Evaluate $\int_{{S}}^{{}} F\cdot n$ where n is outward pointing unit normal I am having trouble simplifying the following calculation:
Let $S=\{(x,y,z)|x^2+y^2+z^2=25,-4\leq x,y,z \leq 4\}$ and $F=(x^3,y^3,z^3)$. I am asked to evaluate the surface integral $\int_{{S}}^{{}}  F\cdot n$ where $n$ is the outward pointing unit normal.
Attempt I 'managed' to calculate the integral in the following way: first I completed S to a ball, used the divergence theorem, then subtracted the leftovers. However this led to a lot of what seems to be needless complexity in the calculation, and the integrals turned out to be very complex. I think my end result was wrong, too; I got $$15000 \pi -6 \times 3/5 \times 512 \pi -6 \times 9 \times 4^3 \times \pi$$
So I guess I'm looking for an easier, more straightforward way to do this which wouldn't involve as many complex calculations.
Thanks!
 A: The method you suggested (using divergence) is actually quite fast. The divergence is $3r^2$ (where $r^2=x^2+y^2+z^2$). Now take a sphere of radius $r$, centred at the origin. If $r\leq4$ then the sphere is contained in $S$, with area $4\pi r^2$. If $4\leq r\leq5$, the area of the intersection of the sphere with $S$ is 
$$4\pi r^2 -6\times2\pi r (r-4)=-8\pi r^2+48\pi r$$
 The integral of the divergence over $S$ is thus
$$\int_0^4 3r^2\times4\pi r^2 dr+\int_4^53r^2\times(-8\pi r^2+48\pi r)dr=28284\pi/5.$$
edit: I misunderstood the question, and computed the integral over the boundary of $x^2+y^2+z^2\leq25$, $-4\leq x,y,z\leq4$ (i.e. my $S$ was closed). To get the correct answer, we need to subtract the integral over the $6$ discs. There we integrate just the constant $4^3$ over $6$ disks with total area $6\times 3^2\pi$, hence the correct answer is
$$28284\pi/5-4^3\times6\times 3^2\pi=11004\pi/5$$
(confirming the accepted result).
A: It is obvious the flux of the whole sphere is $\iiint_V 3r^2 dV = \frac{12}5\pi R^5 = 7500\pi$. I guess there's no problem you getting this.
Then we subtract the flux in the region with $|x|,|y|,|z|\ge 4$ from $7500\pi$. We notice that this will create 6 circular holes from the sphere. By symmetry the flux of all 6 holes should be the same, so we only need to compute one. 
I believe there is no other way than computing $\iint_S \mathbf F\cdot \hat{\mathbf n}\,dS$ directly.
This method does not really use divergence theorem, but I doubt there is any simpler methods. 
The extra cap can be described in spherical coordinates as $\left\{r=5\wedge\theta \in \left[0, \tan^{-1}\frac34\right)\right\}$. The vector field can be rewritten as
\begin{align}
\mathbf F
&= x^3\hat{\mathbf x} + y^3\hat{\mathbf y} + z^3 \hat{\mathbf z} \\
&= (r\sin\theta\cos\phi)^3 (\sin\theta\cos\phi \hat{\mathbf r}+\dotsb) 
 + (r\sin\theta\sin\phi)^3 (\sin\theta\sin\phi \hat{\mathbf r}+\dotsb)
 + (r\cos\theta)^3 (\cos\theta \hat{\mathbf r}+\dotsb) \\
&= r^3  \left( \sin^4\theta(\cos^4\phi + \sin^4\phi) + \cos^4\theta \right) \hat{\mathbf r} + \dotsb,
\end{align}
because $\hat{\mathbf n}dS=r^2 \sin\theta\, d\phi \, d\theta\hat{\mathbf r}$ on the spherical surface, we are left with
\begin{align}
\iint_S \mathbf F\cdot \hat{\mathbf n} \,dS
&= \int_0^{2\pi}\int_0^{\tan^{-1}(3/4)} r^3  \left( \sin^4\theta(\cos^4\phi + \sin^4\phi) + \cos^4\theta \right) \cdot r^2 \sin\theta\, d\theta \, d\phi  \\
&= 5^5 \int_0^{\tan^{-1}(3/4)}\int_0^{2\pi} \left( \sin^5\theta(\cos^4\phi + \sin^4\phi) + \cos^4\theta\sin\theta \right) d\phi \, d\theta \\
&= 5^5\pi \int_0^{\tan^{-1}(3/4)} \left( \frac32 \sin^5\theta + 2\cos^4\theta\sin\theta\right)d\theta \\
&= 5^5\pi \left( \frac32 \cdot \frac{428}{46875} + 2\cdot \frac{2101}{15625} \right) \\
&= \frac{4416}5 \pi,
\end{align}
hence the final answer is
$$ 7500\pi - 6\times\frac{4416}5 \pi = \frac{11004}5\pi \approx 6914.0171. $$
As "verification", I did arrive at the same answer by numerical integration. 
You should need to show yourself how to carry out the integrals of $\sin^5\theta$ etc.
