Question

Evaluate the integral i = z s f ds when f(x, y, z) = z 2 + 3xy and s is the portion of the plane x + 2y + 2z = 0 above the unit disk x 2 + y 2 ≤ 1 in the xy-plane

Answer 1

Parameterize $S$ by

$\vec r(u,v)=\left(u\cos v,u\sin v,-\dfrac{u(\cos v+2\sin v)}2\right)$

with $0\le u\le1$ and $0\le v\le2\pi$. The surface element is

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

Then the integral is

$\displaystyle\iint_S(z^2+3xy)\,\mathrm dS$

$\displaystyle=\int_0^{2\pi}\int_0^1\left(3u^2\cos v\sin v+\frac{u^2}4(\cos v+2\sin v)^2\right)\frac{3u}2\,\mathrm du\,\mathrm dv=\boxed{\frac{15\pi}{32}}$