<s>[INST] Compute $\displaystyle I=\iiint_V(x+y+z)^2 \, dx \, dy \, dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$ Compute $\displaystyle I=\iiint_V(x+y+z)^2 \,dx \,dy \,dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$.
What I did was I found the intersection of the two surfaces, I got that it is the circle $x^2+y^2=2a^2$ in the plane $z=a$, so I think that $D=\{(x, y)\mid x^2+y^2\le 2a^2\}$ is the projection of $V$ on the $xy$ plane. Then everything is simple because I can easily see that $0\le z\le a\sqrt{3}$, so I should have $$I=\iint_D\left(\int_0^{a\sqrt 3}(x+y+z)^2\,dz\right)\,dx\,dy$$ which is easy to compute by polar coordinates.
However, my book does the following: after finding that the intersection is the circle $x^2+y^2=2a^2$ in the plane $z=a$, it says that $0\le z\le a\sqrt 3$ and it goes on to find the section with the plane $z=\text{constant}$, finding this section to be $x^2+y^2\le 2az$ for $z\in [0,a]$ and $x^2+y^2\le 3a^2-z^2$ for $z\in [a, a\sqrt 3]$. It basically goes on to use the method that I asked about here A question about the cross section method for triple integrals.
I am not really sure if my approach is correct. I kind of understand their approach, but I don't understand why mine would be wrong.
 [/INST] As Ninad Munshi put in his comments, the way question reads, it could mean two different regions. I am considering the region as,
$x^2+y^2+z^2 \leq 3a^2$ and $2az \geq  x^2+y^2$.
First find the radius at the intersection of both surfaces.
$x^2+y^2+z^2=3a^2$ and $2az = x^2+y^2$.
$z^2+2az = 3a^2 \implies z = a$. So using cylindrical coordinates, at intersection,
$r^2 = 2az = 2a^2 \implies r = a\sqrt2$.
So integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{a\sqrt2} \int_{r^2/(2a)}^{\sqrt{3a^2-r^2}} f(x,y,z) \ r \ dz \ dr \ d\theta$
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