Surface integral of hyperboloid using polar coordinates fails? I am trying to find the surface area of the hyperboloid $x^2 + y^2 − z^2 = 1$ where $0\le z \le 1 $. My book goes ahead making hyperbolic substitutions, however I don't understand why the simple approach fails.
$$\mathbf n= \langle 2x, 2y, -2z \rangle$$
$$=\langle x, y, -z \rangle$$
$$=\langle x, y, -\sqrt{x^2+y^2-1} \rangle$$
$$||\mathbf n||=\sqrt{2x^2+2y^2-1}$$
$$=\sqrt{2r^2-1}$$
Now, since $z=\sqrt{x^2+y^2-1}$ and $0\le z \le 1 $, so if $x^2+y^2=r^2$, then $1 \le r \le \sqrt{2}$. So I continue
$$\int_0^{2\pi}\int_1^\sqrt{2}r\sqrt{2r^2-1}\ dr \ d\theta$$
$$\approx 4.3942$$
My book comes out with $7.9665$, what did I do wrong?
 A: When doing a surface integral, as is mentioned above in the comments, it is necessary to choose the "correct" normal vector. You start by choosing a parameterization of your surface, of the form $$r(u,v) = (x(u,v), \, y(u,v), \, z(u,v)) $$
In other words, express the $x,y,$ and $z$ coordinates as functions of two parameters. Next, you compute the tangent vectors to the surface given by the partial derivative vectors:
$$ r_u = (x_u,y_u,z_u) $$
$$ r_v = (x_v,y_v,z_v) $$
The "correct" normal vector to use for the surface integral is the cross product $r_u \times r_v$. The area differential is then the magnitude $|r_u \times r_v|$.
One particular common situation is when the surface is a graph (i.e. it is given by $z=f(x,y)$). In this case, you can choose your parameterization to simply be $u=x$ and $v=y$, with $z=f(x,y)$. So it looks like
$$ r(x,y) = (x,y,f(x,y)) $$
If you take the cross product, you get
$$ r_x \times r_y = (-f_x,-f_y,1) $$
Therefore the area differential is $\sqrt{1+f_x^2+f_y^2}$.
