobian does not imply an inverse mapping exists. However, we do have the result holding “locallyâ€?: T HEOREM 12.6 [I NVERSE M APPING T HEOREM ]. Let đ??š : â„?2 → â„?2 be a mapping defined by (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘“ (đ?‘Ľ, đ?‘Ś), đ?‘”(đ?‘Ľ, đ?‘Ś)). Suppose đ??š has continuous partial derivatives in some neighborhood of (đ?‘Ž, đ?‘?) and ∂(đ?‘˘,đ?‘Ł) ∂(đ?‘Ľ,đ?‘Ś) ∕= 0 at −1 (đ?‘Ž, đ?‘?). Then there is a neighborhood of (đ?‘Ž, đ?‘?) in which đ??š has an inverse mapping đ??š , which has continuous partial derivatives.

E XERCISE 24. Show that the mapping (đ?‘Ľ, đ?‘Ś) = đ??š (đ?‘&#x;, đ?œƒ) = (đ?‘&#x; cos đ?œƒ, đ?‘&#x; sin đ?œƒ) has an inverse mapping away from the pole. Suppose we have a small rectangle of side lengths Δđ?‘Ľ and Δđ?‘Ś, with area Δđ??´đ?‘Ľđ?‘Ś = Δđ?‘ĽÎ”đ?‘Ś. Denote the area of the image of the rectangle under đ??š by Δđ??´đ?‘˘đ?‘Ł . Then

∂(đ?‘˘, đ?‘Ł)

Δđ??´đ?‘Ľđ?‘Ś , Δđ??´đ?‘˘đ?‘Ł ≈

∂(đ?‘Ľ, đ?‘Ś)

where the Jacobian is evaluated in the rectangle. This approximation becomes increasingly accurate as Δ� and Δ� approach 0.

E XAMPLE 21. Calculate the approximate area of the image of a small rectangle of area Δđ?‘ĽÎ”đ?‘Ś, located at the point (đ?‘Ľ, đ?‘Ś), under the mapping đ??š defined by (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘Ľ cos đ?œƒ − đ?‘Ś sin đ?œƒ, đ?‘Ľ sin đ?œƒ + đ?‘Ś cos đ?œƒ), where đ?œƒ is a constant. Explain the result geometrically.

25

WATERLOO SOS E XAM -AID: MATH237 S OLUTION. Differentiating, [ đ??ˇđ??š (đ?‘Ľ, đ?‘Ś) =

cos đ?œƒ sin đ?œƒ

− sin đ?œƒ cos đ?œƒ

] ,

so the Jacobian is given by cos2 đ?œƒ + sin2 đ?œƒ = 1, and Δđ??´đ?‘˘đ?‘Ł ≈ âˆŁ1âˆŁ Δđ??´đ?‘Ľđ?‘Ś = đ??´đ?‘Ľđ?‘Ś . That is, area is preserved regardless of the position. This makes sense, since đ??š is the “rotation by đ?œƒ counterclockwiseâ€? mapping.

E XERCISE 25. Calculate the approximate area of the image of a small rectangle of area Δđ?‘ĽÎ”đ?‘Ś, located at the point (đ?‘Ľ, đ?‘Ś), under the mapping đ??š defined by ) ( 2 2đ?‘ đ?‘Ą 2đ?‘ đ?‘Ą đ?‘Ą2 − đ?‘ 2 đ?‘ − đ?‘Ą2 đ?‘Ľ+ 2 đ?‘Ś, đ?‘Ľ+ 2 đ?‘Ś , (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = đ?‘ 2 + đ?‘Ą2 đ?‘ + đ?‘Ą2 đ?‘ 2 + đ?‘Ą2 đ?‘ + đ?‘Ą2 where đ?‘ , đ?‘Ą are constants. Explain the result geometrically. We generalized the notion of a mapping đ??š : â„?2 → â„?2 to a mapping â„?đ?‘› → â„?đ?‘š earlier; and now we generalize our current notion of the Jacobian, in the case when đ?‘š = đ?‘›. D EFINITION 12.7. For đ??š : â„?đ?‘› → â„?đ?‘› , given by đ?‘˘ = đ??š (đ?‘Ľ) = (đ?‘“1 (đ?‘Ľ), . . . , đ?‘“đ?‘› (đ?‘Ľ)) where đ?‘˘ = (đ?‘˘1 , . . . , đ?‘˘đ?‘› ) and đ?‘Ľ = (đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› ). The Jacobian of đ??š is ⎥ ∂(đ?‘˘1 , . . . , đ?‘˘đ?‘› ) ⎢ = det[đ??ˇđ??š (đ?‘Ľ)] = det ⎣ ∂(đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› )

E XAMPLE 22. Calculate the Jacobian

∂(đ?‘Ľ,đ?‘Ś,đ?‘§) ∂(đ?‘&#x;,đ?œƒ,đ?‘§)

∂đ?‘“1 ∂đ?‘Ľ1

â‹…â‹…â‹…

.. .

∂đ?‘“đ?‘› ∂đ?‘Ľ1

∂đ?‘“1 ∂đ?‘Ľđ?‘›

.. .

â‹…â‹…â‹…

∂đ?‘“đ?‘› ∂đ?‘Ľđ?‘›

⎤ ⎼ ⎌.

of the mapping defined by

đ?‘Ľ = đ?‘&#x; cos đ?œƒ,

đ?‘Ś = đ?‘&#x; sin đ?œƒ,

� = �.

S OLUTION. ⎥ ∂(đ?‘Ľ, đ?‘Ś, đ?‘§) = det ⎣ ∂(đ?‘&#x;, đ?œƒ, đ?‘§)

∂đ?‘Ľ ∂đ?‘&#x; ∂đ?‘Ś ∂đ?‘&#x; ∂đ?‘§ ∂đ?‘&#x;

∂đ?‘Ľ ∂đ?œƒ ∂đ?‘Ś ∂đ?œƒ ∂đ?‘§ ∂đ?œƒ

∂đ?‘Ľ ∂đ?‘§ ∂đ?‘Ś ∂đ?‘§ ∂đ?‘§ ∂đ?‘§

⎤

⎥

cos đ?œƒ ⎌ = det ⎣ sin đ?œƒ 0

−đ?‘&#x; sin đ?œƒ đ?‘&#x; cos đ?œƒ 0

⎤ 0 0 ⎌ = 1(đ?‘&#x; cos2 đ?œƒ + đ?‘&#x; sin2 đ?œƒ) = đ?‘&#x;. 1

Notice this mapping is the change of Cartesian coordinates to cylindrical coordinates.

E XAMPLE 23. Calculate the Jacobian

∂(đ?‘Ľ,đ?‘Ś,đ?‘§) ∂(đ?‘&#x;,đ?œƒ,đ?‘§)

đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ,

of the mapping defined by đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ,

26

đ?‘§ = đ?œŒ cos đ?œ™.

WATERLOO SOS E XAM -AID: MATH237 S OLUTION. ⎥ ∂(đ?‘Ľ, đ?‘Ś, đ?‘§) ⎢ = det ⎣ ∂(đ?œŒ, đ?œƒ, đ?œ™)

∂đ?‘Ľ ∂đ?œŒ ∂đ?‘Ś ∂đ?œŒ ∂đ?‘§ ∂đ?œŒ

∂đ?‘Ľ ∂đ?œƒ ∂đ?‘Ś ∂đ?œƒ ∂đ?‘§ ∂đ?œƒ

∂đ?‘Ľ ∂đ?œ™ ∂đ?‘Ś ∂đ?œ™ ∂đ?‘§ ∂đ?œ™

⎤

⎥ sin đ?œ™ cos đ?œƒ ⎼ ⎌ = det ⎣ sin đ?œ™ sin đ?œƒ cos đ?œ™

−đ?œŒ sin đ?œ™ sin đ?œƒ đ?œŒ sin đ?œ™ cos đ?œƒ 0

⎤ đ?œŒ cos đ?œ™ cos đ?œƒ đ?œŒ cos đ?œ™ sin đ?œƒ ⎌ −đ?œŒ sin đ?œ™

= cos đ?œ™(−đ?œŒ2 sin đ?œ™ cos đ?œ™ sin2 đ?œƒ − đ?œŒ2 sin đ?œ™ cos đ?œ™ cos2 đ?œƒ) − đ?œŒ sin đ?œ™(đ?œŒ sin2 đ?œ™ cos2 đ?œƒ + đ?œŒ sin2 đ?œ™ sin2 đ?œƒ) = −đ?œŒ2 sin đ?œ™ cos2 đ?œ™ − đ?œŒ2 sin đ?œ™ sin2 đ?œ™ = −đ?œŒ2 sin đ?œ™. Notice this mapping is the change of Cartesian coordinates to spherical coordinates. The theorems we developed for Jacobian also generalizes: under similar conditions, ∂(đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› ) = ∂(đ?‘˘1 , . . . , đ?‘˘đ?‘› )

1 ∂(đ?‘˘1 ,...,đ?‘˘đ?‘› ) ∂(đ?‘Ľ1 ,...,đ?‘Ľđ?‘› )

,

∂(đ?‘Ľ1 ,...,đ?‘Ľđ?‘› ) where ∂(đ?‘˘ is the Jacobian of the inverse mapping of đ??š . 1 ,...,đ?‘˘đ?‘› ) Suppose we have a small parallelepiped of side lengths Δđ?‘Ľ1 , . . . , Δđ?‘Ľđ?‘› , with “volumeâ€? Δđ?‘‰đ?‘Ľ1 â‹…â‹…â‹…đ?‘Ľđ?‘› = Δđ?‘Ľ1 â‹… â‹… â‹… Δđ?‘Ľđ?‘› . Denote the “volumeâ€? of the image of the parallelepiped under đ??š by Δđ?‘‰đ?‘˘1 ...đ?‘˘đ?‘› . Then

∂(đ?‘˘1 , . . . , đ?‘˘đ?‘› )

Δđ?‘‰đ?‘Ľ1 â‹…â‹…â‹…đ?‘Ľđ?‘› , Δđ?‘‰đ?‘˘1 ...đ?‘˘đ?‘› ≈

∂(đ?‘Ľ1 , . . . , đ?‘Ľđ?‘› )

where the Jacobian is evaluated in the parallelepiped. This approximation becomes increasingly accurate as �1 , . . . , �� approach 0.

