Evaluate $\iiint_{V} (x^2+y^2+z^2)\,dx\,dy\,dz$ in the common part of $2az > x^2+y^2$ and $x^2+y^2+z^2 < 3a^2$ 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?

You are correct, and the person who said the greatest lower limit is $z=0$ is wrong:

