Expected Geometric Mean

Average all possible values of $\sqrt{xy}$ between $x=0$ to $1$ and $y=0$ to $1$ using limits and sum notation: $\displaystyle\lim_{N\to\infty}\frac{1}{N^2}\displaystyle\sum_{a=0}^{N}\displaystyle\sum_{b=0}^{N}\sqrt{\frac{ab}{N^2}}$

Convert to double integral notation by substituting $x=\frac{a}{N},dx=\frac{1}{N},y=\frac{b}{N},dy=\frac{1}{N}$ : $=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\sqrt{xy}\ dx\ dy$ $=\displaystyle\int_{0}^{1}\left[\frac{2}{3}\sqrt{x^3y}\right]_{0}^{1}\ dy$ $=\displaystyle\int_{0}^{1}\frac{2}{3}\sqrt{y}\ dy$ $=\frac{4}{9}\left[y^3\right]_{0}^{1}$ $=\frac{4}{9}$

The answer is thus $30$ .