1. Exercises from 5.3
Today we’re going to practice doing different types of surface integrals. Briefly explain the concept
of a surface integral.
Problem 1. (Folland 5.3.1) Find the area of the surface z = xy inside the cylinder x2 + y 2 = a2
• We pick a parameterization of the surface
g : R 2 → R3
(x, y) 7→ (x, y, xy)
• Finding the infinitessimal area element:
i j
∂g
∂g
dx dy = 1 0
×
dA = ∂x ∂y 0 1
k p
y dx dy = 1 + x2 + y 2 dx dy
x • The area is given by:
Z Z
dA
Z Z ∂g
∂y =
×
dx dy
∂θ S ∂x
Z Z p
=
1 + x2 + y 2 dx dy
A
=
S
S
2π Z a
Z
=
r
0
p
1 + r2 dr dθ
0
Z
a2
= π
√
1 + u du
0
=
i
2π h
(1 + a2 )3/2 − 1
3
Problem 2. (Folland 5.3.3) Let 0 < a < b, find the area of the torus obtained by revolving the circle
(x − b)2 + z 2 = a2 around the z-axis.
We are given a parameterization of the torus:
g : [0, 2π] × [0, 2π] → R3
(θ, φ) 7→ ((b + a cos φ) cos θ, (b + a cos φ) sin θ, a sin φ)
So we carry through the usual construction:
i
∂g
∂g
×
dA = dθ dφ = −(b + a cos φ) sin θ
∂θ ∂φ −a sin φ cos θ
j
(b + a cos φ) cos θ
−a sin φ sin θ
0
dθ dφ = a(b + a cos φ) dθ dφ
a cos φ k
Now we can compute the area:
Z Z
A
=
dA
Z Z ∂g
∂g ×
dθ dφ
=
∂φ S ∂θ
Z 2π Z 2π
=
a(b + a cos φ) dθ dφ
S
0
=
0
4π 2 ab
1
2
Problem 3. (Folland 5.3.8(a,b,d))
Part a): Let S be the surface z = xy with 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, oriented with the normal pointing
upwards. Integrate the vector field F(x, y, z) = (xz, 0, −xy) over this surface.
• We first need the surface normal. The orientation is pointing UPWARDS in the +z-direction,
so the normal is given by:
i j k ∂g
∂g n=
×
= 1 0 y = (−y, −x, 1)
∂x ∂y 0 1 x • We can now compute the surface integral:
Z Z
Z 1Z 2
∂g
∂g
F(g(x, y)) ·
F · n dA =
×
dy dx
∂x ∂y
0
S
0

 

Z 1Z 2
x(xy)
−y
 0  ·  −x  dy dx
=
0
0
−xy
1
Z 1Z 2
−x2 y 2 − xy dy dx
=
0
0
17
8
= − −1=−
9
9
Part b): F(x, y, z) = x2 i + zj − yk on the unit sphere, oriented so that the normal points outward.
• For fun and practice, we’ll do this problem in spherical polar coordinates. A parameterization
of the sphere is:
g(θ, φ) = (cos φ sin θ, sin φ sin θ, cos θ),
θ ∈ [0, φ], φ ∈ [0, 2π]
• In this coordinate system, the normal vector can be computed explicitly (intuitively, we expect
it to just be the unit vector r̂):
∂g
∂g n = ×
∂θ ∂φ i
j
k
= cos φ cos θ sin φ cos θ − sin θ − sin φ sin θ cos φ sin θ
0
=
sin2 θ cos φ i + sin2 θ sin φ j + (cos2 φ sin θ cos θ + sin2 φ sin θ cos θ) k
=
sin2 θ cos φ i + sin2 θ sin φ j + sin θ cos θ k
• We also need to compute the vector field evaluated along the surface:


cos2 θ sin2 φ

F(g(θ, φ)) = 
cos θ
− sin φ sin θ
• Let’s now compute the surface integral:
Z Z
Z 2π Z π
F · n dA =
sin2 θ cos2 θ sin2 φ cos φ + sin2 θ sin φ cos θ − sin2 θ cos θ sin φ dθ dφ
S
0
Z
0
2π
Z
π
=
sin2 θ cos2 θ sin2 φ cos φ dθ dφ Apply Fubini’s theorem
Z π
Z 2π
2
2
2
sin θ cos θ dθ
sin φ cos φ dφ
=
0
=
0
0
0
0