1. ## Answer check (spherical coords)

Use spherical coords to evaluate:

$\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} (2x^2+2y^2+2z^2)^{-1/2} dzdydx$

My sol'n:

We are dealing with the first octant here, with sphere radius=1. So bounds are:

$0\le r \le 1, ~0\le\theta\le \frac{\pi}{2}, ~ ~0\le\phi\le \frac{\pi}{2}$.

$dV = r^2\sin\phi dr d\theta d\phi$.

So hopefully we have (after some cancelling):

$\frac{\sqrt{2}}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 r\sin\phi\ dr d\theta d\phi$

which is

$-\frac{\sqrt{2}\pi}{8}$

I hope that's right

2. Yes, it looks good to me.

3. I derived

$\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(2x^2+2y^2+2z^2)^{-1/2}\,dz\,dy\,dx$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 (2\rho^2)^{-\frac{1}{2}}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho^{-1}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho\,\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{2\sqrt{2}}\sin \phi\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}} \frac{\pi}{4\sqrt{2}}\sin \phi\,d\phi$
$=\left(-\frac{\pi}{4\sqrt{2}}\cos \phi\right)_0^{\frac{\pi}{2}}$
$=-\frac{\pi}{4\sqrt{2}}\cos \left(\frac{\pi}{2}\right)-\left(-\frac{\pi}{4\sqrt{2}}\cos 0\right)$
$=0-\left(-\frac{\pi}{4\sqrt{2}}\right)=\frac{\pi}{4\sqrt{2}} .$

4. Thank you both very much!

Scott -- thanks for the detailed working I made a silly sign mistake at the end as you can see :/.

Your solution is correct (and equivalent to the negative of mine).