Considering the surface in $\mathbb{R^3}$ given by:
$x^2-y^2=1$; $x>0;-1<y<1;0\leq z \leq 1$
Calculate its area by a double integral via a parametrization of the surface.
Firsly I setted up
$\phi:[0,\frac{\pi}{4}] \cup [\frac{7\pi}{4},2\pi] \times [0,1]$; $\phi(u,v)=(sec(u),tan(u),v)$
So, I have $2\int_{0}^{1} \int_{0}^{\pi/4} ||\phi_u \times \phi_v||du dv$=$2\int_{0}^{1} \int_{0}^{\pi/4} \sqrt{(sec^2(u))^2+(-tan(u)sec(u))^2}du dv$=$2\int_{0}^{1} \int_{0}^{\pi/4} \sqrt{sec^4(u)+(sec^2(u)-1)sec^2(u)}du dv$=$2\int_{0}^{1} \int_{0}^{\pi/4} \sqrt{2sec^4(u)-sec^2(u)}du dv$; but I haven't found an useful way to find the result. Any suggestions?
Actually, the previous answer contained a mistake; the integral is not elementary as I'd initially suggested. Computing the area requires elliptic integrals, specifically the incomplete flavors of the first and second kinds.
$$\begin{align*} & 2 \int_0^{\tfrac\pi4} \sqrt{2\sec^4u - \sec^2u} \, du \\ &= 2 \int_0^{\tfrac\pi4} \sec^2u \sqrt{2 - \cos^2u} \, du \\ &= 2 \int_0^{\tfrac\pi4} \sec^2u \sqrt{1 + \sin^2u} \, du \\ &= \sqrt6 - 2 \int_0^{\tfrac\pi4} \frac{\sin^2u}{\sqrt{1+\sin^2u}} \, du & \text{(by parts)} \\ &= \boxed{\sqrt6 - 2E\left(\frac\pi4,i\right) + 2F\left(\frac\pi4,i\right)} \approx 2.199 \end{align*}$$
The last line follows from the definitions for $E,F$ upon splitting up the integrand as
$$\frac{1+\sin^2u-1}{\sqrt{1+\sin^2u}} = \sqrt{1+\sin^2u} - \frac1{\sqrt{1+\sin^2u}}$$