Answer
$[\ln (4) -1] \pi $
Work Step by Step
Consider $I= \int_{-1}^{1} \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \ln (x^2+y^2+1) \ dx \ dy$
We need to convert rectangular coordinates into polar coordinates.
Therefore, $Average= \int_{0}^{2 \pi} \int_{0}^{1} r \ln (r^2+1) \ dr \ d \theta$
Let $a= r^2+1$ and $da= r dr$
So, $I= \int_{0}^{2 \pi} \int_{1}^2 \dfrac{1}{2} \times \ln a \ da \ d \theta$
or, $=\dfrac{1}{2} \int_{0}^{2 \pi} [ a \ln a - a]_1^2 \ d\theta$
or, $=\dfrac{1}{2} \int_{0}^{2 \pi} (2 \ln 2 -1) )\ d\theta$
or, $=[\ln (4) -1] \pi $
