Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{ - 2\sqrt {1 - {y^2}} }^{2\sqrt {1 - {y^2}} } {2xdx} dy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^1 {\left[ {{x^2}} \right]_{ - 2\sqrt {1 - {y^2}} }^{2\sqrt {1 - {y^2}} }dy} \cr
& {\text{Evaluating the limits}} \cr
& = \int_0^1 {\left[ {{{\left( {2\sqrt {1 - {y^2}} } \right)}^2} - {{\left( { - 2\sqrt {1 - {y^2}} } \right)}^2}} \right]dy} \cr
& = \int_0^1 {\left[ {4\left( {1 - {y^2}} \right) - 4\left( {1 - {y^2}} \right)} \right]dy} \cr
& = \int_0^1 {0dy} \cr
& = 0 \cr} $$
