Answer
$$\frac{{256}}{9}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^4 {\int_{{y^2}}^4 {\sqrt x dzdxdy} } } \cr
& {\text{Integrate with respect to }}z \cr
& = \int_0^2 {\int_0^4 {\left[ {z\sqrt x } \right]_{{y^2}}^4dxdy} } \cr
& = \int_0^2 {\int_0^4 {\left[ {4\sqrt x - {y^2}\sqrt x } \right]dxdy} } \cr
& = \int_0^2 {\int_0^4 {\left( {4 - {y^2}} \right){x^{1/2}}dxdy} } \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^2 {\left( {4 - {y^2}} \right)\left[ {\frac{2}{3}{x^{3/2}}} \right]_0^4dy} \cr
& = \frac{2}{3}\int_0^2 {\left( {4 - {y^2}} \right)\left( 8 \right)dy} \cr
& = \frac{{16}}{3}\int_0^2 {\left( {4 - {y^2}} \right)dy} \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{{16}}{3}\left[ {4y - \frac{1}{3}{y^3}} \right]_0^2 \cr
& = \frac{{16}}{3}\left[ {4\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] \cr
& = \frac{{16}}{3}\left( {\frac{{16}}{3}} \right) \cr
& = \frac{{256}}{9} \cr} $$
