Find \(I=\oint_{C}[(e^{-x^{2}/2}-yz)\,dx+(e^{-y^{2}/2}+xz+2x) \,dy+(e^{-z^{2}/2}+5)\,dz]\), where \(C\) is the circle \(x=\cos t\), \(y=\sin t\), \(z=2\), \(0\leq t\leq 2\pi \).

Solution  Let \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k} \), with \(P=e^{-x^{2}/2}-yz\), \(\ Q=e^{-y^{2}/2}+xz+2x\), and \( R=e^{-z^{2}/2}+5\). It can be verified that \[ {\rm{curl }}\,{\bf{F}} = - x{\bf{i}} - y{\bf{j}}{\rm{ + (2 + 2}}z){\bf{k}} \]

To use Stokes' Theorem, take \(S\) to be a plane region enclosed by the circle \(C\) in the plane \(z\rm{=2}\) so that \(\mathbf{n}=\mathbf{k}\). Then \(\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}=6\) on \(S\), and \[ I=\iint\limits_{\kern-11pt S}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=6\iint \limits_{\kern-11pt S}\,dx\,dy=6\,\pi \]