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

$\vec r_u\times\vec r_v=u\,\vec k$

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

Then

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

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