Answer
$\dfrac{1}{2 \pi} $
Work Step by Step
Consider $Average=(1/\dfrac{\pi}{4}) \times \int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} xy \ dy \ dx$
or, $=\dfrac{4}{\pi} \int_{0}^{1} [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx$
or, $=\dfrac{4}{\pi} \times \int_{0}^{1} [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx$
or, $=\dfrac{2}{\pi} \int_{0}^{1}(x-x^3) \ dx$
or, $=\dfrac{2}{\pi} [\dfrac{x^2}{2}-\dfrac{x^3}{3}]_{0}^{1} \ dx$
or, $=\dfrac{1}{2 \pi} $
