Multivariate Calculus - Bruce E. Shapiro

144 LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS
Making the substitution x = 3 cos u in ∫ (9 − x 2 ) 3/2 dx we have dx = −3 sin udu
hence
∫
∫
(9 − x 2 ) 3/2 dx = (9 − 9 cos 2 u) 3/2 (−3 sin u)du
∫
= 9 3/2 (1 − cos 2 u) 3/2 (−3) sin udu
=
∫
−3(3 2 ) 3/2 (sin 2 u) 3/2 sin udu
∫
= −3(3 3 ) sin 4 udu
∫
= −81 (sin 2 u)(sin 2 u)du
Using the trigonometric relationship
sin 2 u = 1 (1 − cos 2u)
2
gives
∫
∫ 1
(9 − x 2 ) 3/2 dx = −81
2 (1 − cos 2u)1 (1 − cos 2u)du
2
= − 81 ∫
(1 − 2 cos 2u + cos 2 2u)du
4
Use the substitution
to get
∫
(9 − x 2 ) 3/2 dx = − 81
4
= − 81
4
cos 2 u = 1 (1 + cos 2u)
2
[
u − sin 2u + 1 ∫
2
[ 3
2 u − sin 2u − 1 ]
8 sin 4u
]
(1 + cos 4u)du
The limits on the definite integral over x were [0, 3]. With the substitution x =
3 cos u, we have u = π/2 when x = 0 and u = 0 when x = 3. Hence
∫ 3
(9 − x 2 ) 3/2 dx = − 81 [ 3
0
4 2 u − sin 2u − 1 ]∣ ∣∣∣
0
8 sin 4u π/2
= − 81 [ 3
4 2 (0) − sin(0) − 1 8 sin 0 − 3 π
2 2 − sin π − 1 ]
8 sin 2π
= 81 3 π
4 2 2 = 243π
8
and consequently
V = 2 3
∫ 3
0
(9 − x 2 ) 3/2 dx = 2 3
243π
8
= 81π
4
Revised December 6, 2006. Math 250, Fall 2006