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.
 A: 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,
$\displaystyle \int_0^{2\pi} \int_0^{a\sqrt2} \int_{r^2/(2a)}^{\sqrt{3a^2-r^2}} f(x,y,z) \ r \ dz \ dr \ d\theta$
A: Here is a starting point to help simplify the integrand before you choose a coordinate system to integrate in. Expanding we get
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2xz+2yz$$
Now there is an ambiguity in the problem - there are two solid regions bounded by those surfaces and you will need specify which is being used. However, both regions share the same symmetry that allows us to say that the integral of the last three terms all vanish since they are odd functions. Thus we can conclude
$$\iiint_V(x+y+z)^2\:dV=\iiint_V x^2+y^2+z^2\:dV$$
This suggests spherical coordinates would be great for the problem, integrating the angular variables first.
