Question

Evaluate the surface integral:S

(x^2z + y^2z) dS
S is the hemisphere
x2 + y2 + z2 = 9, z ≥ 0

Answer 1

Assuming
`S` does not include the plane
`z=0`, we can parameterize the region in spherical coordinates using


`\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle`

where
`0\le u\le2\pi` and
`0\le v\le\frac\pi/2`. We then have


`x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v`

`(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v`

Then the surface integral is equivalent to


`\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_(u=0)^(u=2\pi)\int_(v=0)^(v=\pi/2)\sin^2v\cos v\left\|(\partial\mathbf r(u,v))/(\partial u)* (\partial\mathbf r(u,v))/(\partial u)\right\|\,\mathrm dv\,\mathrm du`

We have


`(\partial\mathbf r(u,v))/(\partial u)=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle`

`(\partial\mathbf r(u,v))/(\partial v)=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle`

`\implies(\partial\mathbf r(u,v))/(\partial u)*(\partial\mathbf r(u,v))/(\partial v)=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle`

`\implies\left\|(\partial\mathbf r(u,v))/(\partial u)*(\partial\mathbf r(u,v))/(\partial v)\|=9\sin v`

So the surface integral is equivalent to


`\displaystyle243\int_(u=0)^(u=2\pi)\int_(v=0)^(v=\pi/2)\sin^3v\cos v\,\mathrm dv\,\mathrm du`

`=\displaystyle486\pi\int_(v=0)^(v=\pi/2)\sin^3v\cos v\,\mathrm dv`

`=\displaystyle486\pi\int_(w=0)^(w=1)w^3\,\mathrm dw`

where
`w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv`.


`=\frac{243}2\pi w^4\bigg|_(w=0)^(w=1)`

`=\frac{243}2\pi`