Multivariate Calculus - Bruce E. Shapiro

164 LECTURE 20. TRIPLE INTEGRALS
Since we are interested in the first octant, we chose the positive solution.
z = ± 1 8√
4 − y 2
The ellipse where the cylinder crosses the yz-plane intersects the y-axis at y = 2 and
the z-axis at z = 1/4. The base of the object in the xy-plane is a triangle formed
by the lines y = x, y = 2, and the y-axis.Therefore the domain of integration is the
set
{
√ }
4 − y 2
V = (x, y, z) : 0 ≤ z ≤ , x ≤ y ≤ 2, 0 ≤ x ≤ 2
8
The volume of this set is ∭ V
(1) × dV , namely
V =
= 1 8
∫ 2 ∫ 2 ∫ (4−y 2 ) 1/2 /8
0 x 0
∫ 2 ∫ 2
0
x
√
4 − y 2 dydx
dzdydx
From integral formula (54) in the inside back cover
∫ √a 2 − y 2 dy = y 2√
a 2 − y 2 + a2
2 sin−1 y a
Setting a = 2, ∫ √4
− y 2 dy = y 2
√
4 − y 2 + 2 sin −1 y 2
Therefore
∫ 2 √
4 − y 2 dy =
[ y √
4 − y
2
2 + 2 sin −1 y 2
x
=
( 2 √
4 − 2
2
2 + 2 sin −1 2 2
] 2
x
) ( x √
− 4 − x
2
2 + 2 sin −1 x )
2
= π − x 2
√
4 − x 2 − 2 sin −1 x 2
and hence the volume is
V = 1 ∫ 2 ∫ 2 √
4 − y
8
2 dydx = 1 ∫ 2 (
π − x √
4 − x
0 x
8 0 2
2 − 2 sin −1 x )
dx
2
= 1 [ ∫ 2
π dx − 1 ∫ 2
x √ ∫ 2
4 − x
8
2
2 dx − 2 sin −1 x ]
2 dx
= π 4 − 1 16
0
∫ 2
0
0
x √ 4 − x 2 dx − 1 4
∫ 2
0
0
sin −1 x 2 dx
The first integral we can solve with the substitution u = 4 − x 2 , which implies that
du = −2dx. Furthermore, when x = 0, u = 4, and when x = 2, u = 0. Thus
∫ 2
x √ 4 − x 2 dx = − 1 ∫ 0
u 1/2 du = 1 ∫ 4
4
u 1/2 du = u3/2
= (4)3/2
= 8 2
2
3 ∣ 3 3
0
4
Revised December 6, 2006. Math 250, Fall 2006
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