- #1

bananabandana

- 113

- 5

## Homework Statement

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} ##

## Homework Equations

## 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