Answer
$$\frac{9}{5}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\int_0^{2 - x} {4yzdzdydx} } } \cr
& {\text{Integrate with respect to }}z \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\left[ {2y{z^2}} \right]_0^{2 - x}} } dydx \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\left[ {2y{{\left( {2 - x} \right)}^2} - 2y\left( 0 \right)} \right]} } dydx \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {2y{{\left( {2 - x} \right)}^2}} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^1 {\left[ {{y^2}{{\left( {2 - x} \right)}^2}} \right]_0^{\sqrt {1 - {x^2}} }} dx \cr
& = \int_0^1 {\left[ {{{\left( {\sqrt {1 - {x^2}} } \right)}^2}{{\left( {2 - x} \right)}^2}} \right]} dx \cr
& = \int_0^1 {\left[ {\left( {1 - {x^2}} \right){{\left( {2 - x} \right)}^2}} \right]} dx \cr
& = \int_0^1 {\left( {1 - {x^2}} \right)\left( {4 - 4x + {x^2}} \right)} dx \cr
& = \int_0^1 {\left( { - {x^4} + 4{x^3} - 3{x^2} - 4x + 4} \right)} dx \cr
& {\text{Integrate}} \cr
& \left[ { - \frac{{{x^5}}}{5} + {x^4} - {x^3} - 2{x^2} + 4x} \right]_0^1 \cr
& = - \frac{{{{\left( 1 \right)}^5}}}{5} + {\left( 1 \right)^4} - {\left( 1 \right)^3} - 2{\left( 1 \right)^2} + 4\left( 1 \right) \cr
& = \frac{9}{5} \cr} $$
