$\int\limits_0^5 \int\limits_{\sqrt{25-x^2}}^{-\sqrt{25-x^2}} \int\limits_{\sqrt{25-x^2-z^2}}^{-\sqrt{25-x^2-z^2}} \frac{1}{x^2+y^2+z^2} \,\mathrm dy\,\mathrm dz\,\mathrm dx$
triple integral trying to change to spherical coordinates.
$\int\limits_0^5 \int\limits_{\sqrt{25-x^2}}^{-\sqrt{25-x^2}} \int\limits_{\sqrt{25-x^2-z^2}}^{-\sqrt{25-x^2-z^2}} \frac{1}{x^2+y^2+z^2} \,\mathrm dy\,\mathrm dz\,\mathrm dx$
triple integral trying to change to spherical coordinates.
The integrals on $y$ and $z$ have their limits in an unusual way (positive below, negative above) but changing both at the same time won't change the value of the integral (as each contributes a minus sign). So we want $ I=\int_0^5 \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \int_{-\sqrt{25-x^2-z^2}}^{\sqrt{25-x^2-z^2}} \frac{1}{x^2+y^2+z^2} \,dy~dz~dx $ The region is the half solid sphere of radius $5$ centered at the origin, with $x\geq0$. In spherical coordinates, this is $ 0\leq\rho\leq5,\ -\pi/2\leq\theta\leq\pi/2,\ 0\leq\phi\leq\pi. $ So $ I=\int_{\pi/2}^{\pi/2}\int_0^{\pi}\int_0^5\frac1{\rho^2}\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=\int_{\pi/2}^{\pi/2}\int_0^{\pi}\int_0^5\sin\phi\,d\rho\,d\phi\,d\theta=5\pi\,\int_0^\pi\sin\phi\,d\phi=5\pi\,(-\cos\phi)|_0^\pi=10\pi. $