MA22S1: SOLUTIONS TO TUTORIAL 9
1. Find the volume of the region bounded by the sphere
x2 + y 2 + z 2 = 9
and the cone
z=
p
x2 + y 2
Solution: Let S be the region in question. The volume of S is given
by the formula
volume (S) =
ZZZ
1 dV
S
If we switch to spherical coordinates (ρ, φ, θ) where
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ
then
volume (S) =
Z
=
Z
2π
2π Z
0
=
2π
=
Z
3
Z
ρ2 sin φ dρ dφ dθ
0
π
4
0
ρ3
3
ρ=3
sin φ dφ dθ
ρ=0
π
4
9 sin φ dφ dθ
0
0
Z
π
4
0
0
Z
Z
2π
φ= π
9 [− cos φ]φ=04 dθ
0
2π
1
=
9 (− √ ) − (−1) dθ
2
0
1
= 9(1 − √ ) [θ]θ=2π
θ=0
2
1
= 18(1 − √ ) π
2
Z
1
2
MA22S1: SOLUTIONS TO TUTORIAL 9
2. Use the transformation T : (u, v) → (x, y) where x =
√u
2
and y =
√v
3
to find the area bounded by the ellipse
2x2 + 3y 2 = 6
Solution: The area of the region E bounded by the ellipse is
area(E) =
ZZ
1 dA
E
To apply the transformation T : (u, v) 7→ (x, y) we first compute the
Jacobian:
∂(x, y) =
∂(u, v) ∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
√1
2
=
0
0 = √1 .
6
√1 3
Next we compute the preimage of the ellipse in the uv-plane: substituting x =
√u
2
and y =
√v
3
into the equation 2x2 + 3y 2 = 6 we
get
u2 + v 2 = 6.
Thus the preimage of E in the uv-plane is the region D bounded by
√
the circle with centre (0, 0) and radius 6. Now we can compute
the area of E:
area (E) =
ZZ
=
ZZ
D
D
∂(x, y) ∂(u, v) du dv
1
√ du dv
6
1
= √ (area (D))
6
1
= √ (6 π)
6
√
=
6π