Answer
$$\frac{{26}}{5}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{{\left( {{x^2} - {y^2}} \right)}^2}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right): - 1 \leqslant x \leqslant 2,\,\,\,\,0 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^1 {\int_{ - 1}^2 {{{\left( {{x^2} - {y^2}} \right)}^2}} } dxdy \cr
& = \int_0^1 {\left[ {\int_{ - 1}^2 {{{\left( {{x^2} - {y^2}} \right)}^2}dx} } \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_{ - 1}^2 {{{\left( {{x^2} - {y^2}} \right)}^2}dx} \cr
& {\text{expanding}} \cr
& = \int_{ - 1}^2 {\left( {{x^4} - 2{x^2}{y^2} + {y^4}} \right)dx} \cr
& = \left( {\frac{{{x^5}}}{5} - \frac{{2{x^3}{y^2}}}{3} + x{y^4}} \right)_{ - 1}^2 \cr
& {\text{evaluating the limits for the variable }}x \cr
& = \left( {\frac{{{{\left( 2 \right)}^5}}}{5} - \frac{{2{{\left( 2 \right)}^3}{y^2}}}{3} + \left( 2 \right){y^4}} \right) - \left( {\frac{{{{\left( { - 1} \right)}^5}}}{5} - \frac{{2{{\left( { - 1} \right)}^3}{y^2}}}{3} + \left( { - 1} \right){y^4}} \right) \cr
& = \left( {\frac{{32}}{5} - \frac{{16{y^2}}}{3} + 2{y^4}} \right) - \left( { - \frac{1}{5} + \frac{{2{y^2}}}{3} - {y^4}} \right) \cr
& = \frac{{32}}{5} - \frac{{16{y^2}}}{3} + 2{y^4} + \frac{1}{5} - \frac{{2{y^2}}}{3} + {y^4} \cr
& = \frac{{33}}{5} - 6{y^2} + 3{y^4} \cr
& \int_0^1 {\left[ {\int_{ - 1}^2 {{{\left( {{x^2} - {y^2}} \right)}^2}dx} } \right]} dy = \int_0^1 {\left( {\frac{{33}}{5} - 6{y^2} + 3{y^4}} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {\frac{{33}}{5}y - 2{y^3} + \frac{{3{y^5}}}{5}} \right)_0^1 \cr
& {\text{evaluate}} \cr
& = \left( {\frac{{33}}{5}\left( 1 \right) - 2{{\left( 1 \right)}^3} + \frac{{3{{\left( 1 \right)}^5}}}{5}} \right) - \left( {\frac{{31}}{5}\left( 0 \right) - 2{{\left( 0 \right)}^3} + \frac{{3{{\left( 0 \right)}^5}}}{5}} \right) \cr
& = \frac{{33}}{5} - 2 + \frac{3}{5} \cr
& = \frac{{26}}{5} \cr} $$
