Question

Evaluate integral_S integral (x - 2y + z) dS. S: z = 16, x^2 + y^2 le 1

Answer 1

`S` is the disk of radius 1 parallel to the
`x,y`-plane and lying in the plane
`z=16`. Parameterize this surface by

`\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+16\,\vec k`

with
`0\le u\le1` and
`0\le v\le2\pi`. Take the normal vector to
`S` to be

`\vec r_u*\vec r_v=u\,\vec k`

(the orientation doesn't matter because this is a scalar surface integral)

Then

`\displaystyle\iint_S(x-2y+z)\,\mathrm dS=\iint_S(u\cos v-2u\sin v+16)\|u\,\vec k\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^(2\pi)\int_0^1(u^2(\cos v-2\sin v)+16u)\,\mathrm du\,\mathrm dv=\boxed{16\pi}`