Calculus III:

Surface area
Surface area of a solid of revolution

We are going to determine the surface area of a solid of revolution.

The surface area of a solid obtained by rotating a function about the x or y axis is already determined in Calculus II. But here it is about a function of two dimentions.

1. Surface area element

Now, we want to find the area of the surface given by z = f(x,y)

(x,y) is a point from the region D in the xy-plane over we integrate the fuction z.

The differential surface element, in three-dimentional space, is:

dS = √[∂f/∂x)² + (∂f/∂y)² + 1]dA.

dA = dx dy , the differential surface area element.

∂f/∂x = f_x, and ∂f/∂y = f_z . Therefore

dS = √[(f_x)² + (f_y)² + 1]dA

Example 1

Find the triangle surface S area of the part of the plane x + y + z = 1, formed by the vertices (0,1,0), (1,0,0), and (0,0,1) that lies in the first octant.



The surface S correspond to the area D on the xy-plane.

We denote by dA the area element in the the domaine D where we integrate. Here dA = dx dy .

The function in the form z = f(x,y) is, by solving for z the equation of the triangle plane x + y + z = 1: z = 1 - x - y

Taking the partial derivatives gives:

f_x = - 1, and f_y = - 1 . Therefore

dS = √[(f_x)² + (f_y)² + 1] dA = √[(-1)² + (-1)² + 1] dA = √3 dA.
dS = √ 3 dA

The limits defining D are:

0 ≤ x ≤ 1      0 ≤ y ≤ 1 - x

Now Let's evaluate S

S = ∫∫_S f(x,y) dS = ∫∫_D f(x,y) dA =
∫₀¹ ∫₀^1-x (1 - x - y) dy dx =
∫₀¹ (y - xy - (1/2)y²)|₀^1-x dx =
∫₀¹ (1/2)(1 - x)² dx =
(1/6)(x - 1)³|₀¹ = 1/6

Therefore

S = ∫∫_S (1 - x - y) dS = = 1/6

Exercice

Evaluate ∫∫_S × , where = = xy yz + zx , and
S is the upper half of the sphere x² + y² + z² = 4, of base the disk x² + y² = 4 in the plane z = 0.
Consider that S has the positive orientation.

Evaluate result1 and result2 written below:



The surface is composed by S1 (hemi-sphere) and S2 (disk): S = S1∪S2,

1) The hemi-sphere (S1):

S1 is a sphere, we need then to use the parametric representation of the surface according to spherical coordinates:

(θ, φ) = 2 sin θ cos φ + 2 sin θ sin φ + 2 cos θ , with the limits:

0 ≤ θ ≤ π/2     0 ≤ φ ≤ 2π

Now we determine:

= ∂(θ, φ)/∂θ, and
= ∂(θ, φ)/∂φ

and, we determine their cross product:

× =
- 4 sin² θ cos φ - 4 sin² θ sin φ - 4 sin θ cos θ

The component of is negative, so

= - × /| × | =
4 sin² θ cos φ + 4 sin² θ sin φ + 4 sin θ cos θ / | × |

Next

((θ, φ)) =
4 sin θ cos φ sin θ sin φ 4 sin θ sin φ cos θ + 4 sin θ cos φ cos θ .

We need also to calculate the scalar product :

((θ, φ)) . ( × ) = result 1

Therefore

∫∫_S1 . = ∫∫_S1 . dS

Since dS = | × | dA,
∫∫_S1 . =
= ∫∫_D . ( × /| × |) | × | dA =
= ∫∫_D . ( × ) dA = ∫₀^2π ∫₀^π/2 . ( × ) dθ dφ =
∫₀^2π ∫₀^π/2 result 1 dθ dφ = result 2


2) The disk (S2)

For the disk at the bottom must be equal to - .

∫∫_S2 . = ∫∫_S2 . (- ) dS = ∫∫_S2 . (- ) dS

We have:

. (- ) = 〈xy, yz, zx 〉 . 〈0, 0, -1 〉 = - xz. Then

∫∫_S2 . (- ) dS = - z ∫∫_D x dA at z = 0 . So ∫∫_S2 . = 0

Therefore

∫∫_S . = ∫∫_S1 . + ∫∫_S2 . =
result 2 + 0 = result 2.