Volume of $\left \{(x,y,z) \in \mathbb{R}^3 \ s.t. \ |x| \leq y \leq 2,\sqrt{x^2+y^2} \geq 2, 0 \leq z \leq \frac{1}{\sqrt{x^2+y^2}} \right\}$ Evaluate the volume of $A=\left \{(x,y,z) \in \mathbb{R}^3 \ \text{ s.t. } \ |x| \leq y \leq 2,\sqrt{x^2+y^2} \geq 2, 0 \leq z \leq \frac{1}{\sqrt{x^2+y^2}} \right\}$.
I've tried to use cylindrical coordinates: $x=r \cos t$, $y=r \sin t$, $z=z$ with $r\geq0$ and $0\leq t \leq 2\pi$.
So the set $A$ is transformed in
$$B=\left \{(r,t,z) \in \mathbb{R}^3 \ \text{ s.t. } \ r|\cos t| \leq r \sin t \leq 2,r \geq 2, 0 \leq z \leq \frac{1}{r} \right\}$$
With a drawing of $A$ it is easy to see that $\pi/4 \leq t \leq 3\pi/4$ and that $2\leq r \leq 2\sqrt{2}$, but I would like to find these condition algebraically; I understand that it is way harder, but I would like to improve this part of my abilities. Can someone check my work and find eventual mistakes?
I've tried this: $0 \leq r|\cos t| \leq r \sin t \implies r \sin t \geq 0 \implies 0 \leq t \leq \pi$; since $r \sin t \geq 0$ it is $r \sin t= r|\sin t|$ and so $r |\cos t| \leq r \sin t \iff r |\cos t| \leq r|\sin t| \iff |\tan t| \geq 1$.
Since I have already the bound $0 \leq t \leq \pi$, it is $|\tan t| \geq 1 \iff \frac{\pi}{4} \leq t < \frac{\pi}{2} \vee \frac{\pi}{2}<t \leq \frac{3\pi}{4}$.
It must be $r \sin t \leq 2 \iff r \leq \frac{2}{\sin t}$; the function $f(t)=\frac{2}{\sin t}$ has maximum $2\sqrt{2}$ at $x=\frac{\pi}{4}$, hence $\rho \leq 2\sqrt{2}$.
Finally, it must be $r|\cos t| \leq 2 \iff r \leq \frac{2}{|\cos t|}$; here I don't know how to proceed, because $g(t)=\frac{2}{|\cos t|}$ is unbounded and has minimum $2\sqrt{2}$ in $\frac{\pi}{4} \leq t < \frac{\pi}{2} \vee \frac{\pi}{2} < t \leq \frac{3\pi}{4}$.
If I'm not wrong, this should not affect the integration interval on $r$ because from this I obtain that $r$ is smaller than something in $[2\sqrt{2},\infty)$ and, from the study of $r \leq \frac{2}{\sin t}$, I get $r \leq 2\sqrt{2}$ and so both these condition leads to $r \leq 2\sqrt{2}$, right?
Since $t=\frac{\pi}{2}$ is only a point, it doesn't affect the volume in the integral and so I can consider $\frac{\pi}{4} \leq t \leq \frac{3\pi}{4}$.
So I get that $B=\{(r,t,z) \in \mathbb{R}^3 \text{ s.t. } 0 \leq z \leq \frac{1}{r},2 \leq r\leq 2\sqrt{2}, \pi/4 \leq t \leq 3\pi/4 \}$; hence finally I get
$$\operatorname{Volume}(A)=\iiint_B r \,  dz \, dr \, dt = \int_{\pi/4}^{3\pi/4} \left(\int_2^{2\sqrt{2}} \left(\int_0^{1/r} r \, dz \right) \, dr \right) \, dt = \frac{\pi}{2}(2\sqrt{2}-2) = \pi(\sqrt{2}-1)$$
In particular, I'm not sure when I say that a point doesn't affect the integral and all the study on $r \leq \frac{2}{|\cos t|}$. Thanks for your help.
 A: Your bounds for $z$ and $t$ are correct but your bounds for $r$ is incorrect.
As you already mentioned, $y \leq 2 \implies r \leq \dfrac{2}{\sin (t)}$
So you cannot take upper bound of $r$ as $2\sqrt2$ as the upper bound changes based on $t$.
So integral to find volume should be,
$\displaystyle \int_{\pi/4}^{3\pi/4} \int_2^{2 \csc(t)} \int_0^{1/r} r \ dz \ dr \ dt$
A: from the equations we get:
$$r|\cos t|\le r\sin t\le 2\tag{1}$$
$$r\ge 2\tag{2}$$
$$0\le z\le\frac1r\tag{3}$$
from $(1)$ and $(2)$ we get that:
$$2\le r\le 2|\sec t|\,\text{  and  }\,r\le 2|\csc t|$$
which you could rewrite as:
$$2\le r\le 2\min\left\{|\sec t|,|\csc t|\right\}$$
as for $z$ we can keep it the same and cant be simplified as it is dependent upon $r$. for $t$:
$$|\cos t|\le \sin t$$ which leaves us with:
$$\pi/4\le t\le 3\pi/4$$
but now notice that for:
$$t\in[\pi/4,3\pi/4],\min\left\{|\sec t|,|\csc t|\right\}=\csc t$$

Finally, this leaves us with:
$$B=\left\{0\le z\le\frac1r,2\le r\le2\csc t,\frac\pi4\le t\le \frac{3\pi}{4}\right\}$$
which makes your integral:
$$\int\limits_{t=\pi/4}^{3\pi/4}\int\limits_{r=2}^{2\csc t}\int\limits_{z=0}^{1/r}r\,\mathrm dz\,\mathrm dr\,\mathrm dt$$
which it appears simplifies down nicely to:
$$2\int_{\pi/4}^{3\pi/4}\csc\theta-1\,\mathrm d\theta$$
