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 + 6 k s is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 2

Answer 1

Since $\mathcal S$ is closed, use the divergence theorem:

$\displaystyle\iint_{\mathcal S}\mathbf f(x,y,z)\cdot\,\mathrm d\mathbf S=\iiint_{\mathcal V}\nabla\times\mathbf f(x,y,z)\,\mathrm dV$

where $\mathcal V$ is the solid with boundary $\mathcal S$. Now,

$\nabla\times\mathbf f=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial6}{\partial z}=1+1+0=2$

so the volume integral, converting to cylindrical coordinates via

$\begin{cases}x(r,u,v)=r\cos u\\y(r,u,v)=v\\z(r,u,v)=r\sin u\end{cases}$

is

$\displaystyle2\iiint_{\mathcal V}\,\mathrm dV=2\int_{u=0}^{u=2\pi}\int_{r=0}^{r=1}\int_{v=0}^{v=2-r\cos u}r\,\mathrm dv\,\mathrm dr\,\mathrm du=4\pi$