Evaluate the Integral $ \iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz$ I am attempting to evaluate the following triple integral using spherical coordinated:
$$
\iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz \\
W = \{(x,y,z) \in \mathbb{R}^3 : 1 \le x^2+y^2+z^2 \le 9; 0 \le x \le y, z \ge 0\}
$$
This tells me that I have the following regions:
$$
x^2 + y^2 + z^2 = 1 \rightarrow \text{Sphere of radius 1}\\
x^2 + y^2 + z^2 = 9 \rightarrow \text{Sphere of radius 3} \\
x=y \rightarrow \text{line in the xy plane}
$$
I have also determined that:
$$
1\le r \le 3
$$
But I am not sure how to determine which part of the intersection of the regions I am to integrate and to solve for $\theta$ and $\phi$. How do I proceed?

$x\geq0, y\geq0, z\geq0$ show $0\leq\theta\leq\dfrac{\pi}{2}$ and $0\leq\phi\leq\dfrac{\pi}{2}$. Also $x\leq y$ states $r\cos\theta\sin\phi\leq r\sin\theta\sin\phi$ or $\tan\theta\geq1$ so $\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{2}$ then
$$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _1^3\frac{r^2 \sin (\varphi )}{r}drd\varphi d\theta=\color{blue}{\pi}$$
