Evaluating $\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} \,dx \,dy \,dz$

Question

I am trying to solve this triple integral problem , but I am having some issues. $$\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} dx dy dz$$

I tried with the 2 different approaches.

• I converted the variables to spherical coordinates, and I think it went well, until the moment to evaluate the integral using those limits $(-1 , -2)$. I couldn't pass that.

• Conversion to cylindrical coordinates, but no luck at all there. So, could someone be so kind in giving me some light around this? Please. Not sure if I am follow a right approach.

Answer 1

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, $$\iiint_{-2<x,y,z<-1} \frac{x^2}{x^2+y^2+z^2}\,dx\,dy\,dz=\frac{1}{3}$$