Answer
$\frac{1}{120}$
Work Step by Step
$\int_0^{\frac{1}{2}}\int_{\sqrt 3y}^{1-y^2}xy^2dxdy$
$x$ = $\sqrt {1-y^2}$
$x^2$ = $1-y^2$
$x^2+y^2$ = $1$
$r^2$ = $1$
$r$ = $1$
$x$ = $\sqrt 3y$
$r\cosθ$ = $\sqrt 3r\sinθ$
$\tanθ$ = $\frac{1}{\sqrt 3}$
$θ$ = $\frac{\pi}{6}$
= $\int_0^{\frac{\pi}{6}}\int_0^1r\cosθ(r\sinθ)^2rdrdθ$
= $\int_0^{\frac{\pi}{6}}\int_0^1r^4\cosθ\sin^2θdrdθ$
= $\frac{1}{5}\int_0^{\frac{\pi}{6}}\cosθ\sin^2θ[r^5]_0^1dθ$
= $\frac{1}{5}\int_0^{\frac{\pi}{6}}\cosθ\sin^2θdθ$
= $\frac{1}{15}[\sin^3θ]|_0^{\frac{\pi}{6}}$
= $\frac{1}{120}$
