$ \text{ Consider a strip parallel to x- axis} \\ $

$ I = \int_{y=0}^{2} \int_{x=\frac{y}{4}}^{3-y} (x^2 + y^2) dx dy \\ $

$ = \int_{y=0}^{2} \left [\frac{x^3} {3} + y^2 x) \right]_\frac{y}{4}^{3-y} dy \\ $

$ = \int_{y=0}^{2} \left[\frac{(3-y)^3} {3} + y^2(3-y) - \frac{ (\frac{y}{4})^3 } {3} - y^2 \frac{y}{4} \right] dy \\ $

$ = \int_{y=0}^{2} \left[\frac{1} {3} (27 - 27y + 9y^2 - y^3 ) + 3y^2 - y^3 - \frac{y^3 } {3 ( 64)} - \frac{y^3}{4} \right] dy \\ $

$ = \int_{y=0}^{2} \left[9 - 9y + 3y^2 - \frac{1}{3}y^3 + 3y^2 - y^3 - \frac{y^3 } {3 ( 64)} - \frac{y^3}{4} \right] dy \\ $

$ = \left[9y - \frac{9y^2}{2} + \frac{3y^3}{3} - \frac{1}{3}\frac{y^4}{4} + \frac{3y^3}{3} - \frac{y^4}{4} - \frac{y^4 } {(3)(4)( 64)} - \frac{y^4}{16} \right]_0^2 \\ $

$ = \frac {463}{48} \\ $