Volume cut sphere radius $= 4$ plane $z=2$ I have to calculate the upper volume of a sphere:
$$x^{2}+y^{2}+z^{2}=16$$
cut by the plane $z = 2$
This is my approach: 
$$\int_{2}^{4}\int_{-\sqrt{16-z^{2}}}^{\sqrt{16-z^{2}}}\int_{-\sqrt{16-z^{2-y^{2}}}}^{\sqrt{16-z^{2}-y^{2}}}dxdydz= \int_{2}^{4}\int_{-\sqrt{16-z^{2}}}^{\sqrt{16-z^{2}}}2\sqrt{16-z^{2}-y^{2}}dydz$$
$$=\int_{2}^{4}\pi (16-z^{2})dz = \pi \left [ 16z-\frac{z^{3}}{3} \right ]_{2}^{4}=\pi \left [ \left ( (16)(4)-\frac{64}{3} \right ) - \left ( (16)(2)-\frac{8}{3} \right )\right ]$$
$$=\pi \left [ 64-\frac{64}{3}-32+\frac{8}{3} \right ]=\pi \left [ 32-18.666 \right ] = 41.887$$
is my result correct?

It's easier in cylinder coordinates:
$$\int_2^4\int_0^{\sqrt{16-z^2}}\int_0^{2\pi}r\,d\theta\,dr\,dz
  =2\pi\int_2^4\tfrac12(16-z^2)\,dz=2\pi(16-\tfrac16(4^3-2^3))=\tfrac{40}3\pi
$$
so it looks like you got it right.
