Answer
$\frac{1}{3}$
Work Step by Step
$\int ^{1}_{0} \int ^{x}_{0} \sqrt {1-x^{2}}dydx=\int ^{1}_{0}y]^{x}_{0}\sqrt {1-x^{2}}dx=\int ^{1}_{0}\sqrt {1-x^{2}}xdx=-\dfrac {1}{3}\left( 1-x^{2}\right) ^{ \frac{3}{2}}]^{1}_{0}=0-\left( \dfrac {-1}{3}\right) =\dfrac {1}{3}$
