Winter 2004 Triple Integrals in Spherical Coordinates – Review MATH 200
1. Evaluate Z ZEpx2+y2+z2dV where Eis bounded below by the cone φ=π/6 and above by the sphere
ρ= 2.
Comment: Look at Z2π
0Zπ/6
0Z2
0
ρ(ρ2sin φ)dρdφdθ.
2. Evaluate ZZEpx2+y2dV where Eis bounded below by the cone φ=π/6 and above by the sphere ρ= 2.
Comment: Look at Z2π
0Zπ/6
0Z2
0
(ρsin φ) (ρ2sin φ)dρdφdθ.
Evaluate ZZE
x dV where Eis bounded below by the cone φ=π/6 and above by the sphere ρ= 2.
Comment: Look at Z2π
0Zπ/6
0Z2
0
(ρsin φcos θ) (ρ2sin φ)dρdφdθ.
3. Find the mass of a solid hemisphere Hof radius 2 if the density at any point is proportional to its distance
from the center of the base.
Comment: Look at Z2π
0Zπ/2
0Z2
0
ρ(ρ2sin φ)dρdφdθ.
4. Find the volume of the solid bounded above by the sphere x2+y2+z2= 4, laterally by the cone z=px2+y2,
and below by the sphere x2+y2+z2= 1.
Comment: Look at Z2π
0Zπ/4
0Z2
1
ρ2sin φ dρdφdθ.
5. Convert the given triple integral into spherical coordinates:
(a) Z3
−3Z√9−x2
−√9−x2Z√9−x2
−y2
0
zpx2+y2+z2, dzdydx,
(b) Z3
0Z√9−y2
0Z√18−x2
−y2
√x2+y2
(x2+y2+z2)dzdxdy.
Comment: (a) Z2π
0Zπ/2
0Z3
0
(ρcos φ)ρ(ρsin φ)dρdφdθ.
(b) The solid is the portion of the the sphere x2+y2+z2= 18 that lies both inside the cylinder x2+y2= 9
and the cone z=px2+y2that all lies in the first octant. Zπ/2
0Zπ/4
0Z2√3
3/sin φ
ρ2sin φ dρdφdθ.
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