Answer
$\dfrac{1}{5}$
Work Step by Step
The region of integration can be expressed as: $R=${$ (x,y) | \sqrt y \leq x \leq 2-\sqrt y, 0 \leq y \leq 1$}
Consider $I=\int_{0}^{1} \int_{\sqrt y}^{2-\sqrt y} xy \ dx \ dy$
or, $=\dfrac{1}{2} \times \int_{0}^{1} [x^2]_{\sqrt y}^{2-\sqrt y}\ dy$
or, $=\dfrac{1}{2} \times \int_{0}^{1} (2-\sqrt y)^2 y-(\sqrt y)^2 y \ dy$
or, $=\dfrac{1}{2} \times \int_{0}^{1} (4-4\sqrt y+y) y- y^2 \ dy$
or, $=\int_0^1 2y-2y^{3/2} \ dy $
or, $=[y^2-\dfrac{4y^{5/2}}{5}]_0^1$
or, $=\dfrac{1}{5}$