13

Creating Mappings (13.3)

For one of our techniques in multiple integration, it will be extremely useful to be able to transform regions to simpler ones via an invertible mapping.

E XAMPLE 24. Create an invertible mapping đ??š which will transform the region đ??ˇđ?‘Ľđ?‘Ś bounded by the lines đ?‘Ś = đ?‘Ľ, đ?‘Ś = 4 − đ?‘Ľ, đ?‘Ś = 2 − đ?‘Ľ, and đ?‘Ś = 0, into a square in the đ?‘˘đ?‘Ł-plane. Calculate the Jacobian of đ??š −1 . S OLUTION. The region looks like:

There are many possible solutions for this. One possibility is to notice that we want the lines đ?‘Ś = 4 − đ?‘Ľ and đ?‘Ś = 2 − đ?‘Ľ, or equivalently, đ?‘Ľ + đ?‘Ś = 4 and đ?‘Ľ + đ?‘Ś = 2. We will choose đ?‘˘ = đ?‘Ľ+đ?‘Ś 2 , so that we will get the lines đ?‘˘ = 1 and đ?‘˘ = 2 in hopes of obtaining a unit square. Since we want the line đ?‘Ś = đ?‘Ľ, we are hoping for 27

WATERLOO SOS E XAM -AID: MATH237 đ?‘Ł = đ?‘Ľ − đ?‘Ś = 0, or at least some scalar value. However, our upper bound becomes cumbersome, since if we want đ?‘Ł = 1, we need đ?‘Ľ − đ?‘Ś = 1 which does not hold. Making use of the fact we have đ?‘Ś = 0, we do have đ?‘Ľâˆ’đ?‘Ś đ?‘Ľâˆ’đ?‘Ś đ?‘Ľ+đ?‘Ś = 1, so we try đ?‘Ł = đ?‘Ľ+đ?‘Ś . Notice the line đ?‘Ś = đ?‘Ľ still maps to the line đ?‘Ł = 0, and now the line đ?‘Ś = 0 maps đ?‘Ľ to đ?‘Ł = đ?‘Ľ = 1. Thus, define the mapping ( ) đ?‘Ľ+đ?‘Ś đ?‘Ľâˆ’đ?‘Ś (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = , . 2 đ?‘Ľ+đ?‘Ś This mapping takes đ??ˇđ?‘Ľđ?‘Ś to the square đ??ˇđ?‘˘đ?‘Ł defined by the points (1, 0), (2, 0), (2, 1), (1, 1). Indeed, đ??ˇđ?‘Ľđ?‘Ś is described by 2 − đ?‘Ľ ≤ đ?‘Ś ≤ 4 − đ?‘Ľ, so that 2 ≤ đ?‘Ľ + đ?‘Ś ≤ 4, and thus 1 ≤ đ?‘Ľ+đ?‘Ś 2 = đ?‘˘ ≤ 2. đ??ˇđ?‘Ľđ?‘Ś also has 0 ≤ đ?‘Ś ≤ đ?‘Ľ, đ?‘Ľâˆ’đ?‘Ś so that đ?‘Ľ − đ?‘Ś ≼ 0 and 2đ?‘Ś ≼ 0. The former gives đ?‘Ł = đ?‘Ľ+đ?‘Ś ≼ 0 since 2đ?‘˘ = đ?‘Ľ + đ?‘Ś ≼ 1 > 0. The latter gives 2đ?‘Ś 1 − đ?‘Ł = 1 − đ?‘Ľâˆ’đ?‘Ś đ?‘Ľ+đ?‘Ś = đ?‘Ľ+đ?‘Ś ≼ 0 for the same reason, so that đ?‘Ł ≤ 1. The Jacobian is ] [ [ ] 1 1 1 1 ∂(đ?‘˘, đ?‘Ł) 2 2 2 2 = det (đ?‘Ľ+đ?‘Ś)−(đ?‘Ľâˆ’đ?‘Ś) −(đ?‘Ľ+đ?‘Ś)−(đ?‘Ľâˆ’đ?‘Ś) = det 2đ?‘Ś 2đ?‘Ľ − (đ?‘Ľ+đ?‘Ś) ∂(đ?‘Ľ, đ?‘Ś) 2 (đ?‘Ľ+đ?‘Ś)2 (đ?‘Ľ+đ?‘Ś)2 (đ?‘Ľ+đ?‘Ś)2 =−

đ?‘Ś đ?‘Ľ+đ?‘Ś 1 đ?‘Ľ − =− =− . (đ?‘Ľ + đ?‘Ś)2 (đ?‘Ľ + đ?‘Ś)2 (đ?‘Ľ + đ?‘Ś)2 đ?‘Ľ+đ?‘Ś

The Jacobian is non-zero on đ??ˇđ?‘Ľđ?‘Ś , and đ??š has continuous partial derivatives on đ??ˇđ?‘Ľđ?‘Ś , so đ??š is invertible by the Inverse Mapping Theorem. By Corollary 12.5, we have ∂(đ?‘Ľ, đ?‘Ś) = ∂(đ?‘˘, đ?‘Ł)

1 ∂(đ?‘˘,đ?‘Ł) ∂(đ?‘Ľ,đ?‘Ś)

=−

1 1 =− , đ?‘Ľ+đ?‘Ś 2đ?‘˘

using the fact that đ?‘˘ = đ?‘Ľ+đ?‘Ś 2 . Alternatively, we can attempt to find the inverse mapping and find its Jacobian directly: Notice đ?‘˘đ?‘Ł = đ?‘Ľâˆ’đ?‘Ś đ?‘Ľ+đ?‘Ś đ?‘Ľâˆ’đ?‘Ś 2 đ?‘Ľ+đ?‘Ś = 2 . A standard trick, notice only the sign of đ?‘Ś has changed from đ?‘˘ to đ?‘˘đ?‘Ł. In particular, we have đ?‘˘ + đ?‘˘đ?‘Ł =

đ?‘Ľ+đ?‘Ś đ?‘Ľâˆ’đ?‘Ś + = đ?‘Ľ, 2 2

đ?‘˘ − đ?‘˘đ?‘Ł =

đ?‘Ľ+đ?‘Ś đ?‘Ľâˆ’đ?‘Ś − = đ?‘Ś. 2 2

Then the inverse mapping is given by (đ?‘Ľ, đ?‘Ś) = đ??š −1 (đ?‘˘, đ?‘Ł) = (đ?‘˘ + đ?‘˘đ?‘Ł, đ?‘˘ − đ?‘˘đ?‘Ł), with Jacobian ∂(đ?‘Ľ, đ?‘Ś) = det ∂(đ?‘˘, đ?‘Ł)

[

1+đ?‘Ł 1−đ?‘Ł

đ?‘˘ −đ?‘˘

] = −đ?‘˘(1 + đ?‘Ł) − đ?‘˘(1 − đ?‘Ł) = −2đ?‘˘.

E XAMPLE 25. Create an invertible mapping đ??š which will transform the paralellogram đ??ˇđ?‘Ľđ?‘Ś with vertices (2, 1), (3, 2), (6, 2), and (5, 1), into a square in the đ?‘˘đ?‘Ł-plane. Calculate the Jacobian of đ??š −1 . S OLUTION. đ??ˇđ?‘Ľđ?‘Ś looks like:

28

WATERLOO SOS E XAM -AID: MATH237

The parallelogram can be defined by the region bounded by the lines đ?‘Ś = đ?‘Ľ − 1, đ?‘Ś = đ?‘Ľ − 4, đ?‘Ś = 1, and đ?‘Ś = 2. Letting đ?‘˘ = đ?‘Ś, we have the bound 1 ≤ đ?‘˘ ≤ 2. To get a square, we want the range of đ?‘Ł to be of unit length. The first two equations can be written as the level curves, đ?‘Ľ − đ?‘Ś = 1, and đ?‘Ľ − đ?‘Ś = 4. Putting √ đ?‘Ł = đ?‘Ľ − đ?‘Ś, we get the desired bound 1 ≤ đ?‘Ł ≤ 2. Thus, define the mapping √ (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘Ś, đ?‘Ľ − đ?‘Ś). This mapping takes đ??ˇđ?‘Ľđ?‘Ś to the square đ??ˇđ?‘˘đ?‘Ł defined by the points (1, 1), (1, 2), (2, 2), (2, 1). The Jacobian of đ??š is given by [ ] 0 1 ∂(đ?‘˘, đ?‘Ł) 1 = det =− √ . √1 √1 − ∂(đ?‘Ľ, đ?‘Ś) 2 đ?‘Ľ −đ?‘Ś 2 đ?‘Ľâˆ’đ?‘Ś 2 đ?‘Ľâˆ’đ?‘Ś The Jacobian is non-zero on đ??ˇđ?‘Ľđ?‘Ś , and đ??š has continuous partial derivatives on đ??ˇđ?‘Ľđ?‘Ś , so đ??š is invertible by the Inverse Mapping Theorem. By Corollary 12.5, we have ∂(đ?‘Ľ, đ?‘Ś) = ∂(đ?‘˘, đ?‘Ł)

1 ∂(đ?‘˘,đ?‘Ł) ∂(đ?‘Ľ,đ?‘Ś)

√ = −2 đ?‘Ľ − đ?‘Ś = −2đ?‘Ł,

√ using the fact that đ?‘Ł = đ?‘Ľ − đ?‘Ś. Alternatively, we can attempt to find the inverse mapping and find its Jacobian directly: We easily get √ đ?‘Ś = đ?‘˘, so đ?‘Ł = đ?‘Ľ − đ?‘Ś yields đ?‘Ł 2 = đ?‘Ľ − đ?‘˘, so that đ?‘Ľ = đ?‘˘ + đ?‘Ł 2 . Then the inverse mapping is given by (đ?‘Ľ, đ?‘Ś) = đ??š −1 (đ?‘˘, đ?‘Ł) = (đ?‘˘ + đ?‘Ł 2 , đ?‘˘), with Jacobian ∂(đ?‘Ľ, đ?‘Ś) = det ∂(đ?‘˘, đ?‘Ł)

[

1 1

2đ?‘Ł 0

] = −2đ?‘Ł.

