Question

g Use the Divergence Theorem to evaluat where F = D xz + x 2 , −2xy, x2 + z 2 E and S is the hemisphere z = p 4 − x 2 − y 2 together with the portion of the xy-plane beneath that hemisphere, oriented outward.

Answer 1

I guess you're supposed to evaluate the integral,

`\displaystyle\iint_S\vec F\cdot\mathrm d\vec S`

where
`\vec F(x,y,z)=\langle xz+x^2,-2xy,x^2+z^2\rangle` and
`S` is the hemisphere
`z=√(4-x^2-y^2)`.

`\vec F` has divergence

`\mathrm{div}\vec F(x,y,z)=(z+2x)+(-2x)+(2z)=3z`

By the divergence theorem,

`\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_V\mathrm{div}\vec F\,\mathrm dV`

where
`V` is the space with
`S` as its boundary.

Convert to spherical coordinates, taking

`\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi`

Then the integral is

`\displaystyle\int_0^(\pi/2)\int_0^(2\pi)\int_0^2(3\rho\cos\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi`

`\displaystyle6\pi\int_0^(\pi/2)\int_0^2\rho^3\sin\varphi\cos\varphi\,\mathrm d\rho\,\mathrm d\varphi`

`\displaystyle3\pi\int_0^(\pi/2)\int_0^2\rho^3\sin2\varphi\,\mathrm d\rho\,\mathrm d\varphi`

`\displaystyle3\pi\left(\int_0^(\pi/2)\sin2\varphi\,\mathrm d\varphi\right)\left(\int_0^2\rho^3\,\mathrm d\rho\right)=\boxed{12\pi}`