Question

Evaluate the surface integral. s z + x2y ds s is the part of the cylinder y2 + z2 = 4 that lies between the planes x = 0 and x = 3 in the first octant

Answer 1

Parameterize the part of the cylinder $\mathcal S$ by

$\mathbf r(u,v)=\begin{cases}x(u,v)=u\\y(u,v)=2\cos v\\z(u,v)=2\sin v\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$. Then

$\|\mathbf r_u\times\mathbf r_v\|=2$

So the surface integral reduces to

$\displaystyle\iint_{\mathcal S}(z+x^2y)\,\mathrm dS=2\int_{v=0}^{v=2\pi}\int_{u=0}^{u=3}(2\sin v+2u^2\cos v)\,\mathrm du\,\mathrm dv$
$=\displaystyle4\int_{v=0}^{v=2\pi}\int_{u=0}^{u=3}(\sin v+u^2\cos v)\,\mathrm du\,\mathrm dv=0$