We were given this question in class and I tried to compute it and it looks to be pretty crazy. Can anyone take a look and let me know if I did it correctly? I would really appreciate it.
Question: Let $O = (0, 0), P = (1, 0), Q = (\cosh\alpha , \sinh\alpha ), \alpha> 0$. Let H be that part of the hyperbola $x^2 - y^2 = 1 $ in the first quadrant. Find the area of the curvilinear triangular region OPQ, where OP is the straight line segment connecting O with P, PQ is the piece of H connecting P and Q, and QO is a straight line segment connecting P with O.
So far this is what I have for the boundaries... $0\le x \le \cosh\alpha$, and $\sqrt{x^2-1} \le y \le \dfrac{\sinh\alpha}{\cosh\alpha}x$
Is this correct?
Then I just integrate $$\int_0^{\cosh\alpha} \int_{\sqrt{x^2-1}}^{\tanh\alpha} \,dy \, dx$$
Which lead me to: $\sinh\alpha - \cosh\alpha \sinh\alpha -\frac12 \ln|\cosh\alpha+\sinh\alpha|$
Does this seem right?
Your setup is not alright, since there is no uniform expression for the $y$-boundaries when $x$ is given. On the other hand, for given $y$ we can integrate with respect to $x$ from $x=y$ to $x=\sqrt{y^2+1}$, whatever $y$. Therefore we should first integrate with respect to $x$. In this way we obtain $$A=\int_0^{\sinh\alpha}\left(\int_y^{\sqrt{y^2+1}} dx\right)\ dy=\int_0^{\sinh\alpha}\bigl(\sqrt{y^2+1}-y)\ dy\ .$$ Substituting $$y:=\sinh t,\quad dy=\cosh t\ dt\qquad(0\leq t\leq\alpha)$$ we get $$A=\int_0^\alpha (\cosh t-\sinh t)\cosh t\ dt={1\over2}\int_0^\alpha e^{-t}(e^t+e^{-t})\ dt={\alpha\over2}+{1-e^{-2\alpha}\over4}\ .$$