Finding the volume of a pseudosphere that has been parametrised in $\theta$ and $t$ I've got a problem calculating the volume of the top half of a pseudosphere.

The pseudosphere is parametrised by 
$$\Phi(t,\theta) = \Big ( \frac{\cos(\theta)}{\cosh(t)}, \frac{\sin(\theta)}{\cosh(t)}, t-\tanh(t)\Big)$$
  with $0\le t$ and $0\le \theta \lt 2\pi$
So the volume I'm trying to find is that between the $x-y$ axis and the surface of the inside of the pseudosphere. Now, we've been given that the formula for any region $V$ in $\mathbb R^3$ is
  $$\iiint_V r\tanh^2(t) \ \mathrm dr \ \mathrm d\theta \ \mathrm d t$$ based on the change of variables $x = r\cos(\theta) \ , \ y = r\sin(\theta) \ $and $ z = t - \tanh(t)$

Since $t\ge 0$ we have $0 \lt r \le 1$ because when $t=0$ is put into $x = \frac{\cos(\theta)}{\cosh(t)}$ we get $x=\cos(\theta)$ and similarly when $t=0$ is put into y = $\frac{\sin(\theta)}{\cosh(t)}$ we get $y = \sin(\theta)$
Then, $$x^2 + y^2 = \cos^2(t) + \sin^2(t) = 1$$ so the radius at $t=0$ is $1$, and as $t \to \infty$ we get $\cosh(t) \to \infty$ so both $x$ and $y$ approach $0$ so the radius approaches $0$.
So the integral becomes 
$$\lim_{b \to \infty}\int_{t=0}^{t=b} \int_{\theta=0}^{\theta = 2\pi} \int_{r=0}^{r=1} r\tanh^2(t) \ \mathrm dr \ \mathrm d\theta \ \mathrm d t$$
$$ = \pi \lim_{b \to \infty}\int_{t=0}^{t=b} \tanh^2(t) \ \mathrm d t$$
$$ = \pi \lim_{b \to \infty}\int_{t=0}^{t=b} 1-\text{sech}^2(t) \ \mathrm d t $$
$$ = \pi \lim_{b \to \infty} \Bigg [ t - \tanh(t)\Bigg]_{t=0}^{t=b}$$
which does not converge.
I got close by multiplying the inside of the integral by $$\lvert \vec T_t \times \vec T_\theta \rvert = \frac{\sinh(t)}{\cosh^2(t)}$$ so the integral becomes
$$\lim_{b \to \infty}\int_{t=0}^{t=b} \int_{\theta=0}^{\theta = 2\pi} \int_{r=0}^{r=1} r\tanh^2(t)\frac{\sinh(t)}{\cosh^2(t)} \ \mathrm dr \ \mathrm d\theta \ \mathrm d t$$
$$ = \lim_{b \to \infty}\int_{t=0}^{t=b} \int_{\theta=0}^{\theta = 2\pi} \int_{r=0}^{r=1} r\frac{\sinh^3(t)}{\cosh^4(t)} \ \mathrm dr \ \mathrm d\theta \ \mathrm d t$$
$$ = \pi \lim_{b \to \infty}\int_{t=0}^{t=b} \frac{\sinh(t)\sinh^2(t)}{\cosh^4(t)}  \ \mathrm d t$$
$$ = \pi \lim_{b \to \infty}\int_{t=0}^{t=b} \frac{\sinh(t)(\cosh^2(t)-1)}{\cosh^4(t)}  \ \mathrm d t$$
make a u-sub: $$u = \cosh(t) \implies \mathrm du = \sinh(t) \mathrm dt$$
with $ u(b) = \cosh(b)$ and $u(0) = 1$
so the integral becomes
$$ \pi \lim_{b \to \infty}\int_{u=1}^{u=\cosh(b)} \frac{u^2-1}{u^4}  \ \mathrm d u $$
$$ = \pi \lim_{b \to \infty} \Bigg [ \frac{-1}{u} + \frac{1}{3u^3}\Bigg]_{u=1}^{u=\cosh(b)}$$
$$ = \pi \lim_{b \to \infty} \Bigg [ -\frac{1}{\cosh(b)} + \frac{1}{3\cosh^3(b)} +1 - \frac{1}{3}\Bigg ]$$
$$ = \frac{2\pi}{3}$$
Which I believe is double the correct answer of $\frac{\pi}{3}$. Can anyone help if possible? 
Cheers heaps!
 A: I think there is a mistake in the change of variable. The new variables
$$ x=r\cos \theta, \quad y=r\sin\theta, \quad z = t-\tanh t$$
are not suitable for this question. 
Take instead the transformation
$$ x=\frac{r\cos \theta}{\cosh t}, \quad y=\frac{r\sin\theta}{\cosh t}, \quad z = t-\tanh t.
$$
The Jacobian is $r\tanh^2 t \,\mathrm{sech}^2 t$. Hence the integral becomes
$$
\begin{align*}
\int_{t=0}^{t=\infty} \int_{\theta=0}^{\theta= 2\pi}\int_{r=0}^{r=1} &r \tanh^2 t \,\mathrm{sech}^2 t\, dr\, d\theta\, dt \\
&= \left(\int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{1}r \,dr\right)\left(\int_0^\infty \tanh^2 t\,\mathrm{sech}^2 t\,dt\right)\\
&= \pi \left[\frac{\tanh^3 t}{3}\right]_0^\infty \\
&= \frac{\pi}{3},
\end{align*}
$$
which is the correct answer.
A: A pseudosphere is generated by rotating a tractrix about its asymptote. Suppose the x-axis is the asymptote, and the tractrix starts at $(0,1)$.
$\frac{dy}{dx}=\frac{-y}{\sqrt{1-y^2}}$
$y^2dx=(1-y^2)\frac{dy}{dx}dy=-y\sqrt{1-y^2}dy$
$\text{Volume}=\pi\int_{x=0}^{x=\infty}{y^2dx}=\pi\int_{x=0}^{x=\infty}{(-y\sqrt{1-y^2})dy}=\pi\int_{y=0}^{y=1}{y\sqrt{1-y^2}dy}=\frac{\pi}{3}$
