# Vector calculus question - surface of ellipsoid

1. Apr 22, 2012

### Froskoy

1. The problem statement, all variables and given/known data
Let $E$ be the ellipsoid

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+z^2=1$$

where $a>\sqrt{2}$ and $b>\sqrt{2}$. Let S be the part of the surface of $E$ defined by $0\le x\le1, 0\le y\le1, z>0$ and let $\mathbf{F}$ be the vector field defined by $\mathbf{F}=(-y,x,0)$. Given that the surface area element of $S$ is given by

$$d \mathbf{S} = \left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 } \right) dxdy$$

find $\int_S\mathbf{F}.d\mathbf{S}$ in the case $a/ne b$

2. Relevant equations
Scalar product

3. The attempt at a solution
$$\int_S\mathbf{F}.d\mathbf{S}=\int_S(-y,x,0).\left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 }\right) dxdy =\int_{y=0}^1\int_{x=0}^1\frac{-xy}{a^2z}+\frac{xy}{b^2z}dxdy =\left({\frac{1}{b^2}-\frac{1}{a^2}}\right)\int_{y=0}^1\int_{x=0}^1\frac{xy}{z}dxdy$$

It's at this point I'm not sure what to do with the parameter $z$. I tried continuing, treating $z$ as constant to get

$$\int_S \mathbf{F}.d\mathbf{S}= \frac{1}{4z} \left({ \frac{1}{b^2}- \frac{1}{a^2}}\right)$$

but don't like the fact there is a $z$ there? Would converting to spherical coordinates help? If so, how would you do it?

With very many thanks,

Froskoy.

Last edited: Apr 22, 2012
2. Apr 22, 2012

### HallsofIvy

Staff Emeritus
Since you know z is not a constant, it makes no sense at all to "treat z as constant".

You are told that
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ z^2= 1$$
so
$$z= \pm\sqrt{1- \frac{x^2}{a^2}- \frac{y^2}{b^2}$$
Do the z> 0 and z< 0 parts separately.

I would NOT use spherical coordinates but a variation on cylindrical coordinates might help. Taking z= 0, we see that the ellipsoid projects to the ellipse $x^2/a^2+ y^2/b^2= 1$. Let $x= a rcos(t)$, $y= b rsin(t)$. The Jacobian is $$\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial t}\end{array}\right|= \left|\begin{array}{cc}a cos(t) & -a rsin(t) \\ b sin(t) & b rcos(t)\end{array}\right|= ab cos^2(t)+ ab sin^2(t)= ab$$
so that $dxdy= ab dr dt$