Question

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of the sphere x2 + y2 + z2 = 1. (Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2 + y2 ≤ 1, oriented downward, and S2 = S1 ∪ S.)

Answer 1

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$, where $R$ is the region enclosed by $S$. Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$) so that $R$ has boundary $S\cup D$.

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

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

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$. Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$