Question

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = x i + y j + 9 k S is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 8

Answer 1

$S$ is a closed surface with interior $R$, so you can use the divergence theorem.

$\vec F(x,y,z)=x\,\vec\imath+y\,\vec\jmath+9\,\vec k\implies\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(x)}{\partial x}+\dfrac{\partial(y)}{\partial y}+\dfrac{\partial(9)}{\partial z}=2$

By the divergence theorem, the flux of $\vec F$ across $S$ is given by the integral of $\nabla\cdot\vec F$ over $R$:

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

Convert to cylindrical coordinates, setting

$x=u\cos v$

$y=y$

$z=u\sin v$

The integral is then

$\displaystyle2\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}\int_{y=0}^{y=8-u\cos v}u\,\mathrm dy\,\mathrm du\,\mathrm dv=\boxed{16\pi}$