Answer
$\dfrac{a^2}{2}+h^2$
Work Step by Step
Our aim is to integrate the integral and find the average using polar coordinates as follows:
$ Average=\dfrac{1}{\pi a^2 }\int^{2\pi}_0 \int^a_0 [(rcos\theta -h)^2+r^2 \sin^2 \theta] \space r \space dr \space d\theta =\dfrac{1}{\pi r^2}\int^{2\pi}_0 \int^a_0 (r^3-2r^2h \cos\theta+rh^2) \space dr \space d\theta $
or, $=\dfrac{1}{\pi a^2} \times \int^{2\pi}_0 (\dfrac{a^4}{4}-\dfrac{2a^3h \cos\theta}{3}+\dfrac{a^2 h^2}{2}) \space d\theta $
or, $=\dfrac{1}{\pi} \times \int^{2\pi}_0 (\dfrac{a^2}{4}-\dfrac{2ah \cos\theta}{3}+\dfrac{h^2}{2}) \space d\theta $
or, $=\dfrac{1}{\pi} \times [\dfrac{a^2\theta}{4}-\dfrac{2ah \sin\theta}{4}+\dfrac{h^2\theta}{2}]^{2\pi}_0$
So, $ Average =\dfrac{1}{2}(a^2+2h^2)=\dfrac{a^2}{2}+h^2$
