Calculus 2nd Edition Rogawski

988 C H A P T E R 17 LINE AND SURFACE INTEGRALS
REMINDER In this example,
G(θ,t)= (t cos θ,tsin θ,t)
Step 1. Compute the tangent and normal vectors.
T θ = ∂G = ⟨−t sin θ,tcos θ, 0⟩ ,
∂θ T t = ∂G = ⟨cos θ, sin θ, 1⟩
∂t
i j k
n = T θ × T t =
−t sin θ t cos θ 0
= t cos θi + t sin θj − tk
∣ cos θ sin θ 1 ∣
The normal vector has length
√
√
∥n∥ = t 2 cos 2 θ + t 2 sin 2 θ + (−t) 2 = 2t 2 = √ 2 |t|
Thus, dS = √ 2|t| dθ dt. Since t ≥ 0 on our domain, we drop the absolute value.
Step 2. Calculate the surface area.
∫∫
∫ 2 ∫ 2π
√ √ 2
Area(S) = ∥n∥ dudv = 2 tdθ dt = 2πt
2
∣ = 4 √ 2π
D
0 0
0
Step 3. Calculate the surface integral.
We express f (x, y, z) = x 2 z in terms of the parameters t and θ and evaluate:
∫ 2π
0
REMINDER
cos 2 θ dθ =
∫ 2π
0
1 + cos 2θ
2
dθ = π
f (G(θ, t)) = f (t cos θ,tsin θ,t)= (t cos θ) 2 t = t 3 cos 2 θ
∫∫
∫ 2 ∫ 2π
f (x, y, z) dS = f (G(θ, t)) ∥n(θ,t)∥ dθ dt
S
t=0 θ=0
=
∫ 2
t=0
∫ 2π
θ=0
(t 3 cos 2 θ)( √ 2t)dθ dt
= √ ( ∫ )(
2 ∫ )
2π
2 t 4 dt cos 2 θ dθ
= √ 2
0
( 32
5
0
)
(π) = 32√ 2π
5
In previous discussions of multiple and line integrals, we applied the principle that
the integral of a density is the total quantity. This applies to surface integrals as well. For
example, a surface with mass density ρ(x, y, z) [in units of mass per area] is the surface
integral of the mass density:
∫∫
Mass of S = ρ(x, y, z) dS
S
Similarly, if an electric charge is distributed over S with charge density ρ(x, y, z), then
the surface integral of ρ(x, y, z) is the total charge on S.
EXAMPLE 5 Total Charge on a Surface Find the total charge (in coulombs) on a
sphere S of radius 5 cm whose charge density in spherical coordinates is ρ(θ, φ) =
0.003 cos 2 φ C/cm 2 .
Solution We parametrize S in spherical coordinates:
G(θ, φ) = (5 cos θ sin φ, 5 sin θ sin φ, 5 cos φ)
By Eq. (2), ∥n∥ = 5 2 sin φ and
∫∫
Total charge =
S
ρ(θ, φ)dS =
∫ 2π ∫ π
θ=0
φ=0
ρ(θ, φ)∥n∥ dφ dθ