Answer
$ \pi $
Work Step by Step
Consider $Average= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} \dfrac{2}{(1+x^2+y^2)} \ dy \ dx$
We need to convert rectangular coordinates into polar coordinates.
Therefore, $Average= \int_{0}^{2 \pi} \int_{0}^{1} \dfrac{2r}{(1+r^2)} \ dr \ d \theta$
or, $= \int_{0}^{2 \pi} [- \dfrac{1}{(1+r^2)}]_{0}^{1} \ d \theta$
or, $=\dfrac{1}{2} \int_{0}^{2 \pi} \ d\theta$
or, $=\dfrac{1}{2} [2\pi -0]$
or, $= \pi $
