Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$ 
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$

My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I can then just apply the definition of the area of a surface
$$A(S) = \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy$$
I consider the parametrization $\Phi (x,y) = (x, y, \sqrt{x^2 + y^2 - 1}) \ $. Then $$\begin{cases} \Phi_x = (1,0,\displaystyle \frac{-x}{\sqrt{x^2 + y^2 - 1}}) \\ \Phi_y = (0,1,\displaystyle \frac{-y}{\sqrt{x^2 + y^2 - 1}})\end{cases} \Longrightarrow \Phi_x \times \Phi_y = (\frac{x}{\sqrt{x^2 + y^2 - 1}},\frac{y}{\sqrt{x^2 + y^2 - 1}},1)$$
Then
$$|| \Phi_x \times \Phi_y||= \displaystyle \sqrt{\frac{x^2}{x^2 + y^2 - 1} + \frac{y^2}{x^2 + y^2 - 1} + 1} = \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1} $$
As we work in a symettric surface, we'll consider $z \in [0, \sqrt 3]$ and simply multiply the result by two. Then, the parametrization goes from $D$ to $\mathbb R^3$, $\Phi : D \subset \mathbb R^2 \rightarrow \mathbb R^3$, being $D$ the following domain
$$D = \lbrace (x,y) \in \mathbb R^2 : 1 \leq x^2 + y^2 \leq 4 \rbrace$$
Thus we get
$$A(S) = 2 \cdot \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy = 2 \cdot \iint_D \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1}\ dx \ dy $$
Using polar coordinates, $\begin{cases} x = r \cdot \cos \theta \\ y = r \cdot \sin \theta \end{cases} : r \in [1,2] \ \& \ \theta \in [0, 2\pi]$ we get the following integral
$$A(S) = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \displaystyle \sqrt{\frac{r^2 \cos^2 \theta + r^2 \sin^2 \theta}{r^2 \cos^2 \theta + r^2 \sin^2 \theta - 1} + 1} \ dr \ d\theta =  2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr \ d\theta$$
$$ = 4 \pi \cdot \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1}  \ dr $$
The problem is that I reach the integral above that I don´t know how to tackle. I think I may have done something wrong along the process since this is a question extracted from an university exam where no computers nor calculators were avilable. Any help?
 A: Continue with
\begin{align}
A= &\ 4\pi \int_{1}^2 r \sqrt{\frac{r^2}{r^2 - 1} + 1} \ \overset{t=r^2-1}{dr }\\
=& \ 2\pi \int_0^3 \sqrt{\frac{2t+1}t}\ dt\overset{ibp}=
2\pi\bigg(\sqrt{t(2t+1)}\bigg|_0^3+\frac12\int_0^3 \frac1{\sqrt{t(2t+1}}dt\bigg)\\
=& \ 2\pi\bigg( \sqrt{21}+\frac1{\sqrt2}\sinh^{-1} \sqrt{2t}\bigg|_0^3\bigg)= 2\pi \bigg( \sqrt{21}+\frac1{\sqrt2}\sinh^{-1} \sqrt{6}\bigg)
\end{align}
A: $x^2+y^2-z^2=1$, $-\sqrt{3}\le z\le\sqrt{3}$
Convert to cylindrical coordinates. Fitting given the symmetry, it's a hyperboloid of 1 sheet.
$z^2=r^2-1\implies r=\sqrt{z^2+1}\implies dr/dz=\frac{z}{\sqrt{z^2+1}}$
Lateral surface area of a cylinder is $2\pi rz$. To generalize, we keep the $2\pi r$ and the $z$ becomes the differential arc length along the curve being rotated about the z axis. So Area=$2\pi r \sqrt{1+(dr/dz)^2}dz$
$2\int_0^\sqrt{3}2\pi \sqrt{z^2+1} \sqrt{1+(dr/dz)^2}dz=4\pi\int_0^{\sqrt{3}}\sqrt{z^2+1}\sqrt{\frac{2z^2+1}{z^2+1}}dz=\int_0^\sqrt{3}4\pi\sqrt{2z^2+1}dz$
Let $\tan{\theta}=z\sqrt{2}$.$\sqrt{2z^2+1}=\sec{\theta}$. $dz=\sec^2{\theta}/\sqrt{2}$.
$\int 2\pi\sqrt{2}\sec^3{\theta}d\theta=2\pi\sqrt{2}(\frac{1}{2}\sec{\theta}\tan{\theta}+\frac{1}{2}\ln|\sec{\theta}+\tan{\theta}|)+C$
$\pi\sqrt{2}(z\sqrt{2}\sqrt{2z^2+1}+\ln|z\sqrt{2}+\sqrt{2z^2+1}|)|_0^\sqrt{3}$
$=\pi\sqrt{2}(\sqrt{42}+\ln|\sqrt{6}+\sqrt{7}|)$
A: The surface is
$ x^2 + y^2 - z^2 = 1 $
Its standard parameterization is
$ P = (x, y, z) = ( \sec t \cos s , \sec t \sin s , \tan t ) $
So the surface area is
$ \text{A} = \displaystyle \int_{t = - \frac{\pi}{3} }^{ \frac{\pi}{3} } \int_{s = 0}^{2 \pi} \| P_t \times P_s \| \ d s \ d t $
And we have
$ P_t = (\sec t \tan t \cos s , \sec t \tan t \sin s , \sec^2 t ) $
$ P_s = (- \sec t \sin s , \sec t \cos s , 0 ) $
So that
$ P_t \times P_s = ( - sec^3 t \cos s , - \sec^3 t \sin s , \sec^2 t \tan t) $
And
$ \| P_t \times P_s \| = | \sec^2 t | \sqrt{ \sec^2 t + \tan^2 t } = \sec^2 t \sqrt{ 2 \tan^2 t + 1 } $
Therefore, the surface area is (using the substituting $u = \tan t $
$ \text{Area} = 2 \pi \displaystyle \int_{u = -\sqrt{3}}^{\sqrt{3}} \sqrt{ 2u^2 + 1} \ du  $
Using the trigonometric substitution $ \sqrt{2} u =  \tan \theta $ , then the integral becomes,
$ \displaystyle \int \dfrac{1}{\sqrt{2}} \sec^3 \theta d \theta $
From the tables,
$ \displaystyle \int \sec^3 \theta \ d \theta = \dfrac{1}{2} \bigg( \sec \theta  \tan \theta + \ln  \bigg| \sec \theta + \tan \theta \bigg| \bigg) $
Evaluating this between $\theta_1 = \tan^{-1}( -\sqrt{6} )$ and $\theta_2 = \tan^{-1}( \sqrt{6} )$ gives
$ \displaystyle \int_{u = -\sqrt{3}}^{\sqrt{3}} \sqrt{ 2u^2 + 1} \ du = \dfrac{1}{2\sqrt{2}} \bigg( 2 \sqrt{42} + \ln\bigg( \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7} - \sqrt{6} } \bigg) \bigg) $
Therefore, the area is
$ \text{Area} = \dfrac{\pi}{\sqrt{2}}\bigg( 2 \sqrt{42} + \ln\bigg( \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7} - \sqrt{6} } \bigg) \bigg) $
