<s>[INST] 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?
 [/INST] You are correct, and the person who said the greatest lower limit is $z=0$ is wrong:

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