Answer
$$\frac{{26}}{{105}}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {2 - {x^2} - {y^2}} \right)}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 1,\,\,\,\,\,\,\,{x^2} \leqslant y \leqslant x \cr
& {\text{replacing the limits for the region }}R \cr
& \int_0^1 {\int_{{x^2}}^x {\left( {2 - {x^2} - {y^2}} \right)dy} dx} \cr
& = \int_0^1 {\left[ {\int_{{x^2}}^x {\left( {2 - {x^2} - {y^2}} \right)dy} } \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& \int_{{x^2}}^x {\left( {2 - {x^2} - {y^2}} \right)dy} \cr
& {\text{integrate using the power rule}} \cr
& = \left[ {2y - {x^2}y - \frac{{{y^3}}}{3}} \right]_{{x^2}}^x \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {2\left( x \right) - {x^2}\left( x \right) - \frac{{{{\left( x \right)}^3}}}{3}} \right] - \left[ {2\left( {{x^2}} \right) - {x^2}\left( {{x^2}} \right) - \frac{{{{\left( {{x^2}} \right)}^3}}}{3}} \right] \cr
& = \left( {2x - {x^3} - \frac{{{x^3}}}{3}} \right) - \left( {2{x^2} - {x^4} - \frac{{{x^6}}}{3}} \right) \cr
& = 2x - {x^3} - \frac{{{x^3}}}{3} - 2{x^2} + {x^4} + \frac{{{x^6}}}{3} \cr
& = 2x - \frac{4}{3}{x^3} - 2{x^2} + {x^4} + \frac{{{x^6}}}{3} \cr
& {\text{then}} \cr
& \cr
& \int_0^1 {\left[ {\int_{{x^2}}^x {\left( {2 - {x^2} - {y^2}} \right)dy} } \right]dx} = \int_1^1 {\left( {2x - \frac{4}{3}{x^3} - 2{x^2} + {x^4} + \frac{{{x^6}}}{3}} \right)dx} \cr
& {\text{integrating}} \cr
& = \left( {{x^2} - \frac{1}{3}{x^4} - \frac{{2{x^3}}}{3} + \frac{{{x^5}}}{5} + \frac{{{x^7}}}{{21}}} \right)_0^1 \cr
& = \left( {{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^4} - \frac{{2{{\left( 1 \right)}^3}}}{3} + \frac{{{{\left( 1 \right)}^5}}}{5} + \frac{{{{\left( 1 \right)}^7}}}{{21}}} \right) - \left( {{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^4} - \frac{{2{{\left( 0 \right)}^3}}}{3} + \frac{{{{\left( 0 \right)}^5}}}{5} + \frac{{{{\left( 0 \right)}^7}}}{{21}}} \right) \cr
& {\text{simplifying}} \cr
& = 1 - \frac{1}{3} - \frac{2}{3} + \frac{1}{5} + \frac{1}{{21}} \cr
& = \frac{{26}}{{105}} \cr} $$
