I suppose "below the sphere" means "inside" it, or below the upper half of it. The integral is
[tex]\displaystyle\iiint_Fz\,\mathrm dV[/tex]
where
[tex]F=\{(x,y,z)\mid0\le x\le2,x\le y\le2,0\le z\le\sqrt{8-x^2-y^2}\}[/tex]
Evaluating the integral in Cartesian coordinates is straightforward enough to not require changing coordinates:
[tex]\displaystyle\int_0^2\int_x^2\int_0^{\sqrt{8-x^2-y^2}}z\,\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\frac12\int_0^2\int_x^2(8-x^2-y^2)\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\frac12\int_0^2\left(16-8x-2x^2+\frac{4x^3-8}3\right)\,\mathrm dx=\boxed{\frac{16}3}[/tex]
