Question

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + tan z j + (x2z + y2)k and S is the top half of the sphere x2 + y2 + z2 = 9. (Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2 + y2 ≤ 9, oriented downward, and S2 = S1 ∪ S.)

Answer 1

By the divergence theorem, the surface integral over $S_2$ is

$\displaystyle\iint_{S_2}\mathbf F\cdot\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dR$

where $R$ denotes the space bounded by $S_2$. Assuming the vector field is given to be

$\mathbf F(x,y,z)=z^2x\,\mathbf i+(y^3+\tan z)\,\mathbf j+(x^2z+y^2)\,\mathbf k$

then

$\nabla\cdot\mathbf F=\dfrac\partial{\partial x}[z^2x]+\dfrac\partial{\partial y}[y^3+\tan z]+\dfrac\partial{\partial z}[x^2z+y^2]=z^2+3y^2+x^2$

Converting to spherical coordinates, we take

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}$

so that the triple integral becomes

$\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=3}(\rho^2\cos^2\varphi+3\rho^2\sin^2\theta\sin^2\varphi+\rho^2\cos^2\theta\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$
$=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4(\cos^2\varphi+2\sin^2\theta\sin^2\varphi+\sin^2\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$
$=162\pi$

Now the integral over $S$ alone will be the difference of the integral over $S_2$ and the integral over $S_1$, i.e.

$\displaystyle\iint_{S_2}=\iint_{S_1\cup S}=\iint_{S_1}+\iint_S\implies\iint_S=\iint_{S_2}-\iint_{S_1}$

We can parameterize the points in $S_1$ by

$\mathbf s(r,\theta)=\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=0\end{cases}$

so that the integral over $S_1$ is

$\displaystyle\iint_{S_1}\mathbf F\cdot\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\mathbf F(x(r,\theta),y(r,\theta),z(r,\theta))\cdot\left(\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_0^{2\pi}\int_0^3(r^3\sin^3\theta\,\mathbf j+r^2\sin^2\theta\,\mathbf k)\cdot(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_0^{2\pi}\int_0^3r^3\sin^2\theta\,\mathrm dr\,\mathrm d\theta$
$=\dfrac{81\pi}4$

So, the integral over $S$ alone evaluates to

$\displaystyle\iint_S=\iint_{S_2}-\iint_{S_1}=162\pi-\dfrac{81\pi}4=\dfrac{567\pi}4$