Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$

Question

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?

Answer 1

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

Answer 2

The conditions $z\ge 0$, $x\ge 1$, $y\ge 0$, $x^2+z^2 \lt 2x$, and $x^2+z^2 \lt y^2$ define the intersection of two solids. Solid one is the quarter of the cylinder $(x-1)^2 + z^2 = 1$ to the right of the plane $x=1$ and above the plane $z=0$. Solid two is the portion of the cone $x^2 + z^2 = y^2$ to the right of the plane $x=1$ and above the plane $z=0$. For $y\ge 2$, the cone completely contains the cylinder, so $A$ is just the cylinder in this portion of $A$. The integral is indeed $1/6$ over this subregion $A_1$.

You can split the remainder of $A$ into two subregions $A_2$ and $A_3$. The first is the subregion where the boundary of the cone is intersecting the slice of the plane $x=1$ between $z=0$ and $z=1$. The second is the subregion where the boundary of the cone is intersecting the upper boundary of the cylinder between $x=1$ and $x=2$. The integral over $A_2$ is $$ \begin{align} \int_{A_2} z/y^2 &= \int_1^{\sqrt{2}} \frac{1}{y^2} \int_0^{\sqrt{y^2-1}} z \int_1^{\sqrt{y^2-z^2}} \,dx\,dz\,dy \\ &= \int_1^{\sqrt{2}}\frac{1}{y^2}\int_0^{\sqrt{y^2-1}} (z\sqrt{y^2-z^2} - z) \, dz\, dy \\ &= \int_1^{\sqrt{2}} \frac{1}{y^2} \left.\left(-(y^2-z^2)^{3/2}/3 - z^2/2\right) \right|_{z=0}^{z=\sqrt{y^2-1}} \,dy \\ &= \int_1^{\sqrt{2}} \frac{1}{y^2}\left(\frac{1}{6} - \frac{y^2}{2} + \frac{y^3}{3}\right) \,dy \\ &= \left. -\frac{1}{6y} - \frac{y}{2} + \frac{y^2}{6}\right|_1^{\sqrt{2}} \\ &= \frac{5}{6} - \frac{7}{12}\sqrt{2}. \end{align} $$

The integral over $A_3$ is $$ \int_{A_3} z/y^2 = \int_{\sqrt{2}}^{2}\frac{1}{y^2}\int_0^1 z \int_1^{\min(\sqrt{1-z^2}+1, \sqrt{y^2-z^2})} \,dx\,dz\,dy. \\ $$ The cylinder boundary intersects the cone boundary at $z = y\sqrt{1-y^2/4}$. This means that $\sqrt{y^2-z^2} \le \sqrt{1-z^2}+1$ for $0\le z \le y\sqrt{1-y^2/4}$ and $\sqrt{1-z^2}+1 \le \sqrt{y^2-z^2}$ for $y\sqrt{1-y^2/4} \le z \le 1$, so we can write $$ \int_0^1 z \left({\min(\sqrt{1-z^2}+1, \sqrt{y^2-z^2})} - 1\right) \, dz $$ as the sum $$ \int_0^{y\sqrt{1-y^2/4}} \left(z\sqrt{y^2-z^2} - z\right) \, dz + \int_{y\sqrt{1-y^2/4}}^1 z\sqrt{1-z^2} \, dz. $$ This sum of two integrals simplifies to $$ -\frac{y^4}{8} + \frac{y^3}{3} - \frac{1}{3}. $$ So $$ \begin{align} \int_{A_3} z/y^2 &= \int_{\sqrt{2}}^2 -\frac{y^2}{8} + \frac{y}{3} - \frac{1}{3y^2} \, dy \\ &= \frac{1}{6} - \frac{\sqrt{2}}{12}. \end{align} $$

So the integral over the entire region $A$ is $$ \left(\frac{1}{6}\right) + \left(\frac{5}{6} - \frac{7}{12}\sqrt{2}\right) + \left(\frac{1}{6} - \frac{1}{12}\sqrt{2}\right) = \frac{7}{6} - \frac{2}{3}\sqrt{2}. $$