# Thread: Nested integral in spherical coordinates

1. ## Nested integral in spherical coordinates

I am suppose to evaluate this integral in spherical coordinates

f(x,y,z) = 1/ (x^2 + y^2 + z^2)^(1/2) over the bottom half of the sphere of radius 5 centered at the origin.

I believe I am to take the integral of 1/ roe (roe^2 sin phi) d roe d phi d theta
so it is just roe times sin phi as the function but the bounds are hard for me

theta of course is from 0 to two pi
I know z goes from -5 to 0

but how to turn this into bounds in spherical coordinates???? Frostking

2. Originally Posted by Frostking
I am suppose to evaluate this integral in spherical coordinates

f(x,y,z) = 1/ (x^2 + y^2 + z^2)^(1/2) over the bottom half of the sphere of radius 5 centered at the origin.

I believe I am to take the integral of 1/ roe (roe^2 sin phi) d roe d phi d theta
so it is just roe times sin phi as the function but the bounds are hard for me

theta of course is from 0 to two pi
I know z goes from -5 to 0

but how to turn this into bounds in spherical coordinates???? Frostking
$\rho$ is the new "radius". It goes from 0 to 5.

$\theta$ remains the same.

But $\phi$ goes from $\frac{\pi}{2}$ to $\pi$ if we are considering the sphere below the z=0 plane.

Now, the function is the same as $\frac{1}{\rho}$.

So you will be integrating $\int_0^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\int_0^5\fr ac{1}{\rho}\rho^2\sin\phi\,d\rho\,d\phi\d,\theta=\ int_0^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\int_0^5\rho \sin\phi\,d\rho\,d\phi\d,\theta$

Can you take it from here?

3. ## nested integral in spherical terms

Yes! Thanks very much for your effort. Frostking