<s>[INST] 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?
 [/INST] 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}
</s>