<s>[INST] Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$ I want to calculate the following improper integral:
$$\int_A \frac{z}{y^2}\\
A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$$
First I noticed that the conditions imply $x^2<2x\rightarrow x<2;1<x^2<y^2\rightarrow y>1$, and thus $B=A\cap \{y>2\}=\{z\in(0,2),x\in(1,2), x^2+z^2<2x\}$. Thus this part of the integral is fairly easy:
$$\int_B\frac{z}{y^2}=\int_2^\infty\int_1^2\int_0^\sqrt{2x-x^2}\frac{z}{y^2}dzdxdy=\int_2^\infty\frac1{y^2}dy\int_1^2\frac{2x-x^2}{2}dx=\frac{1}{6}$$
We are now left with a "proper" integral (i.e. the region on which we are integration is finite, and the integrand is bounded):
$$\int_{A\cap\{y\in(1,2)\}}\frac{z}{y^2}$$
I tried to split the domain in two regions: $A'=\{x\in[1,2);y\in(x,\sqrt{2x});z\in [0,\sqrt{y^2-x^2})\}, A''=\{x\in[1,2);y\in[\sqrt{2x},2];z\in[0,\sqrt{2x-x^2})$
Is my approach correct (I'm not sure, since the computations that follow bring me to an uncorrect result)? Is there any other approach to computing this integral, perhaps less messy?
 [/INST] We can partition the domain $A$ into regions $A_1,A_2,A_3$ . . .

The diagram below shows the projections of $A_1,A_2,A_3$ onto the $xy$-plane.



Region $A_1$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{y^2-x^2}\\[4pt]
x\le\,&y\le\sqrt{2x}\\[4pt]
1\le\,&x\le 2\\[4pt]
\end{align*}
\right.
$$
so letting $a_1$ denote the integral for region $A_1$, we get
\begin{align*}
a_1&=\int_1^2\int_x^{\sqrt{2x}}\int_0^{\sqrt{y^2-x^2}}\!\frac{z}{y^2}\;dz\,dy\,dx\\[4pt]
&=\frac{19-13\sqrt{2}}{30}\\[4pt]
\end{align*}
Region $A_2$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{2x-x^2}\\[4pt]
1\le\,&x\le\frac{y^2}{2}\\[4pt]
\sqrt{2}\le\,&y\le 2\\[4pt]
\end{align*}
\right.
$$
so letting $a_2$ denote the integral for region $A_2$, we get
\begin{align*}
a_2&=\int_\sqrt{2}^2\int_1^{\frac{y^2}{2}}\int_0^{\sqrt{2x-x^2}}\!\frac{z}{y^2}\;dz\,dx\,dy\\[4pt]
&=\frac{11-7\sqrt{2}}{30}\\[4pt]
\end{align*}
Region $A_3$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{2x-x^2}\\[4pt]
1\le\,&x\le 2\\[4pt]
2\le\,&y < \infty\\[4pt]
\end{align*}
\right.
$$
so letting $a_3$ denote the integral for region $A_3$, we get
\begin{align*}
a_3&=\int_2^\infty \int_1^2\int_0^{\sqrt{2x-x^2}}\!\frac{z}{y^2}\;dz\,dx\,dy\\[4pt]
&=\int_2^\infty\!\frac{1}{3y^2}\;dy\\[4pt]
&=\frac{1}{6}
\end{align*}
Combining the results, we get
$$
\int_{\Large{A}}\,\frac{z}{y^2}=a_1+a_2+a_3=\frac{7-4\sqrt{2}}{6}
$$
</s>