Question

Evaluate the surface integral. s x2z + y2z ds s is the hemisphere x2 + y2 + z2 = 9, z ≥ 0

Answer 1

Parameterize the hemisphere $\mathcal S$ by

$\mathbf s(u,v)=\langle3\cos u\sin v,3\sin u\sin v,3\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi2$. Then the surface integral becomes

$\displaystyle\iint_{\mathcal S}(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\cos v\sin^2v\|\mathbf s_u\times\mathbf s_v\|\,\mathrm dv\,\mathrm du$

We have

$\mathbf s_u\times\mathbf s_v=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$

$\implies\|\mathbf s_u\times\mathbf s_v\|=9\sin v$

The integral reduces to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\cos v\sin^3v\,\mathrm dv\,\mathrm du$

Substitute $t=\sin v$, $\mathrm dt=\cos v\,\mathrm dv$:

$\displaystyle486\pi\int_{t=0}^{t=1}t^3\,\mathrm dt=\frac{243\pi}2$