Answer
$$\frac{{1792}}{{27}}$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_{\frac{1}{2}y}^{\sqrt y } {{x^2}} {y^2}} dxdy \cr
& \int_0^4 {\int_{\frac{1}{2}y}^{\sqrt y } {{x^2}} {y^2}} dxdy = \int_0^4 {\left[ {\int_{\frac{1}{2}y}^{\sqrt y } {{x^2}} {y^2}dx} \right]} dy \cr
& {\text{Integrating with respect to }}x \cr
& \int_{\frac{1}{2}y}^{\sqrt y } {{x^2}} {y^2}dx = \left[ {\frac{{{x^3}{y^2}}}{3}} \right]_{\frac{1}{2}y}^{\sqrt y } \cr
& = \left[ {\frac{{{{\left( {{y^{1/2}}} \right)}^3}{y^2}}}{3} - \frac{{{{\left( {\frac{1}{2}y} \right)}^3}{y^2}}}{3}} \right] \cr
& = \frac{{{y^{7/2}}}}{3} + \frac{{{y^5}}}{{24}} \cr
& \int_0^4 {\left[ {\int_{\frac{1}{2}y}^{\sqrt y } {{x^2}} {y^2}dx} \right]} dy = \int_0^4 {\left( {\frac{{{y^{7/2}}}}{3} + \frac{{{y^5}}}{{24}}} \right)} dy \cr
& = \left[ {\frac{{{y^{9/2}}}}{{3\left( {9/2} \right)}} + \frac{{{y^6}}}{{144}}} \right]_0^4 \cr
& = \frac{{2{{\left( 4 \right)}^{9/2}}}}{{27}} + \frac{{{{\left( 4 \right)}^6}}}{{144}} - 0 \cr
& = \frac{{1024}}{{27}} + \frac{{4096}}{{144}} \cr
& = \frac{{1792}}{{27}} \cr} $$
