1. ## Messy Surface Integral

Hello all I'm trying to do this surface integral and I'm getting really nasty numbers.

This was taken from an old test and the results on these tests usually work out to a reasonable clean number compared to what I've gotten as an answer.

Did I make any mistakes? Am I doing this the hard way? Anything wrong?

Thanks again!

2. If you are trying to integrate over the cured surface then a change of variables will make this much easier.

Choose x = 2*Cos phi, y=2*sin phi. I.e. change to cylindrical coordinates.

Thenyou must integrate phi from 0 to pi/4 and z from 0 to 2.

ie
$\displaystyle \int^2_0 \int^{pi/4}_0 4Cos(\phi)Sin(\phi)d \phi dz$

This integral is quite easy to solve.

3. Originally Posted by Kiwi_Dave
If you are trying to integrate over the cured surface then a change of variables will make this much easier.

Choose x = 2*Cos phi, y=2*sin phi. I.e. change to cylindrical coordinates.

Thenyou must integrate phi from 0 to pi/4 and z from 0 to 2.

ie
$\displaystyle \int^2_0 \int^{pi/4}_0 4Cos(\phi)Sin(\phi)d \phi dz$

This integral is quite easy to solve.
No, the boundary is the parabola $y= 4- x^2$, not the circle $y^2= 4- x^2$ as you seem to think.

Jegues, your result is what I get.