Answer
$$\frac{4}{{11}}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{{\left( {{x^5} - {y^5}} \right)}^2}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 1,\,\,\,\, - 1 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^1 {\int_{ - 1}^1 {{{\left( {{x^5} - {y^5}} \right)}^2}} } dydx \cr
& = \int_0^1 {\left[ {\int_{ - 1}^1 {{{\left( {{x^5} - {y^5}} \right)}^2}} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_{ - 1}^1 {{{\left( {{x^5} - {y^5}} \right)}^2}} dy \cr
& {\text{expanding}} \cr
& = \int_{ - 1}^1 {\left( {{x^{10}} - 2{x^5}{y^5} + {y^{10}}} \right)} dy \cr
& = \left( {{x^{10}}y - \frac{{{x^5}{y^6}}}{3} + \frac{{{y^{11}}}}{{11}}} \right)_{ - 1}^1 \cr
& {\text{evaluating the limits for the variable }}x \cr
& = \left( {{x^{10}}\left( 1 \right) - \frac{{{x^5}{{\left( 1 \right)}^6}}}{3} + \frac{{{{\left( 1 \right)}^{11}}}}{{11}}} \right) - \left( {{x^{10}}\left( { - 1} \right) - \frac{{{x^5}{{\left( { - 1} \right)}^6}}}{3} + \frac{{{{\left( { - 1} \right)}^{11}}}}{{11}}} \right) \cr
& = \left( {{x^{10}} - \frac{{{x^5}}}{3} + \frac{1}{{11}}} \right) - \left( { - {x^{10}} - \frac{{{x^5}}}{3} - \frac{1}{{11}}} \right) \cr
& = {x^{10}} - \frac{{{x^5}}}{3} + \frac{1}{{11}} + {x^{10}} + \frac{{{x^5}}}{3} + \frac{1}{{11}} \cr
& = 2{x^{10}} + \frac{2}{{11}} \cr
& \int_0^1 {\left[ {\int_{ - 1}^1 {{{\left( {{x^5} - {y^5}} \right)}^2}} dy} \right]} dx = \int_0^1 {\left( {2{x^{10}} + \frac{2}{{11}}} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{2{x^{11}}}}{{11}} + \frac{2}{{11}}x} \right)_0^1 \cr
& {\text{evaluate}} \cr
& = \left( {\frac{{2{{\left( 1 \right)}^{11}}}}{{11}} + \frac{2}{{11}}\left( 1 \right)} \right) - \left( {\frac{{2{{\left( 0 \right)}^{11}}}}{{11}} + \frac{2}{{11}}\left( 0 \right)} \right) \cr
& = \frac{2}{{11}} + \frac{2}{{11}} \cr
& = \frac{4}{{11}} \cr} $$
