Calculus III:

Stokes' theorem

Green’s Theorem links the line integral to a double integral over the region delimited by an oriented curve.

In this section we are going to to work on Stokes' theorem that likes a line integral to a surface integral.

Stokes' theorem is a higher dimensional version of Green’s Theorem.

1. Definitions

Lets's consider the following surface with the indicated orientation , delimited by the oriented contour C.

This curve C, around the edge of the surface S is called the boundary curve.

The orientation of the surface S induces the positive orientation of C. To determine the positive orientation of C related to the orientation of the unit normal vector, we use the right-hand rule, or think of walking along the curve, while the surface is on the left, so the direction of the unit normal vectors is upwards.

2. Stokes’ Theorem

Here is Stokes' theorem: S is any oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. is a vector field:

Example 1

Evaluate ∫∫_S curl() . by using Stokes’ Theorem.
where = xy + y , and S is the part of z = 6 - x² - y² above the plane z = 2 .
Consider that S is oriented upwards.



The boundary curve C is the intersection of the surface S and the plane z = 2. So

6 - x² - y² = 2. That is:

x² + y² = 4, at z = 2.

Therefore, the boundary curve C of S is the circle of radius 2, centred at the origin on the plane z = 2. The parameterization of this curve S is:

(t) = 2 cos t + 2 sin t + 2 , with 0 ≤ t ≤ 2π .

The vector field evaluated on the curve is:

((t)) = 4 sin t cos t + 0 + 2 sin t

We have :

/dt = - 2 sin t + 2 cos t
and dr = (dr/dt) dt. Then

∫_c . = ∫₀^2Ï€ ((t)). (/dt) dt =
∫₀^2Ï€ 〈 4 sin t cos t , 0 , 2 sin t 〉 . 〈 - 2 sin t , 2 cos t , 0 〉 dt =
∫₀^2Ï€ (- 8 sin² t) dt = (- 4) ∫₀^2Ï€ (1 - cos (2t)) dt =
(- 4) ∫₀^2Ï€ (t - (1/2) sin (2t)) dt = - 8 Ï€

∫_c . = - 8 π

Example 2

Evaluate ∫_C . by using Stokes’ Theorem.
where = z² + y + z ,
C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) with counter-clockwise rotation. .

curl =
		∂/∂x	∂/∂y	∂/∂z
		z2	y	z

= 〈 0, 2z, 0 〉 = 2z .

The triangle C is the boundary curve for the surface S that we need to use in the surface integral. S is any surface that has this boundary. So Let’s use the plane formed by the triangle, with upwards orientation for this surface.

The equation of this plane S is :
x + y + z = 1 . or z = g(x,y) = 1 - x - y.

The related function of the surface f is:
f(x,y,z) = 0 = x + y + z - 1. Hence ∇f = 〈 1, 1, 1 〉

The region D over we integrate is bouded by
0 ≤ x ≤ 1 and 0 ≤ y≤ 1 - x

Let’s use Stokes’ Theorem :

∫_C . = ∫∫_S curl() .

∫∫_S curl() . =
∫∫_S 2z . dS = ∫∫_S 2z . (∇f/|∇f|) dS =
∫∫_D 2z . ∇f dA = ∫∫_D 2z 〈 0, 1, 0 〉 . 〈 1, 1, 1 〉 dA =
∫∫_D 2z dA = ∫₀ ¹ ∫₀^1-x 2(1 - x - y) dy dx =
2∫₀ ¹ (y - xy - y²/2) |₀^1-x dx = 2∫₀ ¹ (1/2)(1 - x )² dx =
∫₀ ¹ (1 - x )² dx = - (1/3) (1 - x )³|₀ ¹ = 1/3

∫_c . = 1/3