E XAMPLE 26. Create an invertible mapping which will transform the ellipse unit circle đ?‘˘2 + đ?‘Ł 2 = 1. S OLUTION. Using đ?‘˘ = đ?‘Ľâˆ’â„Ž đ?‘Ž and đ?‘Ł = equation. The desired mapping is

đ?‘Śâˆ’đ?‘˜ đ?‘?

(đ?‘Ľâˆ’â„Ž)2 đ?‘Ž2

+

(đ?‘Śâˆ’đ?‘˜)2 đ?‘?2

= 1 into the

gives the desired transformation, by examining the form of the (

(đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) =

đ?‘Ľâˆ’â„Ž đ?‘Śâˆ’đ?‘˜ , đ?‘Ž đ?‘?

It is easy to see that (đ?‘Ľ, đ?‘Ś) = (đ?‘Žđ?‘˘ + â„Ž, đ?‘?đ?‘Ł + đ?‘˜) is the inverse mapping. 29

) .

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 26. Find an invertible mapping which will transform the region đ??ˇđ?‘Ľđ?‘Ś in the first quadrant bounded by the curves đ?‘Ś − đ?‘Ľ2 = 1, đ?‘Ś − đ?‘Ľ2 = 2, đ?‘Ľđ?‘Ś 2 = 1, and đ?‘Ľđ?‘Ś 2 = 4 into a rectangle in the đ?‘˘đ?‘Ł-plane.

14

Double Integrals (14.1)

We will define integrability of 2 variable functions in an analogous way to single variables. Let đ??ˇ be a closed and bounded set in â„?2 whose boundary is a piecewise smooth closed curve. Let đ?‘“ : â„?2 → â„? be a function which is bounded on đ??ˇ. Subdivide đ??ˇ into rectangles using lines parallel to the axes, forming a partition đ?‘ƒ . Label the đ?‘› rectangles which lie completely inside đ??ˇ in some order, and denote their areas by Δđ??´đ?‘– , đ?‘– = 1, . . . , đ?‘›. Choose a point (đ?‘Ľđ?‘– , đ?‘Śđ?‘– ) in the đ?‘–-th rectangle, and form the Riemann sum đ?‘› ∑

đ?‘“ (đ?‘Ľđ?‘– , đ?‘Śđ?‘– )Δđ??´đ?‘– .

đ?‘–=1

Let âˆŁÎ”đ?‘ƒ âˆŁ denote the length of the longest side out of the rectangles in đ?‘ƒ . D EFINITION 14.1. A function đ?‘“ : â„?2 → â„? which is bounded on a closed bounded set đ??ˇ ⊂ â„?2 is integrable on đ??ˇ if all Riemann sums approach a common value as âˆŁÎ”đ?‘ƒ âˆŁ → 0. In this case, we define the double integral of đ?‘“ on đ??ˇ as âˆŤâˆŤ đ?‘› ∑ đ?‘“ (đ?‘Ľđ?‘– , đ?‘Śđ?‘– )Δđ??´đ?‘– . đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´ = lim âˆŁÎ”đ?‘ƒ âˆŁâ†’0

đ??ˇ

đ?‘–=1

We can interpret the double integral as: (a) Area. If we fix đ?‘“ (đ?‘Ľ, đ?‘Ś) = 1 for all (đ?‘Ľ, đ?‘Ś) ∈ đ??ˇ, then the double integral can be interpreted as the area of the set đ??ˇ: âˆŤâˆŤ đ??´(đ??ˇ) = 1 đ?‘‘đ??´. đ??ˇ

(b) Volume. If đ?‘“ (đ?‘Ľ, đ?‘Ś) ≼ 0 for all (đ?‘Ľ, đ?‘Ś) ∈ đ??ˇ, then the double integral can be interpreted as the volume of the 3-D set defined by đ?‘† = {(đ?‘Ľ, đ?‘Ś, đ?‘§) : 0 ≤ đ?‘§ ≤ đ?‘“ (đ?‘Ľ, đ?‘Ś), (đ?‘Ľ, đ?‘Ś) ∈ đ??ˇ}, which is the solid region of height đ?‘“ (đ?‘Ľ, đ?‘Ś) and base đ??ˇ. âˆŤâˆŤ đ?‘‰ (đ?‘†) = đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´. đ??ˇ

(c) Mass. If đ?‘“ (đ?‘Ľ, đ?‘Ś) gives mass per unit area at position (đ?‘Ľ, đ?‘Ś) (i.e. density), then the total mass of đ??ˇ will be given by: âˆŤâˆŤ đ?‘€=

đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´. đ??ˇ

30

WATERLOO SOS E XAM -AID: MATH237 (d) Probability. If đ?‘“ (đ?‘Ľ, đ?‘Ś) is the probability density function of a continuous random variable (đ?‘‹, đ?‘Œ ), then the probability that (đ?‘‹, đ?‘Œ ) ∈ đ??ˇ is given by âˆŤâˆŤ đ?‘ƒ ((đ?‘‹, đ?‘Œ ) ∈ đ??ˇ) = đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´. đ??ˇ

(e) Average value. Analogously to single variables, the average value of đ?‘“ over đ??ˇ is given by âˆŤâˆŤ 1 đ?‘“đ?‘Žđ?‘Łđ?‘” = đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´. đ??´(đ??ˇ) đ??ˇ

The basic properties of single integrals can be generalized to double integrals. T HEOREM 14.2. Suppose đ??ˇ is a closed and bounded set. (a) Linearity. For any integrable functions on đ??ˇ, đ?‘“ and đ?‘”, and any constant đ?‘?, âˆŤâˆŤ âˆŤâˆŤ âˆŤâˆŤ (đ?‘“ + đ?‘”) đ?‘‘đ??´ = đ?‘“ đ?‘‘đ??´ + đ?‘” đ?‘‘đ??´, đ??ˇ

đ??ˇ

âˆŤâˆŤ

đ??ˇ

âˆŤâˆŤ đ?‘?đ?‘“ đ?‘‘đ??´ = đ?‘?

đ??ˇ

đ?‘“ đ?‘‘đ??´. đ??ˇ

(b) Monotonicity. If đ?‘“ and đ?‘” are integrable functions on đ??ˇ with đ?‘“ (đ?‘Ľ, đ?‘Ś) ≤ đ?‘”(đ?‘Ľ, đ?‘Ś) for all (đ?‘Ľ, đ?‘Ś) ∈ đ??ˇ, then âˆŤâˆŤ âˆŤâˆŤ đ?‘“ đ?‘‘đ??´ ≤ đ?‘” đ?‘‘đ??´. đ??ˇ

đ??ˇ

(c) Triangle Inequality. If đ?‘“ is an integrable functions on đ??ˇ, then

âˆŤ âˆŤ

âˆŤâˆŤ

đ?‘“ đ?‘‘đ??´

≤ âˆŁđ?‘“ âˆŁ đ?‘‘đ??´.

đ??ˇ

đ??ˇ

(d) Decomposition. If đ??ˇ is decomposed into two closed and bounded subsets đ??ˇ1 and đ??ˇ2 by a piecewise smooth curve, then âˆŤâˆŤ âˆŤâˆŤ âˆŤâˆŤ đ?‘“ đ?‘‘đ??´ = đ?‘“ đ?‘‘đ??´ + đ?‘“ đ?‘‘đ??´. đ??ˇ

15

đ??ˇ1

đ??ˇ2

2-fold Iterated Integrals (14.2)

It is clearly not a simple task in general to calculate a double integral using Riemann sums. Instead, we may usually write the double integral as a 2-fold iterated integral. T HEOREM 15.1. Let đ??ˇ ⊂ â„?2 be defined by đ?‘Śđ?‘™ (đ?‘Ľ) ≤ đ?‘Ś ≤ đ?‘Śđ?‘˘ (đ?‘Ľ),

�� ≤ � ≤ �,

where đ?‘Śđ?‘™ (đ?‘Ľ) and đ?‘Śđ?‘˘ (đ?‘Ľ) are continuous for đ?‘Ľđ?‘™ ≤ đ?‘Ľ ≤ đ?‘Ľ. If đ?‘“ (đ?‘Ľ, đ?‘Ś) is continuous on đ??ˇ, then ) âˆŤâˆŤ âˆŤ đ?‘Ľđ?‘˘ (âˆŤ đ?‘Śđ?‘˘ (đ?‘Ľ) đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ. đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´ = đ??ˇ

đ?‘Ľđ?‘™

đ?‘Śđ?‘™ (đ?‘Ľ)

31

WATERLOO SOS E XAM -AID: MATH237

E XAMPLE 27. Evaluate

âˆŤâˆŤ đ?‘Ľ đ?‘‘đ??´, đ??ˇ

where đ??ˇ is the region bounded by đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘Ľ2 + đ?‘Ś 2 = 9. Use this to find the volume of the solid bounded by the cylinder đ?‘Ľ2 + đ?‘Ś 2 = 9, and the planes đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘§ = đ?‘Ľ. S OLUTION. đ??ˇ looks like:

√ Notice the lines đ?‘Ś = Âąđ?‘Ľ and the curve đ?‘Ľ2 + đ?‘Ś 2 = 9 intersect at đ?‘Ľ2 + đ?‘Ľ2 =√9, or đ?‘Ľ = 3/ 2. Then đ??ˇ can be decomposed into two parts: √ the isosceles triangle obtained by 0 ≤ đ?‘Ľ ≤ 3/ 2, and the segment of the circle of radius 3 obtained by 3/√ 2 ≤ đ?‘Ľ ≤ 3. For √ the first region, say đ??ˇ1 , we have −đ?‘Ľ ≤ đ?‘Ś ≤ đ?‘Ľ. For the second region, say đ??ˇ2 , we have − 9 − đ?‘Ľ2 ≤ đ?‘Ś ≤ 9 − đ?‘Ľ2 . Then âˆŤâˆŤ

