# 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!

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.

• Fair enough, seems like there really is no way to simplify this. I calculated it slightly differently by using divergence on the "cups" and then subtracting the base circles of the cups (this is the calculation you see in the question post), and in retrospect I actually got the right answer too except I accidentally multiplied the flux around the ball by 2. – ro44 Jul 1 '12 at 17:46

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).

• Thanks for your help! One thing I might be missing here is the jump from the triple integral (over the volume of the body) to just one integral. I did the calculation from the triple integral to the single integral in the case where we're talking about $r\leq 4$, so I see how that is justified, but how does it work when talking about the intersection of spheres? – ro44 Jun 30 '12 at 0:16
• I guess another point I'm unclear about is, did you use divergence directly 'on' S here? As I mean, it's a surface with holes... – ro44 Jun 30 '12 at 0:26
• @ro44 Finally I noticed that I misread the question, so I corrected my answer – user8268 Jul 2 '12 at 12:46