No, the version that you doubt is absolutely correct. $\rho$ signifies the distance between the points in your integration domain and the origin. As such, it can only be positive. You get the "lower" points for $\phi \in (\frac \pi 2, \pi)$. ($\phi$ is the angle formed by your point $(x,y,x)$ and the vertical axis.)
I think is simpler to describe your region in cylindrical coordinates .
The cone with $\phi = \frac{\pi}{6} $ has equation $z = \sqrt{3(x^2+y^2)}$ and the other one $z = \dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2}.$
So if we fix a point $(x,y)$ in the $xy$-plane and move straight up from that point we will 'hit' first the cone with $\phi = \dfrac{\pi}{3}$ and then the cone with $\phi = \dfrac{\pi}{6}.$ So our $z$ coordinate must be all the time between
$$\dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2} \leq z \leq \sqrt{3(x^2+y^2)}.$$
So we are done with $z-$coordinate.
Now the bounds for $(x,y)$ must be between the two circles of radius 1 and 2 , so $$1\leq \sqrt{x^2+y^2} \leq 2.$$
So our region in cartesian coordinates must be
$$W = \biggl\{(x,y,z) : 1\leq \sqrt{x^2+y^2} \leq 2 \text{ and } \dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2} \leq z \leq \sqrt{3(x^2+y^2)} \biggr\}.$$
Now converting in cylindrical coordinates our regions becomes
$$W = \biggl\{(r,\theta,z) : 1\leq r \leq 2 \ , 0\leq \theta \leq 2\pi \text{ and } \dfrac{1}{\sqrt{3}}r \leq z \leq \sqrt{3}r \biggr\}.$$
So your integral must be
$$\mathcal{V}(W) = \int_{\theta = 0}^{2\pi}\int_{r=1}^{2}\int_{\frac{1}{\sqrt{3}}r}^{\sqrt{3}r} r \ dz\, dr\, d \theta$$
Ps.The result must be $2\times\mathcal{V}(W)$ since the region W is for $z>0$ but because the whole region is symmetrical to the $xy-$plane the case $z<0$ gives also volume equal to $\mathcal{V}(W).$
Best Answer
Following @kimchilover's suggestion, suppose $X,Y,Z$ are independent and identically distributed random variables having the uniform distribution on $(-2,-1)$.
So the pdf of $(X,Y,Z)$ is just $$f(x,y,z)=\mathbf1_{-2<x,y,z<-1}$$
Now,
$$E\left(\frac{X^2+Y^2+Z^2}{X^2+Y^2+Z^2}\right)=1$$
Or,
$$E\left(\frac{X^2}{X^2+Y^2+Z^2}\right)+E\left(\frac{Y^2}{X^2+Y^2+Z^2}\right)+E\left(\frac{Z^2}{X^2+Y^2+Z^2}\right)=1\tag{*}$$
By symmetry, $(X,Y,Z),(Y,Z,X)$ and $(Z,X,Y)$ have the same distribution, so that in turn $\frac{X^2}{X^2+Y^2+Z^2}, \frac{Y^2}{X^2+Y^2+Z^2}$ and $\frac{Z^2}{X^2+Y^2+Z^2}$ also have the same distribution.
Therefore $(*)$ reduces to
$$3E\left(\frac{X^2}{X^2+Y^2+Z^2}\right)=1$$
That is, $$\iiint_{-2<x,y,z<-1} \frac{x^2}{x^2+y^2+z^2}\,dx\,dy\,dz=\frac{1}{3}$$