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So I have to compute the triple integral of this: $\int\int\int \frac{1}{1+x^2+y^2+z^2}$ and it says the equation of the sphere is $ x^2 + y^2 + z^2 = z$ which is just an elongated sphere running along the z-axis. But I don't understand how to setup the triple integral, the z on the ride side is totally throwing me off.

1) What would the bounds for $\rho$ be?
2) Why are the bounds for $\phi$ always 0 to $\frac{\pi}{2}$?

Any help would be appreciated, thanks.

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Notice $x^2 + y^2 + z^2 = z$ is just $x^2 + y^2 + (z-1/2)^2 = 1/4$, which means you now have a sphere centered at $(0,0,1/2)$ instead of $(0,0,0)$. – Shuhao Cao Apr 26 '13 at 2:04
@ShuhaoCao okay so that means the center of the sphere is at $(0,0,\frac{1}{2})$ and the radius is $\frac{1}{2}$....but how can that give me an elongated sphere like wolfram is showing me? – Richard Apr 26 '13 at 2:06
I just checked alpha, and the scales of axis are different, so it looks elongated. – Shuhao Cao Apr 26 '13 at 2:08
up vote 2 down vote accepted

This is not an elongated sphere, but just displaced so that it sits atop the plane $z=0$. The equation of the sphere in spherical coordinates is

$$\rho^2=z = \rho \cos{\phi} \implies \rho=\cos{\phi}$$

where $\phi \in [0,\pi/2]$ because the sphere is entirely in the half-space $z \ge 0$. The triple integral then takes the form

$$\int_0^{\pi/2} d\phi \, \sin{\phi} \: \int_0^{\cos{\phi}} d\rho \frac{\rho^2}{1+\rho^2} \: \int_0^{2 \pi} d\theta$$

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so if we were taking the triple integral of the WHOLE sphere then would $ \phi $ go from 0 to $2pi$? 0 to $\frac{pi}{2}$ tells me you're integrating a fourth of a sphere...why? – Richard Apr 26 '13 at 4:57
@Richard: No. This is over the whole sphere because the sphere lies entirely in $z \ge 0$. Note that the limit of $\rho$ is not constant - this does not integrate like a sphere centered at the origin. Sketch a picture and see. – Ron Gordon Apr 26 '13 at 5:26

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