Find the surface area of the following hyperboloid parameterized by
$$\begin{align}x&=(\cosh{t})(\cos{\theta}) \\ y&=(\cosh{t})(\sin{\theta}) \\ z&=\sinh{t} \end{align}$$
for
$$\begin{align}0\leq \,&\theta \leq 2\pi \\ -1\leq \,&t\leq 1 \end{align}$$
My Work
Using the following formula
$$\iint_D\left |\vec{r}_u\times\vec{r}_v\right |du dv.$$
where $u=t$ and $v=\theta$
I wind up with the following cross product:
$$\begin{align} &(\sinh(t) \cos(\theta), \sinh(t) \sin(\theta), \cosh(t))\times(-\sin(\theta) \cosh(t), \cos(\theta) \cosh(t), 0)\\ =&-\cos{\theta} \cosh^2{t}, -\cosh^2{t} \sin{\theta}, \cos^2{\theta} \cosh{t} \sinh{t}+\cosh{t} \sin^2{\theta} \sinh{t} \end{align}$$
so that the surface integral becomes
$$\begin{align} &\int_{0}^{2\pi}\int_{-1}^{1}\sqrt{\left|\cos{\theta} \cosh^2{t} \right|^2+\left|\cosh^2{t} \sin{\theta}\right|^2+\left|\cosh{t} \sinh{t} \cos^2{\theta}+\cosh{t} \sin^2{\theta} \sinh{t}\right|^2} dt \, d\theta \\ =&\int_{0}^{2\pi}\int_{-1}^{1}\cosh{t} \sqrt{\cosh^2{t}+\sinh^2{t}} \,dt \, d\theta \\\approx &20.0153 \end{align}$$
Is this the correct approach?