Volume from iterated integrals and two regions The prompt is to find the volume of the solid which is described the equations and is bounded.
$$x^2+y^2+z^2=9 $$
$$x^2-3x+y^2=0 $$
The first one is a sphere with radius 3, the shadow is on the y-x plane.
For the second on I tried using completing the squares.
$$x^2-3x + y^2 =0  $$
$$x^2-3x+ 1/25 + y^2 = 1/25 $$
$$(x-1/5)^2 + y^2 = 1/25 $$
i dont know how to procede now. I also tried...
$$x^2+y^2-3x = 0$$
$$r^2-3x = 0 $$
$$ r^2 = 3x$$
$$ r^2 = 3cos\theta$$
$$ r = \sqrt{3cos\theta} $$
$$ \int_0^3\int_0^{2\pi} \int_0^{\sqrt{3cos\theta}}x^2+y^2+z^9rdrd\theta dz$$ 
Please correct me if the method to get the radius, if its wrong? Im kinda new to calculus.
 A: the first is indeed a sphere of radius $3$ centered at the origin. But the second is a cylinder of radius $3/2$ centered at $(3/2,0,0)$ You have to fix the completion of squares:
$$x^2-3x+y^2=0\;;x^2-3x+9/4+y^2=9/4\;;(x-3/2)^2+y^2=9/4$$
But you try too to use cylindrical coordinates and it seems a better way. The equation for the cylinder is ok:
$r^2=3\cos\theta\;;r=\sqrt{3\cos\theta}$ (dropping the minus sign as it has to be $r\geq0$)
The equation for the sphere is $r^2+z^2=9$. Isolating $z$ to use for the integration limits, $z=\pm\sqrt{9-r^2}$
We are calculating the volume, so, we integrate only for the volume element $rdrd\theta dz$ (in fact, you tried to integrate the value of the square of the distance to the origing all over the volume; if we had to interpret that integral, it is the mean over the points of the region of their squared distance to the origin times the volume)
For the integration limits:
$-\pi/2\lt\theta\lt\pi/2\;;0\lt r\lt\sqrt{3\cos\theta}\;;-\sqrt{9-r^2}\lt z\lt\sqrt{9-r^2}$
$$V=\int_{-\pi/2}^{\pi/2}\int_0^\sqrt{3\cos\theta}\int_{-\sqrt{9-r^2}}^\sqrt{9-r^2}rdzdrd\theta$$
A: If you want integrate in cartesian coordinates, note that:
substituting $x^2+y^2=3x$ in the first equation we find that the limiting values for $z$ are given by $3x+z^2=9 $ that gives 
$$
-\sqrt{3(3-x)}<z<\sqrt{3(3-x)} 
$$
The circle in $x-y$ plane  $x^2+y^2-3x=0$ gives: $y=\pm \sqrt{3x-x^2}$
 so the limits for $y$ are:
$$
-\sqrt{x(3-x)} <y<\sqrt{x(3-x)}
$$ 
and, since $y$ must be real we have also the limits ofr $x$
$$
0<x<3
$$
So, using the symmetry withe respect to the $x-y$ plane, the volume can be calculates as:
$$
V=2\int_0^3\int_{-\sqrt{x(3-x)}}^{\sqrt{x(3-x)}}\int _0^{\sqrt{3(3-x)}}dzdydx
$$
