Answer
$$\frac{2}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\int_0^{\sqrt {1 - {x^2} - {y^2}} } {2xzdzdydx} } } \cr
& {\text{Integrate with respecto to }}z \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\left[ {x{z^2}} \right]_0^{\sqrt {1 - {x^2} - {y^2}} }} dy} dx \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {x{{\left( {\sqrt {1 - {x^2} - {y^2}} } \right)}^2}} dy} dx \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {x\left( {1 - {x^2} - {y^2}} \right)} dy} dx \cr
& = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {\left( {x - {x^3} - x{y^2}} \right)} dy} dx \cr
& {\text{Integrate with respecto to }}y \cr
& = \int_0^1 {\left[ {xy - {x^3}y - \frac{1}{3}x{y^3}} \right]_0^{\sqrt {1 - {x^2}} }} dx \cr
& = \int_0^1 {\left[ {x\sqrt {1 - {x^2}} - {x^3}\sqrt {1 - {x^2}} - \frac{1}{3}x{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \right]} dx \cr
& = - \frac{1}{2}\int_0^1 {\left( { - 2x} \right)\sqrt {1 - {x^2}} } dx - \int_0^1 {{x^3}\sqrt {1 - {x^2}} dx} + \frac{1}{6}\int_0^1 {\left( { - 2x} \right){{\left( {1 - {x^2}} \right)}^{3/2}}dx} \cr
& {\text{Integrate}} \cr
& = \left[ { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{3/2}}} \right]_0^1 + \left[ {\frac{1}{3}{{\left( {1 - {x^2}} \right)}^{3/2}} - \frac{1}{5}{{\left( {1 - {x^2}} \right)}^{5/2}} + \frac{1}{{15}}{{\left( {1 - {x^2}} \right)}^{5/2}}} \right]_0^1 \cr
& = \left[ { - \frac{1}{5}{{\left( {1 - {x^2}} \right)}^{5/2}} + \frac{1}{{15}}{{\left( {1 - {x^2}} \right)}^{5/2}}} \right]_0^1 \cr
& {\text{Evaluate}} \cr
& = - \frac{1}{5}{\left( {1 - {1^2}} \right)^{5/2}} + \frac{1}{{15}}{\left( {1 - {1^2}} \right)^{5/2}} + \frac{1}{5}{\left( {1 - {0^2}} \right)^{5/2}} - \frac{1}{{15}}{\left( {1 - {0^2}} \right)^{5/2}} \cr
& = \frac{1}{5} - \frac{1}{{15}} \cr
& = \frac{2}{{15}} \cr} $$