âˆŤ

√ 3/ 2

âˆŤ

đ?‘Ľ

âˆŤ

đ?‘Ľ đ?‘‘đ??´ =

đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ + âˆŤ

√ 3/ 2

= 0

âˆŤ =

âˆŤ

đ?‘Ľ

đ?‘Ľđ?‘ŚâˆŁâˆ’đ?‘Ľ đ?‘‘đ?‘Ľ + √ 3/ 2

2đ?‘Ľ2 đ?‘‘đ?‘Ľ +

âˆŤ

0

3 √

3/ 2

3 √

âˆŤ

√ 3/ 2

−đ?‘Ľ

0

đ??ˇ

√

3

9−đ?‘Ľ2

√ − 9−đ?‘Ľ2 √

9−đ?‘Ľ2

đ?‘Ľđ?‘ŚâˆŁâˆ’√9−đ?‘Ľ2 đ?‘‘đ?‘Ľ

√ 2đ?‘Ľ 9 − đ?‘Ľ2 đ?‘‘đ?‘Ľ

3/ 2

√

3 2 3/ 2 2

= đ?‘Ľ3 0 − (9 − đ?‘Ľ2 )3/2 √ 3 3 3/ 2 ( )3/2 2 27 2 9 √ −0−0+ = 32 2 3 2 √ ( ) √ 9 2 2 27 √ = + = 9 2. 2 3 2 2

32

đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

Alternatively, đ??ˇ is defined by âˆŁđ?‘ŚâˆŁ ≤ đ?‘Ľ ≤

WATERLOO SOS E XAM -AID: MATH237 √ √ 9 − đ?‘Ś 2 and −3/ 2 ≤ đ?‘Ś ≤ 3/ 2. Then âˆŤ √ âˆŤ √ 2

√

3/ 2

âˆŤâˆŤ đ?‘Ľ đ?‘‘đ??´ = đ??ˇ

=

đ?‘Ľ đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś

√ −3/ 2

âˆŤ =

9−đ?‘Ś

√ 3/ 2

âˆŁđ?‘ŚâˆŁ

√ 1 2 9−đ?‘Ś2 đ?‘Ľ âˆŁâˆŁđ?‘ŚâˆŁ đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś 2

√ −3/ 2 âˆŤ 3/√2

1 2

√ −3/ 2

9 − đ?‘Ś 2 − đ?‘Ś 2 đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś

) 2 3 3/√2 9đ?‘Ś − đ?‘Ś âˆŁâˆ’3/√2 đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś 3 ( ) ( ) √ 3 2 27 √ =9 √ − = 9 2. 3 2 2 2 1 = 2

(

To find the volume of the solid bounded by the cylinder đ?‘Ľ2 + đ?‘Ś 2 = 9, and the planes đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘§ = đ?‘Ľ, we note that the solid is described by 0 ≤ đ?‘§ ≤ đ?‘Ľ and (đ?‘Ľ, đ?‘Ś) ∈ đ??ˇ, so this volume is simply the double integral we calculated above. By the last example, we can sometimes reverse the order of integration, which makes sense since our theorem for expressing the double integral as an integrated integral is symmetric in đ?‘Ľ and đ?‘Ś. However, it is sometimes easier to use one order than the other. There are two factors which can determine which method is preferable, or even required: (1) The shape of the region đ??ˇ. (2) The form of the integrand đ?‘“ (đ?‘Ľ, đ?‘Ś).

E XAMPLE 28. Evaluate

âˆŤ1âˆŤ1 0

đ?‘Ľ

2

đ?‘’đ?‘Ś đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ.

2

S OLUTION. Notice đ?‘’đ?‘Ś does not have a nice antiderivative (with respect to đ?‘Ś) to work with, so we will try to iterate the integration in another way. We can define a region đ??ˇđ?‘Ľđ?‘Ś by đ?‘Ľ ≤ đ?‘Ś ≤ 1 and 0 ≤ đ?‘Ľ ≤ 1, and đ?‘Ľ, 1 are continuous for this range of đ?‘Ľ. đ?‘’đ?‘Ś is continuous everywhere and hence on đ??ˇ.

Notice � ≤ � and 0 ≤ � gives the equation 0 ≤ � ≤ �, while � ≤ � ≤ 1 and 0 ≤ � gives 0 ≤ � ≤ 1. (We can

33

WATERLOO SOS E XAM -AID: MATH237 check these relationships are if and only if.) This corresponds with our diagram of đ??ˇđ?‘Ľđ?‘Ś . We get âˆŤ 0

1

âˆŤ

1

2

đ?‘’đ?‘Ś đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ =

âˆŤ

= 0

âˆŤâˆŤ

đ?‘Ś

âˆŤ

2

đ?‘’đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś =

âˆŤ

1

1

0

0

0

đ?‘Ľ

âˆŤ

E XERCISE 27. Evaluate

1

2

đ?‘Ś

đ?‘’đ?‘Ś đ?‘Ľ âˆŁ0 đ?‘‘đ?‘Ś

1 2 1 1 đ?‘’ đ?‘Ś đ?‘‘đ?‘Ś = đ?‘’đ?‘Ś âˆŁ0 = (đ?‘’ − 1). 2 2 đ?‘Ś2

6đ?‘Ľđ?‘Ś đ?‘‘đ??´ where đ??ˇ is the region bounded by đ?‘Ś = 0, đ?‘Ľ = 2, and đ?‘Ś = đ?‘Ľ2 .

đ??ˇ

E XERCISE 28. Reverse the order of integration in

16

âˆŤ 1 âˆŤ đ?‘’đ?‘Ś 0

1

đ?‘“ (đ?‘Ľ, đ?‘Ś) đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś.

Change of Variable Theorem (2D) (14.3)

Double integrals

âˆŤâˆŤ

đ??ť(đ?‘Ľ, đ?‘Ś) đ?‘‘đ??´ can be made simpler by changing either the integrand đ??ť(đ?‘Ľ, đ?‘Ś) or deforming

đ??ˇđ?‘Ľđ?‘Ś

the set đ??ˇđ?‘Ľđ?‘Ś using a mapping đ??š : â„?2 → â„?2 . This process is known as a change of variables in the double integral. To clarify the process, we typically use đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś in place of đ?‘‘đ??´ when working in the đ?‘Ľđ?‘Ś-plane, and đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł when working in the đ?‘˘đ?‘Ł-plane. For a mapping đ??š : â„?2 → â„?2 defined by (đ?‘Ľ, đ?‘Ś) = đ??š (đ?‘˘, đ?‘Ł) = (đ?‘“ (đ?‘˘, đ?‘Ł), đ?‘”(đ?‘˘, đ?‘Ł)), recall the geometric interpretation of the Jacobian gives

∂(đ?‘Ľ, đ?‘Ś)

Δđ??´đ?‘˘đ?‘Ł , Δđ??´đ?‘Ľđ?‘Ś ≈

∂(đ?‘˘, đ?‘Ł)

for Δđ?‘˘, Δđ?‘Ł sufficiently small. (We interchanged (đ?‘Ľ, đ?‘Ś) and (đ?‘˘, đ?‘Ł) from our earlier equation to match with our new notation in the double integral.) We now present a theorem related to this concept. T HEOREM 16.1 [C HANGE OF VARIABLE ]. Let đ??ˇđ?‘˘đ?‘Ł and đ??ˇđ?‘Ľđ?‘Ś be closed and bounded sets whose boundaries are piecewise smooth curves. Let (đ?‘Ľ, đ?‘Ś) = đ??š (đ?‘˘, đ?‘Ł) = (đ?‘“ (đ?‘˘, đ?‘Ł), đ?‘”(đ?‘˘, đ?‘Ł)) be a one-to-one mapping of đ??ˇđ?‘˘đ?‘Ł onto đ??ˇđ?‘Ľđ?‘Ś , with đ?‘“, đ?‘” ∈ đ??ś 1 , and then âˆŤâˆŤ âˆŤâˆŤ

∕= 0 on đ??ˇđ?‘˘đ?‘Ł . If đ??ť(đ?‘Ľ, đ?‘Ś) is continuous on đ??ˇđ?‘Ľđ?‘Ś ,

∂(đ?‘Ľ, đ?‘Ś)

đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł. đ??ť(đ?‘“ (đ?‘˘, đ?‘Ł), đ?‘”(đ?‘˘, đ?‘Ł))

∂(đ?‘˘, đ?‘Ł)

đ??ť(đ?‘Ľ, đ?‘Ś) đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś = đ??ˇđ?‘Ľđ?‘Ś

∂(đ?‘Ľ,đ?‘Ś) ∂(đ?‘˘,đ?‘Ł)

đ??ˇđ?‘˘đ?‘Ł

Note that usually, we have a mapping đ??š from đ??ˇđ?‘Ľđ?‘Ś to đ??ˇđ?‘˘đ?‘Ł to transform our set; then the Change of Variable Theorem requires the Jacobian of the inverse mapping đ??š −1 .

34

WATERLOO SOS E XAM -AID: MATH237 E XAMPLE 29. Evaluate âˆŤ âˆŤ exp

(

đ?‘Ľâˆ’đ?‘Ś đ?‘Ľ+đ?‘Ś

) đ?‘‘đ??´,

đ?‘Ľ+đ?‘Ś đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the set bounded by the lines đ?‘Ś = đ?‘Ľ, đ?‘Ś = 4 − đ?‘Ľ, đ?‘Ś = 2 − đ?‘Ľ, and đ?‘Ś = 0. S OLUTION. In example 24, we found that the mapping ( (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) =

đ?‘Ľ+đ?‘Ś đ?‘Ľâˆ’đ?‘Ś , 2 đ?‘Ľ+đ?‘Ś

)

maps đ??ˇđ?‘Ľđ?‘Ś onto đ??ˇđ?‘˘đ?‘Ł , a unit square defined by 1 ≤ đ?‘˘ ≤ 2, 0 ≤ đ?‘Ł ≤ 1. We also showed in that example that ∂(đ?‘Ľ, đ?‘Ś) = −2đ?‘˘. ∂(đ?‘˘, đ?‘Ł) We have âˆŤ âˆŤ exp

(

đ?‘Ľâˆ’đ?‘Ś đ?‘Ľ+đ?‘Ś

)

đ?‘Ľ+đ?‘Ś

đ?‘’đ?‘Ł âˆŁ2đ?‘˘âˆŁ đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł = 2đ?‘˘

âˆŤâˆŤ đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś =

0

đ??ˇđ?‘˘đ?‘Ł

đ??ˇđ?‘Ľđ?‘Ś

âˆŤ

1

âˆŤ

= 0

E XAMPLE 30. Evaluate

3

�� �� �� =

âˆŤ

2

1

âˆŤ

3

âˆŤ 2

đ?‘’đ?‘Ł âˆŁ2đ?‘˘âˆŁ đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł 2đ?‘˘

1

đ?‘’đ?‘Ł đ?‘‘đ?‘Ł = đ?‘’ − 1.

