By changing to polar coordinates, evaluate the integral (where a>0):
int[a,0]int[sqrt(a^2 - x^2), 0](x^2 + y^2)dydx
May aswell post here!
Draw a picture of the region of the limits.
The lower limits are x = 0 and y = 0, which tells you that both x > 0 and y > 0, which means you're dealing with the 1st quadrant of the cartesian axis system.
Sketching the upper y limit $\displaystyle y = \sqrt{a^2-x^2} $
gives: $\displaystyle y^2 + x^2 = a^2 $
Circle with centre (0,0), radius a.
So from that sketch you should see that, in polar, the $\displaystyle \theta $ limits range from $\displaystyle 0 \to \frac{\pi}{2} $.
And the radius ranges from $\displaystyle 0 \to a $. Which converts to:
$\displaystyle I = \int_0^{\frac{\pi}{2}} \int_0^{a} ((rcos(\theta))^2+(rsin(\theta))^2).|\frac{d(x,y)} {d(r,\theta)}| dr d\theta $
$\displaystyle |\frac{d(x,y)}{d(r,\theta)}| = r$
$\displaystyle I = \int_0^{\frac{\pi}{2}} \int_0^{a} r^3 dr d\theta $