Answer
$$\frac{{64}}{3}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\int_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } {\sqrt {4 - {y^2}} } dydx} \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right| - \sqrt {4 - {y^2}} \leqslant x \leqslant \sqrt {4 - {y^2}} ,{\text{ }} - 2 \leqslant y \leqslant 2} \right\} \cr
& {\text{Therefore,}} \cr
& \int_{ - 2}^2 {\int_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } {\sqrt {4 - {y^2}} } dydx} = \int_{ - 2}^2 {\int_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } {\sqrt {4 - {y^2}} } dxdy} \cr
& = \int_{ - 2}^2 {\left[ {\int_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } {\sqrt {4 - {y^2}} } dx} \right]} dy \cr
& {\text{Integrate with respect to }}x \cr
& = \int_{ - 2}^2 {\left[ {x\sqrt {4 - {y^2}} } \right]} _{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} }dy \cr
& = \int_{ - 2}^2 {\left[ {\sqrt {4 - {y^2}} \sqrt {4 - {y^2}} - \left( { - \sqrt {4 - {y^2}} } \right)\sqrt {4 - {y^2}} } \right]} dy \cr
& = \int_{ - 2}^2 {\left[ {\left( {4 - {y^2}} \right) + \left( {4 - {y^2}} \right)} \right]} dy \cr
& = 2\int_{ - 2}^2 {\left( {4 - {y^2}} \right)} dy \cr
& {\text{By symmetry}} \cr
& = 4\int_0^2 {\left( {4 - {y^2}} \right)} dy \cr
& = 4\left[ {4y - \frac{{{y^3}}}{3}} \right]_0^2 \cr
& = 4\left[ {4\left( 2 \right) - \frac{{{{\left( 2 \right)}^3}}}{3}} \right] - 4\left[ 0 \right] \cr
& = \frac{{64}}{3} \cr} $$
