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**vincisonfire** The definition of the center of mass is $\displaystyle c_x = \frac{\int\int\int x\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz} $

$\displaystyle c_y = \frac{\int\int\int y\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

$\displaystyle c_z = \frac{\int\int\int z\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

In your case $\displaystyle f(x,y,z) = 1 $ and $\displaystyle \rho(x,y,z) = x^2+y^2+z^2 $

Of course, because of symmetry $\displaystyle c_x=c_y=0 $

EG : $\displaystyle c_z = \frac{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} z(x^2+y^2+z^2) dxdydz}{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} (x^2+y^2+z^2) dxdydz} \simeq 0.749$