Answer
$$\frac{8}{3}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} + {y^2} \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 2,{\text{ }}0 \leqslant y \leqslant 2} \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \left( {2 - 0} \right)\left( {2 - 0} \right) \cr
& A = 4 \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{4}\int_0^2 {\int_0^2 {\left( {{x^2} + {y^2}} \right)} dydx} \cr
& = \frac{1}{4}\int_0^2 {\left[ {\int_0^2 {\left( {{x^2} + {y^2}} \right)} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{1}{4}\int_0^2 {\left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_0^2dx} \cr
& = \frac{1}{4}\int_0^2 {\left( {2{x^2} + \frac{8}{3}} \right)dx} \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\left[ {\frac{{2{x^3}}}{3} + \frac{8}{3}x} \right]_0^2 \cr
& = \frac{1}{4}\left[ {\frac{{2{{\left( 2 \right)}^3}}}{3} + \frac{8}{3}\left( 2 \right)} \right] - \frac{1}{4}\left[ {\frac{{2{{\left( 0 \right)}^3}}}{3} + \frac{8}{3}\left( 0 \right)} \right] \cr
& = \frac{8}{3} \cr} $$
