The original problem was to find the surface area of hyperbolic paraboloid $$2z = x^2 - y^2$$
that lies inside the cylinder $x^2 + y^2 \leq 1$.
My attempt was to parametrize $(x, y, z) = (\cos u, \sin u, v)$ and get $2v = \cos2u$ from the first equation. Then I found $|[r_u, r_v]| = 1$. After this I made assumption (don't know whether it was correct or not) that area of the whole figure is $4$ times area of the part of paraboloid that lies in $y > 0, z > 0$. Now we can calculate $$\mu S = 4\iint_S dudv = 4\int_{-\pi/4}^{\pi/4}du\int_0^{\frac{1}{2}\cos2u} dv = 2$$
However I've found similar problem in my textbook (but the problem there was to find area for $z > 0$) and the answer for that was $\frac{\pi}{3}(2\sqrt2 - 1)$ which is far from what I've got.
My question is: how to get the correct answer with the parametrization that I made? I've read that the correct parametrization should be simple polar coordinates after setting $|[r_u, r_v]| = \sqrt{1 + z_x^2 + z_y^2}$, but is there a method to compute the area with the parametrization that was made above?
For the surface area of $z>0$, it can be integrated via
$$A=\int_{x^2+y^2\le1} \sqrt{1+z_x^2 + z_y^2} \>dxdy = 4\int_0^{\frac\pi4}\int_0^1 \sqrt{1+r^2}rdr d\theta= \frac\pi3(2\sqrt2-1) $$