Evaluating integral...

iwonder · #1 $Old$ 11-19-2008, 01:09 AM

How to use cylindrical or spherical coordinates as appropriate to evaluate the integral:

∫∫∫ z^2/(x^2+y^2+z^2) dV
E

where E is the top half of a sphere of radius a>0 that is centered at the origin.

Thankyou for your help!!

iwonder · #2 $Old$ 11-20-2008, 11:58 AM

I came up with this...

Is this correct?

Moo · #3 $Old$ 11-20-2008, 12:25 PM

Hello,

Hmmm I'm finding $\cos^2(\phi)$ instead of $\cos(\phi)$

Also, there is that coefficient, 2, disturbing me. We one calculated the dxdydz, and my friend found this coefficient. I wasn't able to find the mistake in either of our two computations.
But in the wikipedia, there isn't this coefficient.

As for the boundaries of your integral, it depends on how you define $\theta$ and $\phi$

elizsimca · #4 $Old$ 11-20-2008, 10:54 PM

Quote:

Originally Posted by Moo $View Post$

As for the boundaries of your integral, it depends on how you define $\theta$ and $\phi$

$\theta$ is the polar $\theta$ and $\phi$ is the angle of opening from the z-axis. So $\phi$ from 0 to $\frac{\pi}{2}$ would be like a coffee filter completely closed up along the z-axis and then blossoming outward and to rest on the xy-plane.

In regards to the question, I'm confused about where the 2 is coming from as well...I will ponder this a little more. Any insight into how you came up with the 2? I'm just not seeing it.

elizsimca · #5 $Old$ 11-20-2008, 11:13 PM

Quote:

Originally Posted by iwonder $View Post$

I came up with this...

Is this correct?

Your limits of integration are fine, but remember that from rectangular to polar $z=\rho\cos{\phi}$ and $x^2+y^2+z^2=\rho^2$ . So $\frac{z^2}{x^2+y^2+z^2}=\frac{\rho^2\cos^2{\phi}}{\rho^2}=\cos^2{\phi}$ .

So, then the integral becomes $\int_0^{2\pi}\int_0^a\int_0^\frac{\pi}{2}\rho^2\cos^2\phi\sin\phi d\phi d\rho d\theta$

Perhaps your two came from a trig identity, I'm too tired to think about it..if it came from an identity then it's fine.

Also, I have my integration in a little bit different order than you, but since we are in spherical coordinates and the sphere is centered at the origin we can flip the integration limits arbitrarily.

I think this is all correct...good luck

iwonder · #6 $Old$ 11-20-2008, 11:31 PM

So, is the first limit of integration from 0 to 2/pi or 2pi?
just wanna make sure, thankyou so much!

elizsimca · #7 $Old$ 11-20-2008, 11:42 PM

Quote:

Originally Posted by iwonder $View Post$

So, is the first limit of integration from 0 to 2/pi or 2pi?
just wanna make sure, thankyou so much!

Should be zero to (2*pi) bad latex! I will edit to avoid confusion. sorry about that!

iwonder · #8 $Old$ 11-20-2008, 11:46 PM

It's okay, thanks elizsimca so much!!!!
i already got the answer (2(pi)a^3)/9, hope this is the right answer.