Compute the volume of the body that is bounded by $x^2+y^2\leq 4x, \quad |z|\leq x^2+y^2.$ Compute the volume of the body that is bounded by $x^2+y^2\leq 4x, \quad  |z|\leq x^2+y^2.$
The first inequality describes a plane cutting through a paraboloid. The second inequality describes the $z$-range of the paraboloid, since there is an absolute value, it means that $$-(x^2+y^2)\leq |z|\leq x^2+y^2.$$ This in turn means that the plane cuts through the paraboloid in two equally large sections, one below the $xy$-plane and one above it. Due to symmetry, let's just instead use that $$0\leq\ z \leq x^2+y^2$$
and multiply the obtained volume integral by 2.
I want to find the $xy$-projection of the intersection surface. I have that $$4x=x^2+y^2\Leftrightarrow(x-2)^2+y^2=2^2.$$
So we have that $D=$ a circle disk with center at $(2,0)$ and radius $ 0\leq r \leq 2=(x-2)^2+y^2\leq 4.$
My volume integral then becomes 
$$V_I=2\iint_D[4x-(x^2+y^2)] \, dx \, dy=2\iint_D[(x-2)^2+y^2-4] \, dx \, dy.$$
Polar coordinates by $x=2+r\cos(\theta)$ and $y=r\sin(\theta)$ transforms the integral to
$$V_I=2\iint_D [r^3-4r] \, dr \, d\theta=4\pi\int_0^2[r^3-4r] \ dr=-16\pi.$$
Correct answer is $V_I=48\pi.$ Can anyone please spot my error?
 A: The inequality $x^2+y^2\le 4x$, which, as you've pointed out, is equivalent to $(x-2)^2+y^2\le 4$, is actually not a plane cutting through a paraboloid. There is no $z$ present in this first inequality, which shows that the shape is cylindrical (in both the broader and the narrower sense). It is a cylinder of radius $2$ centered on the $(2,0,z)$ axis. The second inequality $|z|\le x^2+y^2$, if it had been an equality, would indeed describe a paraboloid centered at the origin. Because of the absolute value, imagine double-napped paraboloids - this inequality describes the volume in-between the double-napped paraboloids. So the volume described by both inequalities is the volume in-between the double-napped paraboloids, and inside the cylinder. I think the switch to cylindrical coordinates centered at $(2,0)$ is a good idea. But your integrand is off, I think. Your integrand needs to describe the $|z|\le x^2+y^2$ inequality, whereas the first inequality simply gets you your region of integration in the $(x,y)$ plane. So, at a first pass, I would say that if $D$ is the circle you've defined, your volume integral should be this:
$$V=\iint_D \left[\left(x^2+y^2\right)-\left(-\left(x^2+y^2\right)\right)\right] \, dA =2\iint_D \left(x^2+y^2\right) \, dA.$$
Using your change of variables, we obtain
$$V=2\int_0^{2\pi}\int_0^2 \left[(2+r\cos(\theta))^2+r^2\sin^2(\theta)\right] \, r \, dr \, d\theta.$$
Can you finish?
Here are the double-napped paraboloids plotted over the region $x^2+y^2\le 4$:

The Wolfram Language command to plot this was 
Plot3D[{x^2 + y^2, -x^2 - y^2}, {x, -2, 2}, {y, -2, 2}, RegionFunction -> Function[{x, y, z}, (x)^2 + y^2 <= 4], BoxRatios -> Automatic]

And here is the same but plotted over the region $(x-2)^2+y^2\le 4$:

The Wolfram Language code to produce this image was
Plot3D[{x^2 + y^2, -x^2 - y^2}, {x, -2, 2}, {y, -2, 2}, RegionFunction -> Function[{x, y, z}, (x-2)^2 + y^2 <= 4], BoxRatios -> Automatic]

