Answer
$\pi$
Work Step by Step
$I_{avg}=\int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} \dfrac{2}{(1+x^2+y^2)} \ dy \ dx ...(1)$
We will write in the above equation (1) polar coordinates.
$$I_{avg}= \int_{0}^{2 \pi} \int_{0}^{1} \dfrac{2r}{(1+r^2)} \ dr \ d \theta\\= \int_{0}^{2 \pi} [- \dfrac{1}{(1+r^2)}]_{0}^{1} \ d \theta \\= \int_{0}^{2 \pi} \dfrac{1}{2}\ d\theta \\=\dfrac{1}{2} [2\pi -0] \\= \pi $$
