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-\frac y2=-u\cos v-\frac{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*\mathbf r_v\|\,\mathrm du\,\mathrm dv=\frac{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-\frac{u\sin v}2\right)\,\mathrm du\,\mathrm dv=-3\pi`