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?

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
$$
\int_{\Large{A}}\,\frac{z}{y^2}=a_1+a_2+a_3=\frac{7-4\sqrt{2}}{6}
$$
