# Thread: Integral over a surface

1. ## Integral over a surface

I am having problem with an integral. I dont know how to write in mathcode so excuse me for my notation.

This is in spherical coordinates:
x = theta
y = psi
r = radial component

the integrand is: (cos(x)/r^2)

I want to integrate this over a solid angle of an arbitrary surface.

da is a differential area. And in spherical coordiantes should be: da = r^2*sin(x)dxdy

I know the integral should be 4*pi but I dont get it right. I want to have the antiderivate as (cos(x)^2)/2 from 0 to pi.

Then I get for the whole integral: 1/2*2pi = pi

So I am missing a factor 4 here. Can anyone help me show what I've missed?

2. Isn't the integrand

$\displaystyle \frac{cos(x)}{r^2}da = \frac{cos(x)}{r^2}r^2sin(x) dxdy = cos(x)sin(x)dxdy = \frac{1}{2}sin(2x) dxdy$

In this case, when you integrate with respect to x, you will get $\displaystyle -\frac{1}{4}cos(2x)$

EDIT: I thought I knew what you meant, but I'm not so sure anymore. You should really give some more description of your problem.

3. Thats exactly the same thing. If you take x from zero to pi you'll end up with 1/2 exactly as i did so it doesnt help to rewrite it.