<s>[INST] 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.
 [/INST] 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:
$$
V=2\int_0^3\int_{-\sqrt{x(3-x)}}^{\sqrt{x(3-x)}}\int _0^{\sqrt{3(3-x)}}dzdydx
$$
</s>