Answer
$$\frac{{20}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{ - 1}^1 {\left( {4 - {x^2} - {y^2}} \right)} dxdy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^1 {\left[ {4x - \frac{1}{3}{x^3} - x{y^2}} \right]_{ - 1}^1dy} \cr
& {\text{Evaluate the limits of integration}} \cr
& = \int_0^1 {\left( {\left[ {4\left( 1 \right) - \frac{1}{3}{{\left( 1 \right)}^3} - \left( 1 \right){y^2}} \right] - \left[ {4\left( { - 1} \right) - \frac{1}{3}{{\left( { - 1} \right)}^3} - \left( { - 1} \right){y^2}} \right]} \right)dy} \cr
& = \int_0^1 {\left( {\frac{{11}}{3} - {y^2} + \frac{{11}}{3} - {y^2}} \right)dy} \cr
& = \int_0^1 {\left( {\frac{{22}}{3} - 2{y^2}} \right)dy} \cr
& {\text{Integrate}} \cr
& = \int_0^1 {\left( {\frac{{22}}{3} - 2{y^2}} \right)dy} \cr
& = \left[ {\frac{{22}}{3}y - \frac{2}{3}{y^3}} \right]_0^1 \cr
& {\text{Evaluate the limits}} \cr
& = \left[ {\frac{{22}}{3}\left( 1 \right) - \frac{2}{3}{{\left( 1 \right)}^3}} \right] - 0 \cr
& = \frac{{20}}{3} \cr} $$
