Answer
$$\frac{1}{{12}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^x {y\sqrt {{x^2} - {y^2}} } dydx} \cr
& = \int_0^1 {\left[ {\int_0^{{x^2}} {y\sqrt {{x^2} - {y^2}} } dy} \right]dx} \cr
& {\text{solve the inner integral and treat }}x{\text{ as a constant}} \cr
& = \int_0^x {y\sqrt {{x^2} - {y^2}} } dy \cr
& = - \frac{1}{2}\int_0^x {\left( { - 2y} \right){{\left( {{x^2} - {y^2}} \right)}^{1/2}}} dy \cr
& {\text{use the power rule}} \cr
& = - \frac{1}{2}\left( {\frac{{{{\left( {{x^2} - {y^2}} \right)}^{3/2}}}}{{3/2}}} \right)_0^x \cr
& = - \frac{1}{3}\left( {{{\left( {{x^2} - {y^2}} \right)}^{3/2}}} \right)_0^x \cr
& {\text{evaluating the limits in the variable }}y \cr
& = - \frac{1}{3}\left( {{{\left( {{x^2} - {x^2}} \right)}^{3/2}}} \right) + \frac{1}{3}\left( {{{\left( {{x^2} - {0^2}} \right)}^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& = - \frac{1}{3}\left( 0 \right) + \frac{1}{3}\left( {{x^3}} \right) \cr
& = \frac{1}{3}{x^3} \cr
& {\text{then}} \cr
& \int_0^1 {\left[ {\int_0^{{x^2}} {y\sqrt {{x^2} - {y^2}} } dy} \right]dx} = \int_0^1 {\frac{1}{3}{x^3}dx} \cr
& {\text{integrating}} \cr
& = \frac{1}{3}\left( {\frac{{{x^4}}}{4}} \right)_0^1 \cr
& = \frac{1}{{12}}\left( {{{\left( 1 \right)}^4} - {{\left( 0 \right)}^4}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{12}} \cr} $$
