$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Stokes' Theorem

## Theorem

Stokes' Theorem is essentially a 3-D version of the curl version of Green's Theorem, which, if you recall, we can write as

$\iint_R | \nabla \times \vec{F} | \ dA = \oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt.$

Instead of a region $$R$$ in the plane, we can use a surface $$S$$ in $$\mathbb{R}^3$$. If we make similar assumptions to the ones from Green's Theorem, we get Stokes' Theorem:

$\ans{\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \ dS = \oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt}$

## Example

Evaluate $$\displaystyle\oint_C \vec{F} \cdot d\vec{r}$$ where $$\vec{F} = \langle z,-z,x^2-y^2 \rangle$$ and $$C$$ consists of the three line segments bounding the plane $$z=8-4x-2y$$ in the first octant.

#### Solution

\begin{align} \oint_C \vec{F} \cdot d\vec{r} &= \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \ dS\\ &= \iint_R (\nabla \times \vec{F}) \cdot (\vec{r}_r \times \vec{r}_v) \ dA\\ \\ \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ z & -z & x^2-y^2 \end{vmatrix}\\ \\ &= \langle 1-2y,1-2x,0 \rangle\\ \\ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 0 & -4\\ 0 & 1 & -2 \end{vmatrix}\\ \\ &= \langle 4,2,1 \rangle\\ \\ \oint_C \vec{F} \cdot d\vec{r} &= \int_0^2 \int_0^{4-2x} \langle 1-2y,1-2x,0 \rangle \cdot \langle 4,2,1 \rangle\ dy\ dx\\ &= \int_0^2 \int_0^{4-2x} 6-4x-8y\ dy\ dx = \ans{-\dfrac{88}{3}} \end{align}