Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_{ - \sqrt {16 - {y^2}} }^{\sqrt {16 - {y^2}} } {2xy} dxdy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^4 {\left[ {{x^2}y} \right]_{ - \sqrt {16 - {y^2}} }^{\sqrt {16 - {y^2}} }dy} \cr
& {\text{Evaluating the limits}} \cr
& = \int_0^4 {\left[ {{{\left( {\sqrt {16 - {y^2}} } \right)}^2}y - {{\left( { - \sqrt {16 - {y^2}} } \right)}^2}y} \right]dy} \cr
& = \int_0^4 {\left[ {\left( {16 - {y^2}} \right)y - \left( {16 - {y^2}} \right)y} \right]dy} \cr
& = \int_0^4 {0dy} \cr
& = 0 \cr} $$
