Find the centre of gravity of geometrical solid $z=1-x^2+y^2, z=0$.
$$ \begin{split} M_{oxy} &= \int_{-1}^{1} dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy \int_0^{1-x^2-y^2}z\rho dz \\ &= \rho \int_0^{2\pi}d\phi\int_0^1 dr \int_0^{1-r^2}rzdz \\ &= \rho \int_0^{2\pi} d\phi \int_0^1 dr \frac{(1-r^2)^2}{2}r dr \\ &= \rho \pi \int_0^1 (r^4-2r^2+1)r dr \\ &= \rho \pi \left[\frac{r^6}{6}-2\frac{r^4}{4}+\frac{r^2}{2} \right]_0^1\\ &= \rho \pi/6. \end{split} $$ and $$ \begin{split} M &= \int_0^{2\pi}d\phi\int_0^1dr\int_0^{1-r^2}r\rho dz \\ &= \int_0^{2\pi}d\phi\int_0^1(1-r^2)r\rho dr \\ &= \rho 2 \pi \left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 \\ &= 2 \pi \rho \left(\frac{1}{2}-\frac{1}{4}\right) \\ &= \pi \rho \end{split} $$ Therefore, $$z_0=\frac{M_{oxy}}{M}=\frac{1}{6},x_0=y_0=0$$
Please correct my mistakes.
As it turns out, three of us (OP and two commenters, including myself) all initially overlooked an error in the calculations. The calculation of mass should conclude as follows: $$ 2 \pi \rho \left(\frac12-\frac14\right) = \frac12 \pi \rho. $$
In the question as originally posted, the right-hand side is given as $\pi\rho$ rather than $\frac12 \pi \rho.$
I'm not sure why none of us caught the mistake at first; possibly the proximity of the symbols $2$ and $\frac12$ on the left side of the equation gave the mistaken idea that the fraction to the right of $\rho$ (which actually comes out to $\frac14,$ not $\frac12$) cancels the $2$ on the left of that expression.
Other than this error, all the calculations appear to be correct. I'm even more than usually confident in this, having searched the rest of the calculations unsuccessfully for the missing factor of $2.$