<s>[INST] Integral on a rotational solid 
Compite the integral $$\int_V\frac{4-z^2}{(x^2+y^2)^3}\mathrm dx\mathrm dy\mathrm dz,$$ where $V$ is the solid enclosed by the paraboloids $x^2+y^2=z,x^2+y^2=2z$ and cones $x^2+y^2=(z-2)^2,x^2+y^2=4(z-2)^2.$

My attempt:
$V$ is a rotational solid with $z$ axis as the axis of the rotation. Two cones have picks at the same point above the vertex of the parabolae. I decided to use cyllindrical coordinates: $$\begin{aligned}x&=r\cos\varphi\\y&=r\sin\varphi\\z&=z\\\varphi&\in[0,2\pi].\end{aligned}$$
The Jacobian of the transformation is then $J_\psi(\varphi,r)=r$ and the function under the integral becomes $\frac{4-z^2}{r^6}.$
I thought I should first compute the integral on the solid enclosed by the parabolae and the cone $x^2+y^2=(z-2)^2$ and subtract the integral on the solid enclosed by the parabolae and the cone $x^2+y^2=4(z-z^2).$
Now, I wanted to find the radii of the circles -intersections of the parabolae with cones $x^2+y^2=(z-2)^2$ and $x^2+y^2=4(z-2)^2$ for the first and second integral respectively. Solving $$\begin{cases}x^2+y^2=z\\x^2+y^2=(z-2)^2\end{cases}\\\begin{cases}x^2+y^2=2z\\x^2+y^2=(z-2)^2\end{cases}$$ yields the bounds $r=1,r=2,r=\sqrt{2(3\pm\sqrt 5)}$ and $$\begin{cases}x^2+y^2=z\\x^2+y^2=4(z-2)^2\end{cases}\\\begin{cases}x^2+y^2=2z\\x^2+y^2=4(z-2)^2\end{cases}$$ yields the bounds $r=\sqrt{\frac{17\pm\sqrt{33}}8},\sqrt{2\frac{9\pm\sqrt{17}}4}
$
The first integral I get is
$\int_0^{2\pi}\int_1^{\sqrt{2(3-\sqrt 5)}}\int_{2-r}^r\frac{4-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi+\int_0^{2\pi}\int_{\sqrt{2(3-\sqrt 5)}}^2\int_{r^2/2}^r\frac{4-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi+\int_0^{2\pi}\int_2^{\sqrt{2(3+\sqrt 5)}}\int_{r^2/2}^{2+r}\frac{4z-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi$ and I would analogously write the second integral, but the numbers I got don't seem quite operable. Am I missing something and is there a more efficient method?
 [/INST] Let $D_-=\{(r,z): r\geq 0, z\leq 2, z\leq r^2\leq 2z, (z-2)^2\leq r^2\leq 4(z-2)^2\}$, and consider the following change of variables: $u=z/r^2$ and $v=(z-2)^2/r^2$, then
$$\left|\det\begin{pmatrix}
\frac{\partial u}{\partial r}& \frac{\partial u}{\partial z}\\
\frac{\partial v}{\partial r}& \frac{\partial v}{\partial z}
\end{pmatrix}\right|=2\frac{4-z^2}{r^5}
.$$
Then
$$\iiint_{V\cap \{z\leq 2\}}\frac{4-z^2}{(x^2+y^2)^3} dx dy dz=2\pi\iint_{D_-}\frac{4-z^2}{r^6}rdrdz=
2\pi\int_{u=1/2}^1\int_{v=1/4}^1\frac{1}{2}dudv=\frac{3\pi}{8}.$$
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