# Calculate the integral over the sphere above the $X$-$Y$ plane

I need to calculate the following integral:

$$\iiint \sqrt{x^{2}+y^{2}+z^{2}} \, dx \, dy\, dz$$

over the $x^{2}+y^{2}+(z-\frac{1}{2})^{2}=\frac{1}{4}$ plane. How can I do this?

I mean, calculating the area of interest is pretty easy:

$$0 < r < \cos\psi$$ $$0 < \phi < 2\pi$$ $$0 < \psi < \frac{\pi}{2}$$

But...what's next?

$$\int_{0}^{2\pi}d\phi \int_{0}^{\frac{\pi}{2}}\sin\psi \, d\psi \int_{0}^{\cos\psi}r^{3} \, dr$$

Is this correct? If so, how should I do this? I can't evaluate this integral...

Also, does this integral equal to:

$$\int_{0}^{2\pi} \, d\phi \int_{0}^{\frac{\pi}{2}} \, d\psi \int_{0}^{\cos\psi}r^{3}\sin\psi \, dr \, ?$$

And finally, just out of curiosity...if I have simply $dx\,dy\,dz$ in the integral, I calculate the volume...but well, what do I calculate here? :P

You are finding a volume in $4$th dimension here. Yes, the limits here is correct. In fact, $\phi$ is independent of two coordinates $r$ and $\psi$, so you can have $$\int_{0}^{2\pi}d\phi \int_{0}^{\frac{\pi}{2}}\sin\psi ~d\psi \int_{0}^{\cos\psi}r^{3}dr=2\pi\int_{0}^{\frac{\pi}{2}}\sin\psi ~d\psi \int_{0}^{\cos\psi}r^{3}dr$$ and $$\int_{0}^{\frac{\pi}{2}}\sin\psi ~d\psi \int_{0}^{\cos\psi}r^{3}dr=\int_{0}^{\frac{\pi}{2}}\sin\psi ~d\psi\times\left(r^4/4\right)|_{0}^{\cos\psi}=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\sin\psi \times\cos^4\psi~d\psi$$ Now for the later integral take $u=\cos\psi$ and do the rest.