0

âˆŤâˆŤ đ?‘Ś đ?‘‘đ??´, đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the paralellogram with vertices (2, 1), (3, 2), (6, 2), and (5, 1). S OLUTION. In example 25, we found that the mapping √ (đ?‘˘, đ?‘Ł) = đ??š (đ?‘Ľ, đ?‘Ś) = (đ?‘Ś, đ?‘Ľ − đ?‘Ś) maps đ??ˇđ?‘Ľđ?‘Ś onto đ??ˇđ?‘˘đ?‘Ł , a unit square defined by 1 ≤ đ?‘˘ ≤ 2, 1 ≤ đ?‘Ł ≤ 2. We also showed in that example that ∂(đ?‘Ľ, đ?‘Ś) = −2đ?‘Ł. ∂(đ?‘˘, đ?‘Ł) We have âˆŤâˆŤ

âˆŤâˆŤ

âˆŤ

đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś = đ??ˇđ?‘Ľđ?‘Ś

2

âˆŤ

đ?‘˘âˆŁ2đ?‘ŁâˆŁ đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł = 2 âˆŤ

đ?‘˘âˆŁđ?‘ŁâˆŁ đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł 1

đ??ˇđ?‘˘đ?‘Ł 2

âˆŤ

=2

2

1

( �� �� �� = 2

1

1

2

1 2 � 2

)

2

âˆŁ1

(

1 2 đ?‘Ł 2

)

2

âˆŁ1 =

9 . 2

*Note that this integral is actually very easy to compute using iterated integrals, without a change of variables. 35

WATERLOO SOS E XAM -AID: MATH237

E XAMPLE 31. Evaluate

âˆŤâˆŤ đ?‘Ľ đ?‘‘đ??´, đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the region bounded by đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘Ľ2 + đ?‘Ś 2 = 9. S OLUTION. Compare to Example 27. Rather than set up an iterated integration in Cartesian coordinates, we will convert the equations from Cartesian coordinates to polar coordinates. đ??ˇđ?‘Ľđ?‘Ś is the region given by đ?‘Ľ ≼ 0, đ?‘Ś ≤ đ?‘Ľ, đ?‘Ś ≼ −đ?‘Ľ, and đ?‘Ľ2 + đ?‘Ś 2 ≤ 9, so in polar coordinates, we require đ?‘&#x; cos đ?œƒ ≼ 0, đ?‘&#x; sin đ?œƒ ≤ đ?‘&#x; cos đ?œƒ, đ?‘&#x; sin đ?œƒ ≼ −đ?‘&#x; cos đ?œƒ, and đ?‘&#x;2 ≤ 9. The first equation gives − đ?œ‹2 ≤ đ?œƒ ≤ đ?œ‹2 , the next two equations combined 5đ?œ‹ đ?œ‹ đ?œ‹ 2 give − đ?œ‹4 ≤ đ?œƒ ≤ đ?œ‹4 or 3đ?œ‹ 4 ≤ đ?œƒ ≤ 4 , which implies we must have − 4 ≤ đ?œƒ ≤ 4 . The bound on đ?‘&#x; yields 0 ≤ đ?‘&#x; ≤ 3. (Alternatively, due to the simple geometric interpretation of the polar coordinates, we can obtain these bounds directly from the set in the đ?‘Ľđ?‘Ś-plane.) Recall the Jacobian of this mapping from Cartesian coordinates to polar coordinates is given by đ?‘&#x;. We thus get âˆŤâˆŤ

âˆŤâˆŤ đ?‘Ľ đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś =

đ??ˇđ?‘Ľđ?‘Ś

âˆŤ

đ?œ‹/4

âˆŤ

đ?‘&#x; cos đ?œƒđ?‘&#x; đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ = −đ?œ‹/4

đ??ˇđ?‘&#x;đ?œƒ

đ?‘&#x;2 cos đ?œƒ đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ

0

đ?œ‹/4 1 3 3 27 đ?‘&#x; âˆŁ0 cos đ?œƒ đ?‘‘đ?œƒ = cos đ?œƒ đ?‘‘đ?œƒ −đ?œ‹/4 3 −đ?œ‹/4 3 √ √ 27 2 2 đ?œ‹/4 = sin đ?œƒ âˆŁâˆ’đ?œ‹/4 = 9 = 9 2. 3 2

âˆŤ

đ?œ‹/4

3

âˆŤ

=

E XERCISE 29. Evaluate

âˆŤâˆŤ

(3đ?‘Ľ + 6đ?‘Ś)2 đ?‘‘đ??´,

đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the region bounded by the lines đ?‘Ľ + 2đ?‘Ś = −2, đ?‘Ľ + 2đ?‘Ś = 2, đ?‘Ľ − 2đ?‘Ś = −2 and đ?‘Ľ − 2đ?‘Ś = 2.

E XERCISE 30. Evaluate

âˆŤâˆŤ

đ?‘Ľ2 − đ?‘Ľđ?‘Ś + đ?‘Ś 2 đ?‘‘đ??´,

đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the region given by đ?‘Ľ2 − đ?‘Ľđ?‘Ś + đ?‘Ś 2 = 2.

17

Triple Integrals (15.1)

We will define integrability of 2 variable functions in an analogous way to one and two variables. Let đ??ˇ be a closed and bounded set in â„?3 whose boundary consists of finitely many surface elements which are smooth except possibly at isolated points. Let đ?‘“ : â„?3 → â„? be a function which is bounded on đ??ˇ. Subdivide đ??ˇ into rectangular prisms using planes parallel to the đ?‘Ľđ?‘Ś-, đ?‘Śđ?‘§-, and đ?‘Ľđ?‘§-planes, forming a partition đ?‘ƒ . Label 36

WATERLOO SOS E XAM -AID: MATH237 the đ?‘› rectangular prisms which lie completely inside đ??ˇ in some order, and denote their volumes by Δđ?‘‰đ?‘– , đ?‘– = 1, . . . , đ?‘›. Choose a point (đ?‘Ľđ?‘– , đ?‘Śđ?‘– , đ?‘§đ?‘– ) in the đ?‘–-th rectangular prism, and form the Riemann sum đ?‘› ∑

� (�� , �� , �� )�� .

đ?‘–=1

Let âˆŁÎ”đ?‘ƒ âˆŁ denote the length of the longest side out of the rectangular prisms in đ?‘ƒ . D EFINITION 17.1. A function đ?‘“ : â„?3 → â„? which is bounded on a closed bounded set đ??ˇ ⊂ â„?3 is integrable on đ??ˇ if all Riemann sums approach a common value as âˆŁÎ”đ?‘ƒ âˆŁ → 0. In this case, we define the triple integral of đ?‘“ on đ??ˇ as âˆŤâˆŤâˆŤ đ?‘› ∑ đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰ = lim đ?‘“ (đ?‘Ľđ?‘– , đ?‘Śđ?‘– , đ?‘§đ?‘– )Δđ?‘‰đ?‘– . âˆŁÎ”đ?‘ƒ âˆŁâ†’0

đ??ˇ

đ?‘–=1

We can interpret the triple integral as: (a) Volume. If we fix đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) = 1 for all (đ?‘Ľ, đ?‘Ś, đ?‘§) ∈ đ??ˇ, then the triple integral can be interpreted as the volume of the set đ??ˇ: âˆŤâˆŤâˆŤ đ?‘‰ (đ??ˇ) = 1 đ?‘‘đ?‘‰. đ??ˇ

(b) Mass. If đ?‘“ (đ?‘Ľ, đ?‘Ś) gives mass per unit area at position (đ?‘Ľ, đ?‘Ś) (i.e. density), then the total mass of đ??ˇ will be given by: âˆŤâˆŤâˆŤ đ?‘€=

đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰. đ??ˇ

(c) Probability. If đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) is the probability density function of a continuous random variable (đ?‘‹, đ?‘Œ, đ?‘?), then the probability that (đ?‘‹, đ?‘Œ, đ?‘?) ∈ đ??ˇ is given by âˆŤâˆŤâˆŤ đ?‘ƒ ((đ?‘‹, đ?‘Œ, đ?‘?) ∈ đ??ˇ) = đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰. đ??ˇ

(d) Average value. Analogously to one and two variables, the average value of đ?‘“ over đ??ˇ is given by âˆŤâˆŤâˆŤ 1 đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰. đ?‘“đ?‘Žđ?‘Łđ?‘” = đ?‘‰ (đ??ˇ) đ??ˇ

The basic properties of single and double integrals can be generalized to triple integrals. In particular, (a)-(d) of Theorem 14.2 generalize in the obvious way (linearity, monotonicity, triangle inequality, and decomposition).

18

3-fold Iterated Integrals (15.2)

As in the 2 variable case, we may usually write the triple integral as a 3-fold iterated integral. T HEOREM 18.1. Let đ??ˇ ⊂ â„?3 be defined by đ?‘§đ?‘™ (đ?‘Ľ, đ?‘Ś) ≤ đ?‘§ ≤ đ?‘§đ?‘˘ (đ?‘Ľ, đ?‘Ś),

(đ?‘Ľ, đ?‘Ś) ∈ đ??ˇđ?‘Ľđ?‘Ś ,

37

WATERLOO SOS E XAM -AID: MATH237 where đ?‘§đ?‘™ (đ?‘Ľ, đ?‘Ś) and đ?‘§đ?‘˘ (đ?‘Ľ, đ?‘Ś) are continuous on đ??ˇđ?‘Ľđ?‘Ś , and đ??ˇđ?‘Ľđ?‘Ś is a closed and bounded set in â„?2 whose boundary is a piecewise smooth curve. If đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) is continuous on đ??ˇ, then ) âˆŤâˆŤ âˆŤ âˆŤ (âˆŤ đ?‘§đ?‘˘ (đ?‘Ľ,đ?‘Ś)

� (�, �, �) ��

đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰ = đ??ˇ

đ?‘‘đ??´.

