Section 15.9 Change of Variables in Multiple Integrals
2. Find the image of the set S under the given transformation. S = {(u, v)|0 ≤ u ≤ 1, 0 ≤ v ≤ 2}; x = u + v, y = −v.
Solution:
1584 ¤ CHAPTER 15 MULTIPLE INTEGRALS
1 ≥ ≥ cos 1. Along 3, = 1, 0 ≤ ≤ 1 ⇒ = cos 1, = sin 1 ⇒ = tan(1), 0 ≤ ≤ cos 1. Along
4, = 0 ⇒ = = 0. Thus, IV is the image of the transformation.
(d) Along 1, = 0, 0 ≤ ≤ 1 ⇒ = − = , = + 2= ⇒ = , 0 ≤ ≤ 1. Along 2, = 1, 0 ≤ ≤ 1 ⇒ = 1 − , = 1 + 2 ⇒ = 1 + (1 − )2, 1 ≥ ≥ 0. Along 3, = 1, 0 ≤ ≤ 1 ⇒
= − 1, = + 1 ⇒ = + 2, −1 ≤ ≤ 0. Finally, along 4, = 0, 0 ≤ ≤ 1 ⇒ = −, = 2 ⇒
= 2, −1 ≤ ≤ 0. Thus, V is the image of the transformation.
(e) Along 1, = 0, 0 ≤ ≤ 1 ⇒ = 2 = 0, and since = + = , 0 ≤ ≤ 1. Along 2, = 1, 0 ≤ ≤ 1 ⇒
= 1 + , = 2 ⇒ = 2 − 2, 1 ≤ ≤ 2. Along 3, = 1, 0 ≤ ≤ 1 ⇒ = 2 = 2, 1 ≤ ≤ 2. Finally, along 4, = 0, 0 ≤ ≤ 1 ⇒ = , = 2 ⇒ = 2, 0 ≤ ≤ 1. Thus, III is the image of the transformation.
(f ) Along 1, = 0, 0 ≤ ≤ 1 ⇒ = = 0, and since = 3− 3= 3, 0 ≤ ≤ 1. Along 2,
= 1, 0 ≤ ≤ 1 ⇒ = , = 1 − 3 ⇒ = 1 − 3, 1 ≥ ≥ 0. Along 3, = 1, 0 ≤ ≤ 1 ⇒
= , = 3− 1 ⇒ = 3− 1, 0 ≤ ≤ 1. Finally, along 4, = 0, 0 ≤ ≤ 1 ⇒ = 0, −1 ≤ ≤ 0.
Thus, II is the image of the transformation.
2. The transformation maps the boundary of to the boundary of the image , so we first look at side 1in the plane. 1is described by = 0, 0 ≤ ≤ 1, so = + = and = − = 0. Therefore, the image is the line segment = 0, 0 ≤ ≤ 1. 2is the line segment = 1, 0 ≤ ≤ 2, so = 1 + and = −. Eliminating , we have = 1 − ,
1 ≤ ≤ 3. 3is the line segment = 2, 0 ≤ ≤ 1, so = + 2 and = −2 ⇒ the image is the line segment = −2, 2 ≤ ≤ 3. Finally, 4is the line segment = 0, 0 ≤ ≤ 2, so = and = − ⇒ the image is the line segment
= −, 0 ≤ ≤ 2. The image of the set is the region shown in the plane, a parallelogram bounded by these four line segments.
3. The transformation maps the boundary of to the boundary of the image , so we first look at side 1in the plane. 1is described by = 0, 0 ≤ ≤ 3, so = 2 + 3 = 2 and = − = . Eliminating , we have = 2, 0 ≤ ≤ 6. 2is the line segment = 3, 0 ≤ ≤ 2, so = 6 + 3 and = 3 − . Then = 3 − ⇒ = 6 + 3(3 − ) = 15 − 3, 6 ≤ ≤ 12. 3is the line segment = 2, 0 ≤ ≤ 3, so = 2 + 6 and = − 2, giving = + 2 ⇒ = 2 + 10, 6 ≤ ≤ 12. Finally, 4is the segment = 0, 0 ≤ ≤ 2, so = 3 and = − ⇒ = −3, 0 ≤ ≤ 6.
[continued]
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21. Use the given transformation to evaluate the integral. RR
RxydA, where R is the region in the first quadrant bounded by the lines y = x and y = 3x and the hyperbolas xy = 1, xy = 3; x = u/v, y = v.
