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?

The lower bound of $z$ is correct. It should not be zero. However the order of $x$ and $y$ should be corrected - seems a typo.
But it is easier to do this in either cylindrical coordinates or spherical coordinates.
In cylindrical coordinates,
Sphere is $r^2+z^2 = 3 a^2$
Paraboloid is $2az = r^2$
At intersection, $2az + z^2 = 3a^2 \implies z = a, -3a$. If $a \gt 0$, the intersection is above xy-plane.
So at intersection, $r^2 = 2az = 2a^2 \implies r = \sqrt2 a$ as you already found.
So integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{\sqrt2 a} \int_{r^2/2a}^{\sqrt{3a^2 - r^2}} (r^2 + z^2) \ r \ dz \ dr \ d\theta $
