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4. Let n be the outward unit normal vector field of the surface Sirr, ) - (rcos)i + (r sin 8j+ (1-)k, 0 STS 1,0 SO S 27, and

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Answer #1

By Stokes theorem, we have

\iint_S\nabla\times {\bf F}\cdot{\bf n}~d\sigma=\int_C{\bf F}\cdot d{\bf r}

where C is the boundary of S . Since S has parametrization {\bf r}(r,\theta)=(r\cos\theta,r\sin\theta,1-r) , its boundary is

\{(\cos\theta,\sin\theta,0):\theta\in [0,2\pi]\}

Thus,

\begin{align*}\iint_S\nabla\times {\bf F}\cdot{\bf n}~d\sigma&=\int_C{\bf F}\cdot d{\bf r}\\ &=\int_0^{2\pi}(0,\cos\theta\cdot\sin\theta,0)\cdot (-\sin\theta,\cos\theta,0)~d\theta \\ &=\int_0^{2\pi}\cos^2\theta\cdot\sin\theta~d\theta\end{align*}

Let \cos \theta=w so that -\sin \theta~d\theta=dw ; then

\begin{align*}\int \cos^2\theta\cdot\sin\theta~d\theta&=-\int w^2~dw\\ &=-\frac {w^3}3\\ &=-\frac 13\cos^3\theta\end{align*}

Thus,

\begin{align*}\iint_S\nabla\times {\bf F}\cdot{\bf n}~d\sigma&=\int_C{\bf F}\cdot d{\bf r}\\ &=\int_0^{2\pi}\cos^2\theta\cdot\sin\theta~d\theta\\ &=-\frac 13\cos^3\theta\mid_0^{2\pi}\\ &=0\end{align*}

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