Evaluate $$\iiint\_{V} x^2+y^2+z^2 \,dx\,dy\,dz$$

Where $V$ (the integration region) is the common part of the paraboloid $x^2 + y^2 \leq 2az$ and the sphere $x^2+y^2+z^2 \leq a^2$.

I first found the intercept of the paraboloid and the sphere: $x^2+y^2 = 2a^2$. In the $z$ direction, the area is bounded by the sphere on top and the paraboloid on the bottom. Hence I setup the integration:

$$4\int\_{0}^{\sqrt{2a^2-x^2}} dy \int\_{0}^{\sqrt{2}a} dx\int\_{(x^2+y^2)/2}^{\sqrt{3a^2-x^2-y^2}} (x+y+z)^2\,dz$$

However, I have been told that the $z$ direction integral should be from $0$ to $\sqrt{3a^2-x^2-y^2}$. I do not understand why this is the case if we are interested in the common part of the circle and paraboloid. Could someone please explain this?