Vector calculus question - surface of ellipsoid

[Froskoy]

Homework Statement

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

Homework Equations

Scalar product

The Attempt at a Solution

<br /> \int_S\mathbf{F}.d\mathbf{S}=\int_S(-y,x,0).\left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 }\right) dxdy<br /> <br /> =\int_{y=0}^1\int_{x=0}^1\frac{-xy}{a^2z}+\frac{xy}{b^2z}dxdy<br /> <br /> =\left({\frac{1}{b^2}-\frac{1}{a^2}}\right)\int_{y=0}^1\int_{x=0}^1\frac{xy}{z}dxdy<br />

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

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

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.

 

[HallsofIvy]

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} &amp; \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial r} &amp; \frac{\partial y}{\partial t}\end{array}\right|= \left|\begin{array}{cc}a cos(t) &amp; -a rsin(t) \\ b sin(t) &amp; b rcos(t)\end{array}\right|= ab cos^2(t)+ ab sin^2(t)= ab
so that dxdy= ab dr dt