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. \[ \iint\limits_{\kern-13ptS}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=\oint_{C}\mathbf{F} \,{\bf\cdot}\, d\mathbf{r}=\int_{0}^{2\pi }4\,\cos (2t) \,dt= 2 \big[ \sin(2t)\big] _{0}^{2\pi }=0 \]