# Homework Help: Flux integral over ellipsoidal surface

1. Jan 19, 2016

### pondzo

1. The problem statement, all variables and given/known data

I am having trouble with part iii) of the following problem:

Verify the divergence theorem for the function $\vec{u} = (xy,- y^2, + z)$ and the surface enclosed by the three parts:
(i) $z = 0, x^2+y^2 < 1$,
(ii) $x^2+y^2 = 1, 0 \le z \le 1$ and
(iii) $(a^2-1)x^2+(a^2-1)y^2+z^2 = a^2 , 1 \le z \le a, a > 1$.

2. Relevant equations

Divergence theorem:

$\int_V\vec{ \nabla}\cdot\vec{u}~dV=\oint_{\partial v}\vec{u}\cdot\vec{dS}$

3. The attempt at a solution

The boundaries in iii) describe part of an ellipsoid. Rewrite the equation as follows:
$\frac{x^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{y^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{z^2}{a^2}=1$

LHS$= \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\vec{ \nabla}\cdot\vec{u}~\rho~ d\rho ~dz ~d\theta$
$~~~~ = \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\rho - \rho^2\sin{\theta}~ d\rho ~dz ~d\theta$
$~~~~=\frac{4\pi}{1-a^2}$

I think that is correct... maybe someone can check my setup.

I'm stuck with the RHS.. I tried to use cylindrical coordinates and it got ugly quickly, I tried to use generalised spherical coordinates but it got ugly equally as fast. I expect the process to be somewhat ugly, but mine is so much I feel for certain i'm doing something wrong. I'll give a brief overview of my strategy:

For cylindrical coordinates I tried to parameterize it as follows: $\vec{r}(\theta,z)=(\frac{z^2-a^2}{1-a^2}\cos{\theta},\frac{z^2-a^2}{1-a^2}\sin{\theta},z)$. For starters im not even sure if this is right. I then get the normal vector to this by computing: $\vec{n}=\frac{\partial \vec{r}}{\partial \theta}\times\frac{\partial \vec{r}}{\partial z}$. Computing this and normalising i get: $\hat{n}=\frac{(\cos{\theta},\sin{\theta},\frac{-2z}{1-a^2})}{\sqrt{1+\frac{4z^2}{(1-a^2)}}}$.Plugging all of this into the flux integral looks extremely messy and I don't think I will be able to do the integral. So I abort and resort to spherical coordinates...

For spherical coordinates I tried to parameterize it as follows: $\vec{r}(\theta,\phi)= (\frac{a}{\sqrt{a^2-1}}\sin{\theta}\cos{\phi},\frac{a}{\sqrt{a^2-1}}\sin{\theta}\sin{\phi},a\cos{\theta})$ and the rest of the story goes like the cylindrical one..Any help would be greatly appreciated!

Last edited: Jan 19, 2016
2. Jan 19, 2016

### LCKurtz

I don't understand your use of $s$. What is part (ii) $s = 1, 0 \le z \le 1$ supposed to describe?

3. Jan 19, 2016

### pondzo

I find the notation confusing too, this is my lecturers not mine. I assumed $s^2=x^2+y^2$ though out and hence ii) describes a cylinder of radius 1 and height 1.

4. Jan 19, 2016

### LCKurtz

If part (ii) is supposed to be $x^2+y^2 = 1,~0\le z \le 1$ then write it that way. Then while you are at it, write part (iii) out so it makes sense. Is part (i) supposed to be $0\le x^2 + y^2 \le 1,~z = 0$ ?

5. Jan 19, 2016

### pondzo

Apologies, I thought I might have been interpreting it in the wrong way so I left it just in case. I have amended the original post. And yes i) is the unit disk in the xy plane.

6. Jan 19, 2016

### LCKurtz

OK, that makes more sense. So for part (iii) if you let $b^2 = \frac{a^2}{a^2-1}$ your equation is$$\frac{x^2}{b^2}+\frac{y^2}{b^2}+\frac{z^2}{a^2} = 1$$I haven't worked it out, but what I would try is$$x = b\sin\phi\cos\theta,~y = b\sin\phi\sin\theta,~z = a\cos\phi$$

Edit, added: @pondzo: I realized after I posted this that it is what you tried for generalized spherical coordinates. For what it's worth, I calculated the integral with Maple and got$$\frac{2\pi}{3}\left(\frac{a^6\sqrt{2a^2-1}}{(a^2-1)(1-2a^2)^2} \right)$$for (iii).

Last edited: Jan 19, 2016