Surface integral examples

Example 1

For the cylinder of radius 3 and height 5 given by x2 + y2 = 32 and 0 ≤ z ≤ 5, let the charge density be proportional to the distance from the xy-plane. Find the total charge on the cylinder.

Solution: Since the distance from the point (x,y,z) to the xy-plane is z, let the charge density be f(x,y,z) = kz for some constant k. The total charge on the cylinder is the surface integral of f over the cylinder S: ∫∫SfdS.

Parameterize the cylinder by

Φ(θ,t) = (3 cos θ, 3 sin θ,t)
for 0 ≤ θ ≤ 2π and 0 ≤ t ≤ 5. The total charge on the cylinder is
∫ ∫SfdS = ∫ 05 ∫ 02πf(Φ(θ,t))∥ ∥ ∥∥∂-Φ- ∂-Φ- ∥∥ ∥ ∂θ (θ,t) × ∂t (θ,t)∥dθdt

Calculate the components of the above integral as follows.

f(Φ(θ,t)) = kt
∂Φ ---- ∂θ(θ,t) = (-3 sin θ, 3 cos θ, 0)
∂Φ-- ∂t(θ,t) = (0, 0, 1)
∂Φ-- ∂θ(θ,t) ×∂-Φ- ∂t(θ,t) = || i j k || || || | - 3sinθ 3 cosθ 0 | | 0 0 1 |
= i3 cos θ - j(-3 sin θ)
= (3 cos θ, 3 sin θ, 0)
∥ ∥ ∥ ∂Φ ∂Φ ∥ ∥∥ ∂θ-(θ,t) × ∂t-(θ,t)∥∥ = ∘ -----2--------2-- 9 cos θ + 9sin θ
= 3
Therefore, the total charge is
∫ ∫SfdS = ∫ 05 ∫ 02πkt3dθdt
= ∫ 05( ) 3tθ|θθ==20πdt
= ∫ 05k6πtdt
= 2|5 k3πt |0 = 3π(25)k = 75πk

Example 2

Let S be the same cylinder as above (x2 + y2 = 9 for 0 ≤ z ≤ 5).

Let F be the vector field F(x,y,z) = (2x, 2y, 2z).

Find the integral of F over S. Use outward pointing normal.

PIC

What is the sign of integral? Since the vector field and unit normal point outward, the integral better be positive.

As above, parameterize the cylinder by

Φ(θ,t) = (3 cos θ, 3 sin θ,t)
for 0 ≤ θ ≤ 2π and 0 ≤ t ≤ 5. We calculated above that
∂Φ-- ∂θ(θ,t) ×∂-Φ- ∂t(θ,t) = (3 cos θ, 3 sin θ, 0)

Is (3 cos θ, 3 sin θ, 0) an outward pointing normal? It is a normal at the point Φ(θ,t) = (3 cos θ, 3 sin θ,t). As shown in the below figure, it is an outward pointing normal.

PIC

Calculate total flux:

∫ ∫SF ⋅ dS = ∫ 05 ∫ 02πF(Φ(θ,t)) ⋅( ) ∂Φ-- ∂Φ-- ∂ θ (θ,t) × ∂t (θ,t)dθdt
= ∫ 05 ∫ 02πF(3 cos θ, 3 sin θ,t) ⋅ (3 cos θ, 3 sin θ, 0)dθdt
= ∫ 05 ∫ 02π(6 cos θ, 6 sin θ, 2t) ⋅ (3 cos θ, 3 sin θ, 0)dθdt
= ∫ 05 ∫ 02π18 cos 2θ + 18 sin 2θdθdt
= ∫ 05 ∫ 02π18dθdt
= ∫ 0518(2π)dt
= 18(2π)(5) = 180π

There are two orientations for the cylinder. If we had instead chose the orientation given by the inward point normal, what must ∫∫SF ⋅ dS be? -180π