Subject: Calculus

# Stokes Theorem

Let be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Let be F a vector field whose components have continuous partial derivatives on an open region in \mathbb{R}^3 that contains S. Then

\int_{C} F\cdot dr =\int \int_{S} \nabla \times F \cdot dS

## Introduction

Stokes� Theorem is named after the Irish mathematical physicist Sir George Stokes (1819�1903). Stokes was a professor at Cambridge University (in fact he held the same position as Newton, Lucasian Professor of Mathematics) and was especially noted for his studies of fluid flow and light.

Stokes� Theorem can be regarded as a higher form of the Green�s theorem. Recall that Green�s theorem relates or transform a line integral around its boundary plane to a double integral of over a plane region. Stokes� theorem, however, relates a surface integral to a line integral around a boundary curve.

no figure yet

The figure (left) shows an oriented surface with a normal vector n. The orientation of S induces a positive orientation of the boundary curve C as shown in the figure. This means that if you walk, in the positive direction of C (as represented by the arrows) with your head pointing at the direction of the normal vector, then the surface will always be on your left.

To better understand this, let us proceed with examples. Let us consider the figure shown below.

## Illustration

Evaluate \int_{C} F \cdot dr, where F(x,y,z) = -y^2\mathbf{\hat{i}} + x\mathbf{\hat{j}} + z^2\mathbf{\hat{k}} and C is the curve of intersection of the plane y + z = 2 and cylinder x^2 + y^2 = 1.

Solution. The curve C (an ellipse) is shown in Figure 3. Although \int_{C} F \cdot dr could be evaluated directly, it�s easier to use Stokes� Theorem. We first compute

\nabla \times F = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y^2 & x & z^2 \end{vmatrix} = (1+2y)\mathbf{\hat{k}}

Although there are many surfaces with boundary C, the most convenient choice is the elliptical region S in the plane y + z = 2 that is bounded by C. If we orient S upward, then C has the induced positive orientation. The projection D of S on the xy-plane is the disk x^2 + y^2 \leq 1.

\int_{C} F\cdot dr =\int \int_{S} \nabla \times F \cdot dS = \int \int_{D}(1+2y)dA
=\int_{0}^{2\pi} \int_{0}^{1}(1+2r\sin{\theta})r dr d\theta
=\int_{0}^{2\pi}\left [\frac{r^2}{2}+\frac{r^3}{3}\sin{\theta}\right ]_{0}^{1}d\theta=\int_{0}^{2\pi}\left (\frac{1}{2}+\frac{2}{3}\sin{\theta}\right )d\theta
=\frac{1}{2}2 \pi +0 =\pi

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