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I'm evaluating $\iiint_V f(x,y,z) dV$ where V is defined by $2x \le x^2+y^2+z^2 \le 4x $

To simplify things I swapped x and z, and moved to spherical coordinates:

$ 0 \le \theta \le 2\pi, 2 \cos \phi \le \rho \le 4 \cos \phi$

and as the last condition implies $ 4 \cos \phi \ge \rho \ge 0$, I added $0 \le \phi \le \pi/2$

The point is: I have no idea if this parametrization is correct, because I can't figure what that disequations represents. From the spherical parametrization it looks like some paraboloid but I'm confused. Could you help in this? And how do I approach such "confused" situations? thanks.

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Note that $x^2+y^2+z^2=2x$ is equivalent to $(x-1)^2+y^2+z^2=1^2$. So we have a sphere of radius $1$ with center $(1,0,0)$. Inequality $x^2+y^2+z^2\geq 2x$ describes all point outside this sphere or on its boundary. Similarly, $x^2+y^2+z^2\leq 4x$ describes all points of the closed ball of radius $2$ with center $(2,0,0)$. If look carefully at this spheres you will see that the small one is contained in the bigger one. Thus the desired volume is a ball of radius 2 with small ball of radius 1 removed. This small ball touches internally the bigger one.

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