I'm trying to show that a pseudosphere with $K_{\Sigma} =-1$ (Gaussian curvature) has a surface area of $4\pi$.
The parameterization is $(sech u \cos v, sech u \sin v, u - \tanh u)^T$, where $u \in (\infty, \infty)$ and $v \in [0, 2\pi)$. Sorry for the italics but the tex won't recognize \sech :S
Anyways, after calculating the metric and finding its determinant, I get that $$A =\int_{0}^{2\pi} \int_{-\infty}^{\infty} \tanh(u) sech(u) dudv$$
However, this gives $0$ as the answer. After confronting Wolfram mathworld, they get that their surface area integral is
$$A =2 \int_{0}^{2\pi} \int_{0}^{\infty} \tanh(u) sech(u) dudv$$
But I do not know why. Any clarifications would be great