�� (�,�)

đ??ˇđ?‘Ľđ?‘Ś

We can then use our earlier theorem to break this up into a 3-fold iterated integral. Notice also that the theorem is symmetric in đ?‘Ľ, đ?‘Ś, đ?‘§, so if đ??ˇ can be defined by đ?‘Ľđ?‘™ (đ?‘Ś, đ?‘§) ≤ đ?‘Ľ ≤ đ?‘Ľđ?‘˘ (đ?‘Ś, đ?‘§), with (đ?‘Ś, đ?‘§) ∈ đ??ˇđ?‘Śđ?‘§ for example, we may integrate with respect to đ?‘Ľ first.

E XAMPLE 32. Find the volume of the solid đ??ˇ bounded by the cylinder đ?‘Ľ2 + đ?‘Ś 2 = 9, and the planes đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘§ = đ?‘Ľ. S OLUTION. We have 0 ≤ đ?‘§ ≤ đ?‘Ľ, and so we are looking for âˆŤâˆŤâˆŤ âˆŤâˆŤ âˆŤ đ?‘Ľ âˆŤâˆŤ âˆŤâˆŤ đ?‘Ľ đ?‘‰ (đ??ˇ) = 1 đ?‘‘đ?‘‰ = đ?‘‘đ?‘§ đ?‘‘đ??´ = đ?‘§ âˆŁ0 đ?‘‘đ??´ = đ?‘Ľ đ?‘‘đ??´, đ??ˇ

đ??ˇđ?‘Ľđ?‘Ś

0

đ??ˇđ?‘Ľđ?‘Ś

đ??ˇđ?‘Ľđ?‘Ś

where đ??ˇđ?‘Ľđ?‘Ś is the region bounded by đ?‘Ľ ≼ 0, đ?‘Ś = đ?‘Ľ, đ?‘Ś = −đ?‘Ľ, and đ?‘Ľ2 + đ?‘Ś 2 = 9. Then by Example 31, we have âˆŤâˆŤâˆŤ âˆŤâˆŤ √ 1 đ?‘‘đ?‘‰ = đ?‘Ľ đ?‘‘đ??´ = 9 2. đ??ˇ

đ??ˇđ?‘Ľđ?‘Ś

E XAMPLE 33. Let đ??ˇ be the “ice-cream coneâ€?, defined by the solid bounded by đ?‘§ 2 = đ?‘Ľ2 + đ?‘Ś 2 , đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 = 1, and đ?‘§ ≼ 0. Express the integral of đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) over đ??ˇ as a 3-fold iterated integral in Cartesian coordinates. S OLUTION. Since đ?‘§ ≼ 0, the other two bounds give √ √ đ?‘Ľ2 + đ?‘Ś 2 ≤ đ?‘§ ≤ 1 − đ?‘Ľ2 − đ?‘Ś 2 . For the range on đ?‘Ľ and đ?‘Ś, we must first find the intersection of the cone đ?‘§ 2 = đ?‘Ľ2 + đ?‘Ś 2 with the sphere đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 = 1 (notice the sign of đ?‘§ does not affect the possible values of đ?‘Ľ and đ?‘Ś). Then we want to solve đ?‘Ľ2 + đ?‘Ś 2 = 1 − đ?‘Ľ2 + đ?‘Ś 2 2đ?‘Ľ2 + 2đ?‘Ś 2 = 1. √ Then the bounds on đ?‘Ś are determined by đ?‘Ś = Âą 1/2 − đ?‘Ľ2 . The region is then further defined by √ √ − 1/2 − đ?‘Ľ2 ≤ đ?‘Ś ≤ 1/2 − đ?‘Ľ2 . Finally, we determine the bound on đ?‘Ľ, by using đ?‘Ľ2 + đ?‘Ś 2 = 21 and noting that the maximal range occurs when đ?‘Ś = 0, so that √ √ −1/ 2 ≤ đ?‘Ľ ≤ 1/ 2. Thus, the integral of đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) over đ??ˇ can be written as: âˆŤâˆŤâˆŤ âˆŤ √ âˆŤ √

1/2−đ?‘Ľ2

1/ 2

đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰ = đ??ˇ

√ −1/ 2

−

√

1/2−đ?‘Ľ2

38

âˆŤ √1−đ?‘Ľ2 −đ?‘Ś2 √

� (�, �, �) �� �� ��. �2 +� 2

WATERLOO SOS E XAM -AID: MATH237

E XERCISE 31. Find the volume of the tetrahedron bounded by the coordinate planes and the plane through (2, 0, 0), (0, 1, 0), and (0, 0, 3) in two different ways.

19

Change of Variable Theorem (3D) (15.3)

Triple integrals

âˆŤâˆŤâˆŤ

đ??ť(đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘‰ can be made simpler by changing either the integrand đ??ť(đ?‘Ľ, đ?‘Ś, đ?‘§) or de-

đ??ˇđ?‘Ľđ?‘Śđ?‘§

forming the set đ??ˇđ?‘Ľđ?‘Śđ?‘§ using a mapping đ??š : â„?3 → â„?3 . This process is known as a change of variables in the triple integral. To clarify the process, we typically use đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘§ in place of đ?‘‘đ?‘‰ when working in the đ?‘Ľđ?‘Śđ?‘§-space, and đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł đ?‘‘đ?‘¤ when working in the đ?‘˘đ?‘Łđ?‘¤-space. T HEOREM 19.1 [C HANGE OF VARIABLE ]. Let đ??ˇđ?‘˘đ?‘Łđ?‘¤ and đ??ˇđ?‘Ľđ?‘Śđ?‘§ be closed and bounded sets whose boundaries consist of finitely many surface elements which are smooth except possibly at isolated points. Let (đ?‘Ľ, đ?‘Ś, đ?‘§) = đ??š (đ?‘˘, đ?‘Ł, đ?‘¤) = (đ?‘“ (đ?‘˘, đ?‘Ł, đ?‘¤), đ?‘”(đ?‘˘, đ?‘Ł, đ?‘¤), â„Ž(đ?‘˘, đ?‘Ł, đ?‘¤)) ∂(đ?‘Ľ,đ?‘Ś,đ?‘§) ∕= 0 on đ??ˇđ?‘˘đ?‘Łđ?‘¤ . If đ??ť(đ?‘Ľ, đ?‘Ś, đ?‘§) is be a one-to-one mapping of đ??ˇđ?‘˘đ?‘Łđ?‘¤ onto đ??ˇđ?‘Ľđ?‘Śđ?‘§ , with đ?‘“, đ?‘”, â„Ž ∈ đ??ś 1 , and ∂(đ?‘˘,đ?‘Ł,đ?‘¤) continuous on đ??ˇđ?‘Ľđ?‘Śđ?‘§ , then

âˆŤâˆŤâˆŤ âˆŤâˆŤâˆŤ

∂(đ?‘Ľ, đ?‘Ś, đ?‘§)

đ?‘‘đ?‘˘ đ?‘‘đ?‘Ł đ?‘‘đ?‘¤. đ??ť(đ?‘Ľ, đ?‘Ś, đ?‘§) đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘§ = đ??ť(đ?‘“ (đ?‘˘, đ?‘Ł, đ?‘¤), đ?‘”(đ?‘˘, đ?‘Ł, đ?‘¤), â„Ž(đ?‘˘, đ?‘Ł, đ?‘¤))

∂(đ?‘˘, đ?‘Ł, đ?‘¤)

đ??ˇđ?‘Ľđ?‘Śđ?‘§

đ??ˇđ?‘˘đ?‘Łđ?‘¤

2 2 2 2 2 2 E XAMPLE 34. Let đ??ˇ be âˆŤâˆŤâˆŤthe “ice-cream coneâ€?, defined by the solid bounded by đ?‘§ = đ?‘Ľ + đ?‘Ś , đ?‘Ľ + đ?‘Ś + đ?‘§ = 1, and đ?‘§ ≼ 0. Evaluate đ?‘§ đ?‘‘đ?‘‰ . đ??ˇ

S OLUTION. Recall from Example 33 that this integral becomes extremely unwieldy when using Cartesian coordinates. Instead, let us try a change of variables.

Cylindrical coordinates: đ??ˇ can be defined by 0 ≤ đ?œƒ ≤ 2đ?œ‹, and since đ?‘Ľ2 + đ?‘Ś 2 = đ?‘&#x;2 , by transforming the √ √ √ 2 2 2 equation đ?‘Ľ2 + đ?‘Ś 2 ≤ đ?‘§ ≤ 1 − √ đ?‘Ľ − đ?‘Ś , we have the bound 2đ?‘&#x; ≤ đ?‘§2 ≤ 1 − đ?‘&#x; . To√find the bounds on đ?‘&#x;, 2 we find the intersection of đ?‘§ = 1 − đ?‘&#x; and đ?‘§ = đ?‘&#x;. Then 1 − đ?‘&#x; = đ?‘&#x; , giving đ?‘&#x; = 1/ 2. Thus, we have the

39

WATERLOO SOS E XAM -AID: MATH237 √ bound 0 ≤ đ?‘&#x; ≤ 1/ 2. Recall the Jacobian for cylindrical coordinates is given by đ?‘&#x;, so that âˆŤâˆŤâˆŤ

2đ?œ‹

âˆŤ

âˆŤ

√ 1/ 2

√

âˆŤ

1−đ?‘&#x; 2

� �� =

đ?‘§đ?‘&#x; đ?‘‘đ?‘§ đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ 0

0

đ??ˇ

âˆŤ

đ?‘&#x;

√ 1/ 2

= 2đ?œ‹ âˆŤ

0 √ 1/ 2

√ 1 1−đ?‘&#x; 2 đ?‘&#x; đ?‘§ 2 âˆŁđ?‘&#x; đ?‘‘đ?‘&#x; 2

đ?‘&#x;(1 − đ?‘&#x;2 − đ?‘&#x;2 ) đ?‘‘đ?‘&#x; (0 ) 1 2 2 4 1/√2 =đ?œ‹ đ?‘&#x; − đ?‘&#x; âˆŁ0 2 4 ( ) 1 1 đ?œ‹ =đ?œ‹ − = . 4 8 8 =đ?œ‹

