Thomas Calculus 13th [Solutions]

1140 Chapter 15 Multiple Integrals
23.
a 0 0
J ( u, v, w) 0 b 0 abc;
for R and G as in Exercise 22, | xyz | dx dy dz
0 0 c
R
G
2 2 2 /2 /2 1
2
8 a b c ( sin cos )( sin sin )( cos ) sin d d d
0 0 0
4a b c
/2 /2 3
/2
sin cos sin cos d d a b c sin cos d a b c
3 0 0 3 0
6
2 2 2 2 2 2 2 2 2
2 2 2
a b c uvw dw dv du
1 0 0
24. u x, v xy , and w 3 z x u, y v , and z 1 w J ( u, v, w) v 1 0 1 ;
u
3 2
u u 3u
0 0 1
3
2 2 1
3 2 2
x y 3xyz dx dy dz u v 3 u v w J ( u, v, w)
du dv dw v vw du dv dw
u u 3 3 0 0 1 u
D
G
3 2 3 2 2 3
2 3
1 ( v vw ln 2) dv dw 1 (1 w ln 2) v dw 2 (1 w ln 2) dw 2 w w ln 2
3 0 0 3 0 2
0
3 0
3 2
0
2 3 9 ln 2 2 3 ln 2 2 ln 8
3 2
2 2
y 2
z 1
2 2 2
25. The first moment about the xy -coordinate plane for the semi-ellipsoid, x
using the
a b c
transformation in Exercise 23 is, M xy z dz dy dx cw | J ( u, v, w)|
du dv dw
D
G
2 2 2 2 2
2
abc w du dv dw abc M of the hemisphere 1, 0 abc
xy
x y z z
;
4
G
the mass of the semi-ellipsoid is
2abc
z abc 3 3 c
3 4 2abc
8
2
26. A solid of revolution is symmetric about the axis of revolution, therefore, the height of the solid is solely a
function of r. That is, y f ( x) f ( r ). Using cylindrical coordinates with x r cos , y y and z r sin ,
b 2 x f ( r) b 2 f ( r)
b 2
we have V r dy d dr r dy d dr r y d dr
a 0 0 a 0 0
a 0
G
r f ( r)
d dr
b 2 b
r f ( r) dr 2
a 0 a
rf ( r) dr . In the last integral, r is a dummy or stand-in variable and as such it can be
replaced by any variable name. Choosing x instead of r we have V
b
2
a
xf ( x) dx , which is the same result
obtained using the shell method.
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