Question

Evaluate the integral i = z s yz ds when s is the part of the plane 2x + y + 2z = 0 shown in enclosed by the cylinder x 2 + y 2 = 4 .

Answer 1

Parameterize the region $\mathcal S$ by

$x=u\cos v$
$y=u\sin v$
$2x+y+2z=0\implies z=-x-\dfrac y2=-u\cos v-\dfrac{u\sin v}2$
$\implies\mathbf r(u,v)=(x(u,v),y(u,v),z(u,v))$

with $0\le u\le2$ and $0\le v\le2\pi$. Then we have surface element

$\mathrm dS=\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv=\dfrac{3u}2\,\mathrm du\,\mathrm dv$

so the surface integral becomes

$\displaystyle\frac32\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u^2\sin v\left(u\cos v-\dfrac{u\sin v}2\right)\,\mathrm du\,\mathrm dv=-3\pi$