Solution:
610 ¤ CHAPTER 15 MULTIPLE INTEGRALS 15. ( )
( ) =
2 1 1 2
= 3and − 3 = (2 + ) − 3( + 2) = − − 5. To find the region in the -plane that
corresponds to we first find the corresponding boundary under the given transformation. The line through (0 0) and (2 1) is
= 12which is the image of + 2 = 12(2 + ) ⇒ = 0; the line through (2 1) and (1 2) is + = 3 which is the image of (2 + ) + ( + 2) = 3 ⇒ + = 1; the line through (0 0) and (1 2) is = 2 which is the image of
+ 2 = 2(2 + ) ⇒ = 0. Thus is the triangle 0 ≤ ≤ 1 − , 0 ≤ ≤ 1 in the -plane and
( − 3) =1 0
1−
0 (− − 5) |3| = −31 0
+522=1−
=0
= −31 0
− 2+52(1 − )2
= −31
22− 133−56(1 − )31
0= −31
2 −13+56
= −3
16. ( )
( ) =
14 14
−34 14
= 1
4, 4 + 8 = 4 · 14( + ) + 8 ·14( − 3) = 3 − 5. is a parallelogram bounded by the lines − = −4, − = 4, 3 + = 0, 3 + = 8. Since = − and = 3 + , is the image of the rectangle enclosed by the lines = −4, = 4, = 0, and = 8. Thus
(4 + 8) =4
−4
8
0(3 − 5)14 = 144
−4
3
22− 5=8
=0
= 144
−4(96 − 40) = 14
96 − 2024
−4= 192 17. ( )
( ) =
2 0 0 3
= 6, 2 = 42and the planar ellipse 92+ 42≤ 36 is the image of the disk 2+ 2≤ 1. Thus
2 =
2+2≤1
(42)(6) =2
0
1
0(242cos2) = 242
0 cos2 1 0 3
= 241
2 +14sin 22
0
1 441
0= 24()1 4
= 6
18. ( )
( ) =
√2 − 23
√2
23
= 4
√3, 2− + 2= 22+ 22and the planar ellipse 2− + 2≤ 2
is the image of the disk 2+ 2≤ 1. Thus
(2− + 2) =
2+2≤1
(22+ 22)
√4
3
=2
0
1 0
√8
33 = √4
3
19. ( )
( ) =
1 −2
0 1
= 1
, = , = is the image of the parabola 2= , = 3 is the image of the parabola
2 = 3, and the hyperbolas = 1, = 3 are the images of the lines = 1 and = 3 respectively. Thus
=
3 1
√3
√
1
=
3 1
ln√
3 − ln√
=3 1 ln√
3 = 4 ln√3 = 2 ln 3.
20. Here =
, = 2
so ( )
( ) =
2 −22
−2 1
= 1
and is the image of the square with vertices (1 1), (2 1), (2 2), and (1 2). So
2 =
2 1
2 1
2
2
1
=
2 1
2 = 3 4
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23. (a) Evaluate RRR
EdV , where E is the solid enclosed by the ellipsoid xa22 + yb22 +zc22 = 1. Use the transformation x = au, y = bv, z = cw.
(b) The earth is not a perfect sphere; rotation has resulted in flattening at the poles. So the shape can be approximated by an ellipsoid with a = b = 6378 km and c = 6356 km. Use part (a) to estimate the volume of the earth.
Solution:
1
SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ¤ 611
21. (a) ( )
( ) =
0 0 0 0 0 0
= and since =
, =
, =
the solid enclosed by the ellipsoid is the image of the
ball 2+ 2+ 2≤ 1. So
=
2+2+2≤ 1
= ()(volume of the ball) = 43
(b) If we approximate the surface of the earth by the ellipsoid 2
63782 + 2
63782 + 2
63562 = 1, then we can estimate the volume of the earth by finding the volume of the solid enclosed by the ellipsoid. From part (a), this is
= 43(6378)(6378)(6356) ≈ 1083 × 1012km3. (c) The moment of intertia about the -axis is =
2+ 2
( ) , where is the solid enclosed by
2
2 +2
2 +2
2 = 1. As in part (a), we use the transformation = , = , = , so
( )
( )
= and
=
2+ 2
=
2+2+2≤ 1
(22+ 22)()
= 0
2
0
1
0(22sin2 cos2 + 22sin2 sin2) 2sin
=
2 0
2
0
1
0(2sin2 cos2) 2sin + 2 0
2
0
1
0(2sin2 sin2) 2sin
= 3
0 sin3 2
0 cos2 1
0 4 + 3
0 sin3 2
0 sin2 1 0 4
= 31
3cos3 − cos
0
1
2 +14sin 22
0
1 551
0+ 31
3cos3 − cos
0
1
2 −14sin 22
0
1 551
0
= 34 3
()1 5
+ 34 3
()1 5
= 154(2+ 2)
22. is the region enclosed by the curves = , = , 14= , and 14= , so if we let = and = 14then is the image of the rectangle enclosed by the lines = , = ( ) and = , = ( ). Now
= ⇒ = ()14= 04 ⇒ 04= −1 ⇒ = (−1)104= −2525and
= −1 = (−2525)−1 = 35−25, so
( )
( ) =
3525−25 −2535−35
−25−3525 25−2515
= 875−1− 625−1= 25−1. Thus the area of , and the work done by the engine, is
=
25−1 = 25
(1) = 25
ln ||
= 25(−)(ln −ln ) = 25(−) ln
.
23. Letting = − 2 and = 3 − , we have = 15(2 − ) and = 15( − 3). Then( )
( ) =
−15 25
−35 15
= 1 5 and is the image of the rectangle enclosed by the lines = 0, = 4, = 1, and = 8. Thus
− 2
3 − =
4 0
8 1
1 5
= 1 5
4 0
8 1
1
= 151 224
0
ln ||8
1= 85ln 8.
