<s>[INST] Calculate the triple integral $\iiint_D \sqrt{x^2+y^2+z^2}\, dV$. 
Calculate the triple integral of
$$\iiint_D \sqrt{x^2+y^2+z^2}\, dV$$ where $D$ is bounded by
(1) $x^2+y^2+z^2=2ay$ and (2) $y=\sqrt{x^2+z^2}$.

So far I was thinking that if I made $y$ turn to $z$ and $z$ turn to $y$ I could use cilyndrical coordinates
So (1)$a^2=r^2+(z-a)^2$ and (2) $z=r$ and both intersect at $z=a$
It is correct to propose the integral$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-(z-a)^2}}r\sqrt{z^2+r^2} \,dr\,dz\,d\theta$$?
And if so I evaluate and get $a^3\pi$ but Im suspicious of this result
 [/INST] Yes! Made $y$ turn to $z$ and $z$ turn to $y$, then $z=\sqrt{x^2+y^2}$ turns one time about $z-$axis. This gives us - with spherical coordinates - $0\leq\varphi\leq2\pi$, also $z=a$ shows $0\leq\theta\leq\dfrac{\pi}{4}$ and with $x^2+y^2+z^2=2az$ we have $\rho\leq2a\cos\theta$, hence
$$\iiint_D \sqrt{x^2+y^2+z^2} dV = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{2a\cos\theta} \rho^3\sin\theta \,d\rho \,d\theta \,d\varphi = \color{blue}{\dfrac{8-\sqrt{2}}{5}\pi a^4}$$
</s>