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I'm working on a problem where $F$ is a vector field $F=\langle x,y,z\rangle $ where the $E$ is the unit ball $ x^2+y^2+z^2 \le 1 $.

This is an easy surface integral to calculate using Divergence Theorem:

$$ \iiint_E {\rm div}(F)\ dV = \iint_{S=\partial E} \vec{F}\cdot d{\bf S}$$

But, to confirm the divergence theorem by calculating the surface integral directly, how would you set up the bounds on the double integral for a unit ball?

By the way, $div(\vec{F}) = 0$ in this case, and then it's a triple integral $\iiint_E 1 dV$. Since E is a solid ball and $1$ with in triple integral results in a volume calculation, the answer is $\frac{4\pi r^3}{3} = \frac{4\pi 1^3}{3} = \frac{4\pi }{3}$. So that's the final numerical answer I'm looking for when doing the surface integral in order to verify the divergence theorem.

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  • $\begingroup$ ${\bf x}(x,y)=(x,y,\sqrt{1-x^2-y^2}), \ {\bf y}(x,y)=(x,y,-\sqrt{1-x^2-y^2})$ where $x^2+ y^2 \leq 1$. How about calculate double integral twice ? but in this case since $F=(x,y,z)$, we can calculate double integral easily. $\endgroup$ – HK Lee May 8 '14 at 5:14
  • $\begingroup$ Alternatively, in spherical coordinates: $r(\theta,\phi)=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)$, $0\leq\theta\leq 2\pi$,$0\leq\phi\leq\pi$ $\endgroup$ – A Nonny May 8 '14 at 5:33
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    $\begingroup$ I'm not really following why there are two parametrization here which look exactly the same. Perhaps I should be calculating the surface integral for a unit sphere? $\endgroup$ – Bob Shannon May 8 '14 at 5:34
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    $\begingroup$ If we cut a sphere along equator, we have two surfaces. So we need two : I edit the second parametrization ${\bf y}(x,y)=(y,x,-\sqrt{1-x^2-y^2})$ $\endgroup$ – HK Lee May 8 '14 at 5:56
  • $\begingroup$ Oh ok, I see now. Spherical coordinates may come in handy here then to prevent having to do two surface integrals. $\endgroup$ – Bob Shannon May 8 '14 at 5:59
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After reading a few examples on the web and in my textbook I think I may be have figured this out. It seems that the hardest (or most tedious) part of this problem is actually performing the calculus. Indeed, I needed to calculate the surface integral of a unit sphere floating around in the given vector field, because a unit sphere is the surface that lines a unit ball that was given in the problem.

A unit sphere can be parametrized using spherical coordinates:

$$ r(\phi,\theta) = \sin(\phi)\cos(\theta)\hat{i} + \sin(\phi)\sin(\theta)\hat{j}+\cos{\phi}\hat{k}$$

By definition, a surface integral: $$ \iint_S \vec{F} \cdot d\vec{S}= \iint_D\vec{F}(\vec{r}(\phi,\theta))\cdot(\vec{r}_u \times \vec{r}_v) dA$$

If you have some time and decide to crank this out and set up the integral:

$$\vec{F}(\vec{r}(\phi,\theta)) = \cos(\phi)\hat{i}+\sin(\phi)\sin(\theta)\hat{j}+\sin (\phi)\cos(\theta)\hat{k} $$ $$ \vec{r}_\phi \times \vec{r}_\theta = \sin^2(\phi)\cos(\theta)\hat{i}+\sin^2(\phi)\sin(\theta)\hat{j}+\sin(\phi)cos(\phi)\hat{k}$$

Then: $$\vec{F}\cdot(\vec{r}_u \times \vec{r}_v) = \cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)$$

The integral is now ready to put in the oven:

$$\iint_D [\cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)] dA$$

$$= \int_0^{2\pi}\int_0^\pi [\cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)] d\phi d\theta$$

$$= 2 \int_0^\pi\sin^2(\phi)\cos(\phi) d\phi \int_0^{2\pi}\cos(\theta)d\theta\int_0^\pi\sin^3(\phi)d\phi\int_0^{2\pi}\sin^2\theta d\theta$$

$$= 0 + \int_0^\pi \sin^3(\phi)d\phi \int_0^{2\pi} \sin^2(\theta)d\theta $$ $$ = \frac{4\pi}{3} $$

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    $\begingroup$ Another way : Over unit sphere, normal vector is just ${\bf F}$ So $\iint_S {\bf F}\cdot d{\bf S} = \iint_S dS = 4\pi$ $\endgroup$ – HK Lee May 8 '14 at 6:39

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