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27. Evaluate the integral by making an appropriate change of variables. RR
Rcosy−x
y+x
dA, where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, 2) and (0, 1).
Solution:
612 ¤ CHAPTER 15 MULTIPLE INTEGRALS
24. Letting = + and = − , we have = 12( + )and = 12( − ). Then ( )
( ) =
12 12 12 −12
= −1
2and is the image of the rectangle enclosed by the lines = 0, = 3, = 0, and = 2. Thus
( + ) 2−2 =3 0
2
0 −12 = 123 0
=2
=0 = 123
0(2− 1)
= 121
22− 3 0= 121
26− 3 −12
= 14(6− 7)
25. Letting = − , = + , we have = 12( + ), = 12( − ). Then ( )
( ) =
−12 12 12 12
= −1
2and is the image of the trapezoidal region with vertices (−1 1), (−2 2), (2 2), and (1 1). Thus
cos
−
+
=
2 1
−
cos
−1 2
= 1 2
2 1
sin
=
=−
= 1 2
2 1
2 sin(1) = 32sin 1
26. Letting = 3, = 2, we have 92+ 42= 2+ 2, = 13, and = 12. Then( )
( ) = 1
6and is the image of the quarter-disk given by 2+ 2 ≤ 1, ≥ 0, ≥ 0. Thus
sin(92+ 42) =
1
6sin(2+ 2) =2 0
1 0
1
6sin(2) =12
−12cos 21
0= 24(1 − cos 1) 27. Let = + and = − + . Then + = 2 ⇒ = 12( + )and − = 2 ⇒ = 12( − ).
( )
( ) =
12 −12 12 12
= 1
2. Now || = | + | ≤ || + || ≤ 1 ⇒ −1 ≤ ≤ 1, and || = |− + | ≤ || + || ≤ 1 ⇒ −1 ≤ ≤ 1. is the image of the square region with vertices (1 1), (1 −1), (−1 −1), and (−1 1).
So
+ = 121
−1
1
−1 = 12
1
−1
1
−1= − −1. 28. Let = + and = , then = − , = , ( )
( ) = 1and is the image under of the triangular region with vertices (0 0), (1 0) and (1 1). Thus
( + ) =1 0
0(1) () =1 0 ()
=
=0 =1
0 () as desired.
15 Review
1. This is true by Fubini’s Theorem.
2. False.1 0
0
+ 2 describes the region of integration as a Type I region. To reverse the order of integration, we
must consider the region as a Type II region:1 0
1
+ 2 .
3. True by Equation 15.1.11.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
31. Let f be continuous on [0, 1] and let R be the triangular region with vertices (0, 0), (1, 0), and (0, 1). Show that RR
Rf (x + y)dA =R1
0 uf (u)du.
Solution:
612 ¤ CHAPTER 15 MULTIPLE INTEGRALS
24. Letting = + and = − , we have = 12( + )and = 12( − ). Then( )
( ) =
12 12 12 −12
= −1
2and is the image of the rectangle enclosed by the lines = 0, = 3, = 0, and = 2. Thus
( + ) 2−2 =3 0
2
0 −12 = 123 0
=2
=0 = 123
0(2− 1)
= 121
22− 3 0= 121
26− 3 − 12
= 14(6− 7)
25. Letting = − , = + , we have = 12( + ), = 12( − ). Then( )
( ) =
−12 12 12 12
= −1
2 and is the image of the trapezoidal region with vertices (−1 1), (−2 2), (2 2), and (1 1). Thus
cos
−
+
=
2 1
−
cos
−1 2
= 1 2
2 1
sin
=
=−
= 1 2
2 1
2 sin(1) = 32sin 1
26. Letting = 3, = 2, we have 92+ 42 = 2+ 2, = 13, and =12. Then ( )
( ) = 1
6and is the image of the quarter-disk given by 2+ 2≤ 1, ≥ 0, ≥ 0. Thus
sin(92+ 42) =
1
6sin(2+ 2) =2 0
1 0
1
6sin(2) = 12
−12cos 21
0= 24(1 − cos 1) 27. Let = + and = − + . Then + = 2 ⇒ = 12( + )and − = 2 ⇒ = 12( − ).
( )
( ) =
12 −12 12 12
= 1
2. Now || = | + | ≤ || + || ≤ 1 ⇒ −1 ≤ ≤ 1, and || = |− + | ≤ || + || ≤ 1 ⇒ −1 ≤ ≤ 1. is the image of the square region with vertices (1 1), (1 −1), (−1 −1), and (−1 1).
So
+ = 121
−1
1
−1 = 12
1
−1
1
−1= − −1. 28. Let = + and = , then = − , = ,( )
( ) = 1and is the image under of the triangular region with vertices (0 0), (1 0) and (1 1). Thus
( + ) =1 0
0(1) () =1 0 ()
=
=0 =1
0 () as desired.
15 Review
1. This is true by Fubini’s Theorem.
2. False.1 0
0
+ 2 describes the region of integration as a Type I region. To reverse the order of integration, we
must consider the region as a Type II region:1 0
1
+ 2 .
3. True by Equation 15.1.11.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
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