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
 A: The domain $D$ is the area inside the intersection between the sphere $$S=\{(x,y,z)\in\Bbb R^3:x^2+(y-a)^2+z^2=a^2,\;a\in \Bbb R\}$$
and the half cone $$C=\{(x,y,z)\in\Bbb R^3:y=x^2+z^2\}$$
The function to integrate is $f(x,y,z)=\sqrt{x^2+y^2+z^2}$.
In spherical coordinates, with radial distance $r$, azimuthal angle $\theta$, and polar angle $\phi$, we have $x=r\sin\phi\cos\theta$, $y=r\sin\phi\cos\theta$, $z=r\cos\phi$ and $\mathrm dV=r^2\sin\phi\,\mathrm d r\,\mathrm d\theta\,\mathrm d\phi$. 
For $x=0$, we have $y\le\sqrt{z^2}=|z|$ and then $\phi\in\left[\frac\pi 4,\frac{3\pi}4\right]$. For $z=0$, we have $y\le\sqrt{x^2}=|x|$ and then $\theta\in\left[\frac\pi 4,\frac{3\pi}4\right]$.
Thaat is we have to integrate the function $f(r\sin\phi\cos\theta,r\sin\phi\cos\theta,r\cos\phi)=\tilde f(r,\phi,\theta)=r$ over the region inside the spherical wedge
$$
W=\left\{(r,\phi,\theta)\in\Bbb R^3,\, a\in\Bbb R: r\le |a|,\,\frac\pi 4\le\phi\le\frac{3\pi}4,\,\frac\pi 4\le\theta\le\frac{3\pi}4\right\}
$$
So the integral becomes
$$
\iiint_D f\mathrm dV=\iiint_W \tilde f\mathrm dW=\int_{\pi/4}^{3\pi/4}\int_{\pi/4}^{3\pi/4}\int_{0}^{|a|}r^3\sin\phi\,\mathrm d r\,\mathrm d\theta\,\mathrm d\phi=\frac{a^4}{4}\frac{\pi}{2}{\sqrt 2}
$$
A: 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
$$\iiint_D \sqrt{x^2+y^2+z^2} dV = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{2a\cos\theta} \rho^3\sin\theta \,d\rho \,d\theta \,d\varphi = \color{blue}{\dfrac{8-\sqrt{2}}{5}\pi a^4}$$
