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Let $${\bf F}(x,y,z)=\begin{bmatrix}f(z)\\f(z)\\f(z)\end{bmatrix}$$ be a smooth vector field that depends on the z-coordinate. Let $R$ be the cilinder in $\mathbb R^3$ bounded by $S=\{(x,y,z):x^2+y^2\le 1\text{ and } -1\le z\le 1\}$, and the closed disks $S_\pm=\{x^2+y^2\le 1, z=\pm1\}$.

Let $\vec N$ be the outward pointing unit normal on $R$. How can I prove $\iint_S{\bf F}\bullet N dS$=0? And is it still true if ${\bf F}$ only depends on $x$?

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This is false. The integral equals $\pi(f(1)-f(-1))$. – Chris Eagle May 23 '12 at 21:55

The bijection $(x,y,z)\mapsto(-x,-y,z)$ of the vertical side $S$ of the cylinder sends $F\cdot N$ to $-F\cdot N$ hence the integral of $F\cdot N$ on $S$ is indeed zero.

The argument extends to any vector field $F$ such that $F(-x,-y,z)=F(x,y,z)$ and, more generally, to any vector field $F$ such that $F(-x,-y,z)\cdot(x,y,0)=F(x,y,z)\cdot(x,y,0)$ for every $(x,y,z)$ in $S$.

Note that the integral of $F\cdot N$ on $S_\pm$ is $\pm\pi f(\pm1)$ hence the integral on the complete boundary $\partial R=S\cup S_+\cup S_-$ is not zero but $\pi(f(1)-f(-1))$.

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