Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 981: 22

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$$\eqalign{ & \int_0^1 {\int_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } {2{x^2}y} dy} dx \cr & {\text{Integrate with respect to }}y \cr & = \int_0^1 {\left[ {{x^2}{y^2}} \right]_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} }dx} \cr & {\text{Evaluating}} \cr & = \int_0^1 {\left[ {{x^2}{{\left( {\sqrt {1 - {x^2}} } \right)}^2} - {x^2}{{\left( { - \sqrt {1 - {x^2}} } \right)}^2}} \right]dx} \cr & {\text{Integrate with respect to }}x \cr & = \int_0^1 {\left[ {{x^2}\left( {1 - {x^2}} \right) - {x^2}\left( {1 - {x^2}} \right)} \right]dx} \cr & = \int_0^1 {\left( {{x^2} - {x^4} - {x^2} + {x^4}} \right)dx} \cr & = \int_0^1 {0dx} \cr & = 0 \cr} $$