Originally Posted by

**Bert** I have maked some mistakes in follow integrals but I can't find them.

Who can help?

$\displaystyle \int \int \int z \ dxdydz \ = \ \int^2_{0} \int^2_0 \int ^{\sqrt{4-y^{2}}} z \ dz dydx $

$\displaystyle =\int^2_0\int^2_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}} = \int^2_0\int^2_0\frac{4-y^2}{2}dydx= \int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^2$

$\displaystyle =\int^2_0[2y- \frac{1}{6}y^3]^2_0\ = \int^2_0\ 4 \ -\frac{8}{6} \ = \frac{8}{3}\int^2_0dx=\frac{8}{3}[x]_0^2=\frac{16}{3}$

$\displaystyle \int \int \int z \ dxdydz \ = \int_0^2\int_0^{2-x}\int_0^{\sqrt{4-y^2}} z \ dx dy dz $

$\displaystyle \int^2_0\int^{2-x}_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}}= \int^2_0\int^{2-x}_0\frac{4-y^2}{2}dydx=$ $\displaystyle \int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^{2-x}$

$\displaystyle =\int^2_0[2y- \frac{1}{6} y^3]^{2-x}_0=4$

Greets.