I am finding the center of mass of sphere $x^2+y^2+z^2=2z$ when $\delta(x,y,z)=\sqrt{x^2+y^2+z^2}$. I did the mass as follows (I hope it is right):
$$M=\int_0^{2\pi}d\theta\int_0^{\pi/2}\int_0^{2\cos{\phi}}\rho^3\sin{\phi}\,d\rho \,d\phi=8\pi/5.$$
Now I did the $x_0$ of the center:
$$x_0=\frac{1}{M} \int_0^{2\pi} \cos(\theta) \, d\theta \int_0^{\pi/2} \int_0^{2\cos\phi} \rho^4\sin^2 \phi \, d\rho \, d\phi=0.$$
Here, I suspect the rest for $y_0$ and $z_0$ is obvious. Indeed, I think that I would get $y_0=0, z_0=1$ after evaluating the corresponding integrals because the center of the sphere is $(0,0,1)$. What do you think?
I think that it is more natural to assume that the sphere is $x^2+y^2+z^2=1$ and that $\delta(x,y,z)=\sqrt{x^2+y^2+(z+1)^2}$. Then\begin{align}M&=\int_0^\pi\int_0^{2\pi}\int_0^1r^2\sin(\varphi)\sqrt{r^2+2r\cos(\varphi)+1}\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi\\&=2\pi\int_0^\pi\int_0^1r^2\sin(\varphi)\sqrt{r^2+2r\cos(\varphi)+1}\,\mathrm dr\,\mathrm d\varphi\end{align}and, yes, it is equal to $\frac{8\pi}5$.
By symmetry, the center of mass is a point of the type $(0,0,z_0)$. In fact\begin{align}z_0&=\frac1M\int_0^\pi\int_0^{2\pi}\int_0^1r^3\sin(\varphi)\cos(\varphi)\sqrt{r^2+2r\cos(\varphi)+1}\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi\\&=\frac{2\pi}M\int_0^\pi\int_0^1r^3\sin(\varphi)\cos(\varphi)\sqrt{r^2+2r\cos(\varphi)+1}\,\mathrm dr\,\mathrm d\varphi\\&=\frac17.\end{align}
Of course, this means that the center of mass of the original version of the problem is $\left(0,0,1+\frac17\right)=\left(0,0,\frac87\right)$.