Question

Evaluate the surface integral ∫∫FS * 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 + 8 kS is the boundary of the region enclosed by the cylinder x^2 + z^2 = 1 and the planes y = 0 and x + y = 6

Answer 1

$\vec F(x,y,z)=x\,\vec\imath+y\,\vec\jmath+8\,\vec k\implies\nabla\cdot\vec F(x,y,z)=2$

so that by the divergence theorem,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=2\iiint_R\mathrm dV$

where $R$ is the interior of $S$. In cylindrical coordinates, the integral is

$\displaystyle2\iiint_R\mathrm dV=2\int_0^{2\pi}\int_0^1\int_0^{6-r\cos\theta}r\,\mathrm dy\,\mathrm dr\,\mathrm d\theta=\boxed{12\pi}$

where we set

$\begin{cases}x=r\cos\theta\\y=y\\z=r\sin\theta\end{cases}$