# Divergence Theorem to evaluate integral where S is a sphere

Use the Divergence Theorem to evaluate

$$\iint_S x^2y^2+y^2z^2+z^2x^2 dS$$

where $S$ is the surface of the sphere $x^2+y^2+z^2=1$.

So the divergence of F is $2y^2x+2z^2y+2x^2z$. Then I tried to integrate using spherical coordinates, but the integral seemed to be very exhausting. Is there a better way to do it?

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What's $\mathbf F$? –  joriki Nov 9 '12 at 21:38
By F I meant F =P(x,y,z)+Q(x,y,z)+R(x,y,z)=x^2 y^2 + y^2 z^2 + z^2 x^2 –  Keksainis Nov 9 '12 at 21:41
That's a scalar; it doesn't have a divergence. –  joriki Nov 9 '12 at 21:45

The normal vector on the surface of the sphere is $(x,y,z)$. Thus you can view this as the integral of $(x,y,z)\cdot(xy^2,yz^2,zx^2)$. The divergence of $(xy^2,yz^2,zx^2)$ is $y^2+z^2+x^2=r^2$. The integral of $r^2$ over the unit sphere is
$$\int_0^1\mathrm drr^4\int_{-1}^1\mathrm d\cos\theta\int_0^{2\pi}\mathrm d\phi=\frac45\pi\;.$$
@Keksainis: I don't understand. What limits were you integrating it with? How did you come up with those other limits? These are the limits of the unit sphere. What do you mean by "integration technique"? I used the divergence theorem, as you had asked us to. Other than that it's just trivial integration of constants with respect to $\cos\theta$ and $\phi$ and of $r^4$ with respect to $r$, which yields $r^5/5$. –  joriki Nov 10 '12 at 12:02
@Keksainis: I also used spherical coordinates. There are different sets of conventions for spherical coordinates; perhaps you're using a different one? Regarding $r^4$, a factor $r^2$ comes from the function we want to integrate and a factor $r^2$ comes from the Jacobian of the spherical coordinates. –  joriki Nov 10 '12 at 12:25