# Calculating Flux through Ellipsoid

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1. Mar 28, 2015

### bananabandana

1. The problem statement, all variables and given/known data
Let $E$ be the ellipsoid:
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+z^{2}=1$$
Let $S$ be the part of the surface of $E$ defined by:
$$0 \leq x \leq 1, \ 0 \leq y \leq 1, \ z > 0$$
Let F be the vector field defined by $$F=(-y,x,0)$$

A) Explain why $\int_{S} \vec{F} \cdot \vec{dS} =0$ in the case $a=b$

B) In the case $a \neq b$, find $\int_{S} \vec{F} \cdot \vec{dS}$
2. Relevant equations

3. The attempt at a solution
Part A)
This really isn't obvious to me. Can we make some kind of symmetry argument? I can see that $\nabla \cdot \vec{F} =0$ but this doesn't help us as the surface $S$ is not closed.

Part B)
It looks like I can do this ,but the answer I get is so ugly that it makes me very suspicious.

Can easily show (by calculating $\nabla E(x,y,z)$ that:

(1) $$\vec{dS} = \begin{pmatrix} \frac{x}{a^{2}z} \\ \frac{y}{b^{2}z} \\ 1 \end{pmatrix} dxdy$$

So:

(2) $$I=\int \vec{F} \cdot \vec{dS} =\frac{a^{2}-b^{2}}{a^{2}b^{2}} \int_{x=0}^{x=1}\int_{y=0}^{y=1} \frac{xy}{z}$$

(3) Substituting in for $z$:
$$I = \frac{a^{2}-b^{2}}{a^{2}b^{2}} \int_{x=0}^{x=1}\int_{y=0}^{y=1} \frac{xy}{\sqrt{1-\bigg(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \bigg)}}$$

If I evaluate the inner integral by making the subtitution:
(4) $$u=\sqrt{1-\bigg(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \bigg)}$$
and then evaluate the outer one with a similar job, I get the result:

(5) $$I = \frac{a^{2}-b^{2}}{3} \bigg[ \bigg(1-(\frac{1}{a^{2}}+\frac{1}{b^{2}})\bigg)^\frac{3}{2} - \bigg(1-\frac{1}{b^{2}}\bigg)^{\frac{3}{2}} - \bigg(1-\frac{1}{a^{2}}\bigg)^{\frac{3}{2}}-1 \bigg]$$

Is this ridiculous? If so where did I go wrong? :P
Realize it is quite a tedious question, but has been very frustrating. Help would be much appreciated!

Thank you :D

2. Mar 28, 2015

### RUber

Notice that the size of the ellipse is all that changes as z goes from zero to one. So you can fix z for one slice at a time.
Your equation 2 should be enough to see why it is zero when a=b.
Fix your bounds on you integrals so z goes from 0 to 1 and bounds on x and y are functions.

3. Apr 4, 2015

### bananabandana

Isn't the region of integration a square? ( I thought the definition of the region as $0 \leq x \leq 1$ and $0 \leq y \leq 1$ implied this?). Hence, no need for writing $y=y(x)$ in the limits...
and yes, I know I can get it from (2), but the question was phrased in a way that it suggested you were meant to see this intuitively before getting to (2), and I didn't know how to do that.

Thanks!

4. Apr 4, 2015

### RUber

It can't be the square, unless a and b are large enough. Look at the case where a=b=1. (1,1) which would be in the square region is not in your region of interest.
You have a few cases. Look at the simple one I just mentioned, then to see how a and b affect the outcome examine when they are both less than 1 (in magnitude), both more than one and mixed.

5. Apr 4, 2015

### bananabandana

Ah, sorry. I forgot to say that the question stated $a > \sqrt{2}$ , $b >\sqrt{2}$ and $a \neq b$. Might this change things? (Though I admit, I don't know how they got these conditions).

6. Apr 4, 2015

### RUber

Okay, then the whole area of the square is in the region.
On ds, what would be the harm in leaving z in the z column?
Then the dot product would eliminate it.

7. Apr 4, 2015

### LCKurtz

That just guarantees your square lies completely inside the ellipse on the xy plane.

8. Apr 4, 2015

### bananabandana

Oh, of course, because the diagonal of the square is $\sqrt{2}$. I see! Thanks. I think I am leaving $z$ in the $z$ column though, but the vector field $\vec{F}$ is $\vec{F}=(-y,x,0)$ - or did you mean something else?

But am I correct in thinking now that the answer I first got is most likely right? :)

9. Apr 4, 2015

### LCKurtz

I didn't check your work any farther than your integral above. It is set up correctly.

10. Apr 6, 2015

### bananabandana

So long as it's set up right, I'm happy. Thanks for the help! :)