Spherical coordinates: đ??ˇ can be defined by 0 ≤ đ?œƒ ≤ 2đ?œ‹, and since đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ, đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ, đ?‘§ = đ?œŒ cos đ?œ™, the inequality đ?‘§ 2 ≼ đ?‘Ľ2 + đ?‘Ś 2 gives đ?œŒ2 cos2 đ?œ™ ≼ đ?œŒ2 sin2 đ?œ™, and the inequality đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 ≤ 1 gives đ?œŒ2 ≤ 1. The former equation puts no restriction on đ?œŒ, so that 0 ≤ đ?œŒ ≤ 1. Assuming đ?œŒ ∕= 0, the latter equation gives cos2 đ?œ™ ≼ sin2 đ?œ™, or cos 2đ?œ™ ≼ 0. Then 0 ≤ đ?œ™ ≤ đ?œ‹4 . Recall the Jacobian for spherical coordinates is given by đ?œŒ2 sin đ?œ™, so that âˆŤâˆŤâˆŤ âˆŤ 2đ?œ‹ âˆŤ đ?œ‹/4 âˆŤ 1 đ?‘§ đ?‘‘đ?‘‰ = (đ?œŒ cos đ?œ™)(đ?œŒ2 sin đ?œ™) đ?‘‘đ?œŒ đ?‘‘đ?œ™ đ?‘‘đ?œƒ đ??ˇ

0

0

0

âˆŤ 1 4 1 đ?œ‹/4 1 đ?œŒ âˆŁ0 sin 2đ?œ™ đ?‘‘đ?œ™ 4 2 0 ) ( 11 đ?œ‹ 1 đ?œ‹/4 = 2đ?œ‹ − cos 2đ?œ™ âˆŁ0 = . 42 2 8 (

)

= 2đ?œ‹

E XAMPLE 35. Find the volume of the solid bounded by the surface (�2 + � 2 + � 2 )2 = 2�(�2 + � 2 ). S OLUTION. We will again try two different change of variables.

Cylindrical coordinates: (Recall Example 14.) Notice (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 ≼ 0 and 2(đ?‘Ľ2 + đ?‘Ś 2 ) ≼ 0, hence đ?‘§ ≼ 0. Using đ?‘&#x;2 = đ?‘Ľ2 + đ?‘Ś 2 , our equation becomes (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 = 2đ?‘§(đ?‘Ľ2 + đ?‘Ś 2 ) (đ?‘&#x;2 + đ?‘§ 2 )2 = 2đ?‘§đ?‘&#x;2 đ?‘&#x;4 + 2(đ?‘§ 2 − đ?‘§)đ?‘&#x;2 + đ?‘§ 4 = 0, 40

WATERLOO SOS E XAM -AID: MATH237 which is a quadratic in đ?‘&#x;2 . Solving, √ √ √ −2(đ?‘§ 2 − đ?‘§) Âą 4(đ?‘§ 2 − đ?‘§)2 − 4đ?‘§ 4 đ?‘&#x;2 = = đ?‘§ − đ?‘§ 2 Âą −2đ?‘§ 3 + đ?‘§ 2 = đ?‘§ − đ?‘§ 2 Âą đ?‘§ 1 − 2đ?‘§. 2 Since đ?‘&#x;2 is real, 1 − 2đ?‘§ ≼ 0, so that đ?‘§ ≤ 21 . Together with our earlier bound, we have 0 ≤ đ?‘§ ≤ 21 . The above √ √ equation gives us the bound on đ?‘&#x; as well, by taking the positive square root (đ?‘&#x; ≼ 0): đ?‘§ − đ?‘§ 2 − đ?‘§ 1 − 2đ?‘§ ≤ √ √ đ?‘&#x; ≤ đ?‘§ − đ?‘§ 2 + đ?‘§ 1 − 2đ?‘§. Recall the Jacobian for cylindrical coordinates is given by đ?‘&#x;, so that √ âˆŤ √ âˆŤâˆŤâˆŤ âˆŤ âˆŤ đ?‘§âˆ’đ?‘§ 2 +đ?‘§ 1−2đ?‘§

1/2

2đ?œ‹

1 �� =

√ 0

0

đ??ˇ

√ đ?‘§âˆ’đ?‘§ 2 −đ?‘§ 1−2đ?‘§

đ?‘&#x; đ?‘‘đ?‘&#x; đ?‘‘đ?‘§ đ?‘‘đ?œƒ

√ 1 2 đ?‘§âˆ’đ?‘§2 +đ?‘§âˆš1−2đ?‘§ = 2đ?œ‹ đ?‘‘đ?‘§ đ?‘&#x; âˆŁâˆš 2 √ đ?‘§âˆ’đ?‘§ −đ?‘§ 1−2đ?‘§ 2 0 [ )2 (√ )2 ] âˆŤ 1/2 (√ √ √ 2 2 đ?‘‘đ?‘§ =đ?œ‹ đ?‘§ − đ?‘§ + đ?‘§ 1 − 2đ?‘§ − đ?‘§ − đ?‘§ − đ?‘§ 1 − 2đ?‘§ âˆŤ

1/2

0

âˆŤ

1/2

= 2đ?œ‹

√ đ?‘§ 1 − 2đ?‘§ đ?‘‘đ?‘§.

0

We now make the change of variables đ?‘˘ = 1 − 2đ?‘§ to get đ?‘‘đ?‘˘ = −2đ?‘‘đ?‘§, and the bounds become 1 to 0. âˆŤâˆŤâˆŤ đ??ˇ

1/2

√

0

) ( 1 − đ?‘˘âˆš 1 đ?‘‘đ?‘˘ 1 đ?‘‘đ?‘‰ = 2đ?œ‹ đ?‘˘ − đ?‘§ 1 − 2đ?‘§ đ?‘‘đ?‘§ = 2đ?œ‹ 2 2 0 1 ) ( âˆŤ 3 đ?œ‹ 1 1 đ?œ‹ 2 3 2 5 1 đ?œ‹ 10 − 6 2đ?œ‹ 2 2 2 2 = đ?‘˘ − đ?‘˘ âˆŁ0 = = . đ?‘˘ − đ?‘˘ đ?‘‘đ?‘˘ = 2 0 2 3 5 2 15 15 âˆŤ

âˆŤ

Spherical coordinates: (Recall Example 16.) Using đ?‘Ľ = đ?œŒ sin đ?œ™ cos đ?œƒ, đ?‘Ś = đ?œŒ sin đ?œ™ sin đ?œƒ and đ?‘§ = đ?œŒ cos đ?œ™, we convert the equation into spherical coordinates. We get đ?œŒ4 = 2đ?œŒ cos đ?œ™(đ?œŒ2 sin2 đ?œ™) = đ?œŒ3 sin 2đ?œ™ sin đ?œ™, which simplifies to đ?œŒ = sin 2đ?œ™ sin đ?œ™ since the pole is included in this equation. đ?œŒ ≼ 0 gives the restriction 0 ≤ đ?œ™ ≤ đ?œ‹2 . Furthermore, đ?œŒ = sin 2đ?œ™ sin đ?œ™ gives the bounds 0 ≤ đ?œŒ ≤ sin 2đ?œ™ sin đ?œ™. Recall the Jacobian for spherical coordinates is given by đ?œŒ2 sin đ?œ™, so that âˆŤâˆŤâˆŤ

âˆŤ

2đ?œ‹

âˆŤ

đ?œ‹/2

âˆŤ

sin 2đ?œ™ sin đ?œ™

đ?œŒ2 sin đ?œ™ đ?‘‘đ?œŒ đ?‘‘đ?œ™ đ?‘‘đ?œƒ

1 �� = 0

đ??ˇ

0 đ?œ‹/2

âˆŤ = 2đ?œ‹ 0

2đ?œ‹ = 3

âˆŤ

2đ?œ‹ 3

âˆŤ

=

0

1 3 sin 2đ?œ™ sin đ?œ™ đ?œŒ sin đ?œ™ âˆŁ0 đ?‘‘đ?œŒ đ?‘‘đ?œ™ 3

đ?œ‹/2

sin4 đ?œ™ sin3 2đ?œ™ đ?‘‘đ?œ™

0

0

đ?œ‹/2

(1 − cos 2đ?œ™)2 (1 − cos2 2đ?œ™) sin 2đ?œ™ đ?‘‘đ?œ™, 4

2đ?œ™ where in the last equality we have used the double angle formula to get sin2 đ?œ™ = 1−cos for the first part, 2 2 2 and the fact that sin 2đ?œ™ + cos 2đ?œ™ = 1 for the second part. We now make the change of variables đ?‘˘ = cos 2đ?œ™

41

WATERLOO SOS E XAM -AID: MATH237 to get đ?‘‘đ?‘˘ = −2 sin 2đ?œ™, and the bounds become 1 to −1. âˆŤâˆŤâˆŤ

đ?œ‹/2

(1 − cos 2đ?œ™)2 (1 − cos2 đ?œ™) sin 2đ?œ™ đ?‘‘đ?œ™, 4 0 ( ) âˆŤ −1 1 đ?œ‹ (1 − đ?‘˘)2 (1 − đ?‘˘2 ) − đ?‘‘đ?‘˘ = 6 1 2 âˆŤ 1 đ?œ‹ (đ?‘˘2 − 2đ?‘˘ + 1)(1 − đ?‘˘2 ) đ?‘‘đ?‘˘ = 12 −1 âˆŤ 1 đ?œ‹ đ?‘˘2 − đ?‘˘4 − 2đ?‘˘ + 2đ?‘˘3 + 1 − đ?‘˘2 đ?‘‘đ?‘˘ = 12 −1

2đ?œ‹ 1 đ?‘‘đ?‘‰ = 3

đ??ˇ

âˆŤ

Notice [−1, 1] is a symmetric interval, so all odd functions (i.e. đ?‘˘, đ?‘˘3 ) integrate to 0. âˆŤâˆŤâˆŤ 1 đ?‘‘đ?‘‰ = đ??ˇ

đ?œ‹ 12

âˆŤ

1

âˆŤ 1 đ?œ‹ −đ?‘˘4 + 1 đ?‘‘đ?‘˘ 12 −1 ( ) đ?œ‹ 2 2đ?œ‹ = . 2− = 12 5 15

−đ?‘˘4 − 2đ?‘˘ + 2đ?‘˘3 + 1 đ?‘‘đ?‘˘ =

−1

) ( đ?œ‹ 1 1 = − đ?‘˘5 + đ?‘˘ âˆŁâˆ’1 12 5

E XAMPLE 36. Find the volume of the solid đ??ˇ bounded by đ?‘§ = đ?‘Ľ2 + đ?‘Ś 2 − 1, đ?‘§ = đ?‘Ľ2 + đ?‘Ś 2 − 4, and đ?‘§ ≤ 0. S OLUTION. We will again try two different change of variables.

