Find the surface integral $\iint\limits_{\kern-8ptS}{\rm{curl} \, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS$, where $\mathbf{F}=\mathbf{F}(x,y,z)=y \mathbf{i}+x\mathbf{j}+z\mathbf{k}$ and $S$ is the surface $z=5-(x^{2}+y^{2}),$ $z\geq 1$, using Stokes' Theorem.

Solution  Figure 73 shows the surface $S$. The boundary curve $C$ is the circle $x^{2}+y^{2}=4$ that lies in the plane $z=1$. The vector form of $C$ is $\mathbf{r}=\mathbf{r}(t)=2\cos t\,\mathbf{i}+2\sin t\,\mathbf{j}+ \mathbf{k}$, $0\leq t\leq 2\pi$. Then $$\begin{eqnarray*} d\mathbf{r}& =&(-2\,\sin t\,\mathbf{i}+2\,\cos t\,\mathbf{j})\,dt \\[4pt] \mathbf{F}& =&y\mathbf{i}+x\mathbf{j}+z\mathbf{k}=2\,\sin t\mathbf{i}+2\,\cos t \mathbf{j}+\mathbf{k} \\[4pt] \mathbf{F}\,{\bf\cdot}\, d\mathbf{r}& =&(-4\,\sin ^{2}t+4\,\cos ^{2}t)\,dt=4\,\cos (2t) \,dt \end{eqnarray*}$$

Now we use Stokes' Theorem. $$\iint\limits_{\kern-13ptS}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=\oint_{C}\mathbf{F} \,{\bf\cdot}\, d\mathbf{r}=\int_{0}^{2\pi }4\,\cos (2t) \,dt= 2 \big[ \sin(2t)\big] _{0}^{2\pi }=0$$