Surface area example

Example

Find the surface area of the cone S

Φ(r,θ) = (r cos θ,r sin θ,r)
for 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 1.

Solution: Recall that the generic formula for surface area is

A = ∫ ∫ ∥∥ ∂Φ ∂Φ ∥∥ ∥∥ ---(u,v) × ----(u, v)∥∥ D ∂u ∂vdudv.
In this case, the variables are r and θ rather than u and v. Hence, we need to integrate
∥ ∥ ∥∥∂Φ-- ∂-Φ- ∥∥ ∥ ∂r (r,θ ) × ∂θ (r,θ)∥
over the region D defined by 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 1.

We calculate the cross product as follows.

∂ Φ ---- ∂r = (cos θ, sin θ, 1)
∂-Φ- ∂θ = (-r sin θ,r cos θ, 0)
∂-Φ- ∂r ×∂Φ-- ∂θ = | | | i j k | || cos θ sinθ 1 || || || - rsinθ r cosθ 0
= i(-r cos θ) - j(r sin θ)
+ kr(cos 2θ + sin 2θ)
= -r cos θi - r sin θj + rk
∥ ∥ ∥ ∂Φ ∂ Φ ∥ ∥∥ ----× ----∥∥ ∂r ∂θ = ∘ ----------------------- r2 cos2θ + r2sin2θ + r2
= √ ---- 2r2 = r√ -- 2

The area of cone S is

A(S) = ∫ 01 ∫ 02π∥ ∥ ∥∥∂Φ-- ∂Φ-∥∥ ∥ ∂r × ∂θ ∥dθdr
= ∫ 01 ∫ 02πr√ -- 2dθdr
= ∫ 012πr√ -- 2dr
= √ -|1 πr2 2|| 0 = π√ -- 2.