Cylindrical coordinates: The region we want is defined by đ?‘Ľ2 + đ?‘Ś 2 − 4 ≤ đ?‘§ ≤ đ?‘Ľ2 + đ?‘Ś 2 − 1 and đ?‘§ ≤ 0. Using đ?‘Ľ2 + đ?‘Ś 2 = đ?‘&#x;2 , this becomes đ?‘&#x;2 − 4 ≤ đ?‘§ ≤ đ?‘&#x;2 − 1, with đ?‘§ ≤ 0. We must split our integral into two parts; we have intersections at đ?‘&#x;2 − 4 = 0, giving đ?‘&#x; = 2, and at đ?‘&#x;2 − 1 = 0, giving đ?‘&#x; = 1. On [0, 1], we have đ?‘&#x;2 − 4 ≤ đ?‘§ ≤ đ?‘&#x;2 − 1. On [1, 2], we have đ?‘&#x;2 − 4 ≤ đ?‘§ ≤ 0. Recall

42

WATERLOO SOS E XAM -AID: MATH237 the Jacobian for cylindrical coordinates is given by đ?‘&#x;, so that âˆŤ 2âˆŤ âˆŤ 2đ?œ‹ (âˆŤ 1 âˆŤ đ?‘&#x;2 −1 âˆŤâˆŤâˆŤ đ?‘&#x; đ?‘‘đ?‘§ đ?‘‘đ?‘&#x; + 1 đ?‘‘đ?‘‰ = đ??ˇ

0

0

(âˆŤ

đ?‘&#x; 2 −4

1

đ?‘&#x; đ?‘‘đ?‘§ đ?‘‘đ?‘&#x; âˆŤ

2

đ?‘&#x;[(đ?‘&#x; − 1) − (đ?‘&#x; − 4)] đ?‘‘đ?‘&#x; +

= 2đ?œ‹

đ?‘‘đ?œƒ

đ?‘&#x; 2 −4

1 2

)

0

0

2 2

)

đ?‘&#x;[0 − (đ?‘&#x; − 4)] đ?‘‘đ?‘&#x; 1

1

2

) 4đ?‘&#x; − đ?‘&#x;3 đ?‘‘đ?‘&#x; 1 ( 0 ) 3 2 1 1 2 2 = 2đ?œ‹ đ?‘&#x; âˆŁ0 + 2đ?‘&#x;2 âˆŁ1 − đ?‘&#x;4 âˆŁ1 2 4 ( ) 3 1 15đ?œ‹ = 2đ?œ‹ +8−2−4+ = . 2 4 2 (âˆŤ

= 2đ?œ‹

âˆŤ

3đ?‘&#x; đ?‘‘đ?‘&#x; +

Spherical coordinates: Using đ?‘Ľ2 + đ?‘Ś 2 = đ?œŒ2 sin2 đ?œ™ and đ?‘§ = đ?œŒ cos đ?œ™, we get the conditions đ?œŒ2 sin2 đ?œ™ − 4 ≤ đ?œŒ cos đ?œ™ ≤ đ?œŒ2 sin2 đ?œ™ − 1,

đ?œŒ cos đ?œ™ ≤ 1.

Since đ?œŒ ≼ 0, the latter gives cos đ?œ™ ≤ 0 so that đ?œ‹2 ≤ đ?œ™ ≤ đ?œ‹. The former equation gives, after using the quadratic formula and some simple arguments, √ √ cos đ?œ™ + cos2 đ?œ™ + 4 sin2 đ?œ™ cos đ?œ™ + cos2 đ?œ™ + 16 sin2 đ?œ™ ≤đ?œŒâ‰¤ . 2 sin2 đ?œ™ 2 sin2 đ?œ™ This bound for đ?œŒ is horrendous, and although it may be possible to get to the answer from here, it is not advised. Here is a prime example of when cylindrical coordinates is much preferred over spherical.

E XAMPLE 37. Evaluate âˆŤâˆŤâˆŤ đ??ˇ

2

đ?‘’đ?‘Ľ √

+� 2 +� 2

đ?‘Ľ2 + đ?‘Ś 2

��,

where đ??ˇ is the solid bounded by the spheres of radii 1 and 2 centered at the origin, and outside the double cone đ?‘§ 2 = đ?‘Ľ2 + đ?‘Ś 2 . S OLUTION. We will again try two different change of variables.

43

WATERLOO SOS E XAM -AID: MATH237 Cylindrical coordinates: Let us ignore what the transformed region đ??ˇâ€˛ looks like and focus on the integrand. Using đ?‘Ľ2 + đ?‘Ś 2 = đ?‘&#x;2 and that the Jacobian is đ?‘&#x;, we get âˆŤâˆŤâˆŤ

2

đ?‘’đ?‘Ľ √

đ??ˇ

+� 2 +� 2

đ?‘Ľ2

+

đ?‘Ś2

đ?‘’đ?‘&#x;

âˆŤâˆŤâˆŤ đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘§ =

2

+� 2

âˆŤâˆŤâˆŤ đ?‘&#x; đ?‘‘đ?‘§ đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ =

đ?‘&#x; đ??ˇâ€˛

2

2

2

đ?‘’đ?‘&#x; đ?‘’đ?‘§ đ?‘‘đ?‘§ đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ.

đ??ˇâ€˛ 2

Notice đ?‘’đ?‘&#x; is a constant with respect to đ?‘§, and đ?‘’đ?‘§ is “impossibleâ€? to integrate. Let us try spherical coordinates instead. Spherical coordinates: Using đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 = đ?œŒ2 , since the region is bounded between the spheres of radii 1 and 2, we get 1 ≤ đ?œŒ ≤ 2. Since the region is outside the double cone đ?‘§ 2 = đ?‘Ľ2 + đ?‘Ś 2 , we have − đ?œ‹4 ≤ đ?œ™ ≤ đ?œ‹4 . Notice there is no restriction on đ?œƒ. Then as đ?‘Ľ2 + đ?‘Ś 2 = đ?œŒ2 sin2 đ?œ™ and the Jacobian is đ?œŒ2 sin đ?œ™, âˆŤâˆŤâˆŤ đ??ˇ

2

đ?‘’đ?‘Ľ √

+� 2 +� 2

đ?‘Ľ2 + đ?‘Ś 2

âˆŤ

2đ?œ‹

âˆŤ

đ?œ‹/4

1

âˆŤ

đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘§ = −đ?œ‹/4

0

âˆŤ

2đ?œ‹ âˆŤ

đ?œ‹/4

0 1

âˆŤ

−đ?œ‹/4

= (2đ?œ‹)

0

(đ?œ‹) (1 2

2

đ?‘’đ?œŒ đ?œŒ đ?‘‘đ?œŒ đ?‘‘đ?‘§ đ?‘‘đ?œƒ

= 0

2

đ?‘’đ?œŒ đ?œŒ2 sin đ?œ™ đ?‘‘đ?œŒ đ?‘‘đ?‘§ đ?‘‘đ?œƒ đ?œŒ sin đ?œ™

2

đ?‘’

đ?œŒ2

)

1

âˆŁ0 =

đ?œ‹2 (đ?‘’ − 1) . 2

R EMARK 19.2. (1) Try to get an understanding of the shape of the region you are integrating over. This will help in deciding whether to try an iterated integration, or attempt a change of variables. The constraints given and the geometry of the region may give clues as to which change of variables is appropriate. (2) Remember, do not get stuck in the thought that đ?‘Ľ2 + đ?‘Ś 2 = đ?‘&#x;2 for cylindrical coordinates or đ?‘§ = đ?œŒ cos đ?œ™ for spherical coordinates. We may interchange our definitions for these coordinate systems if the need arises. For example, suppose we were asked to find the volume of the solid bounded by the surface (đ?‘Ľ2 + đ?‘Ś 2 + đ?‘§ 2 )2 = 2đ?‘Ľ(đ?‘Ś 2 + đ?‘§ 2 ). Now there is a symmetry around the đ?‘Ľ-axis, so we may choose to define our coordinate systems as such, like đ?‘Ś 2 + đ?‘§ 2 = đ?‘&#x;2 for instance. We may also shift our coordinate system, so that the new origin lies at a point other than the old origin. (3) Be careful to check the bounds of the integration. For example, although in cylindrical and spherical coordinates we have 0 ≤ đ?œƒ ≤ 2đ?œ‹, our transformed region could have the constraint 0 ≤ đ?œƒ ≤ đ?œ‹. As another example, we typically have a bound of the form 0 ≤ đ?‘&#x; ≤ đ?‘“ (đ?œƒ, đ?‘§), where đ?‘“ is a function of đ?œƒ and đ?‘§. However, it may be the case that there is a non-zero lower bound.

E XERCISE 32. Find the volume of the solid bounded by the surface (�2 + � 2 + � 2 )2 = �.

E XERCISE 33. Let đ??ˇ be the solid bounded between the surfaces đ?‘§ = đ?‘Ľ2 + đ?‘Ś 2 and đ?‘§ = average value of đ?‘“ (đ?‘Ľ, đ?‘Ś, đ?‘§) = (đ?‘Ľ + đ?‘Ś)2 over đ??ˇ?

44

√

đ?‘Ľ2 + đ?‘Ś 2 . What is the

WATERLOO SOS E XAM -AID: MATH237 EâˆŤâˆŤâˆŤ XERCISE 34. Let đ?‘‰ be the solid bounded by đ?‘§ = 0, đ?‘§ = đ?‘Ľ2 + đ?‘Ś 2 , and đ?‘Ľ2 + (đ?‘Ś − 1)2 = 1. Evaluate √ 2đ?‘Ľ 2 đ?‘‘đ?‘‰ . đ??ˇ

đ?‘Ľ +đ?‘Ś

45

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