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Example Show that Stokes's theorem holds for the following: $\displaystyle{\boldsymbol{F} = -y\boldsymbol{i} + x\boldsymbol{j} + \boldsymbol{k}, \ S: z = (4 - x^2 - y^2)^{1/2}}$.

Answer The boundary $\partial S$ of $S$ is a circle $x^2 + y^2 = 4$．Then the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = 2\cos{t}\boldsymbol{i} + 2\sin{t}\boldsymbol{j}$. Thus, the line integral is given by the followings:

$\displaystyle \oint_{\partial S}\boldsymbol{F}\cdot d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \oint_{\partial S}(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j} + \boldsymbol{k})\cdot(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j})dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}[4\sin^{2}{t} + 4\cos^{2}{t}]dt = 8\pi$  

We next find the surface integral.

$\displaystyle \nabla \times \boldsymbol{F} = \left\vert\begin{array}{ccc} \bol... ...c{\partial}{\partial z}\\ -y & x & 1 \end{array}\right\vert = 2\boldsymbol{k} $

Now normal vector of $S$ is given by

$\displaystyle \boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = \left\vert\begin{a... ...t\vert = \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}$

The unit normal vector is given by

$\displaystyle \boldsymbol{n} = \frac{ \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\b... ... \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}\Vert } $

Thus, we have，
$\displaystyle \iint_{S}(\nabla \times \boldsymbol{F}) \cdot\boldsymbol{n} dS$ $\displaystyle =$ $\displaystyle 2\iint_{S}\boldsymbol{k} \cdot\frac{\boldsymbol{r}_{x} \times \bo... ...bol{r}_{y} \Vert} \Vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} \Vert dx dy$  
  $\displaystyle =$ $\displaystyle 2\iint_{\Omega}\boldsymbol{k} \cdot\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} dx dy$  
  $\displaystyle =$ $\displaystyle \iint_{\Omega}2 dx dy = 2(4\pi) = 8 \pi$  

This shows that Stokes' theorem holds.