I have some trouble calculating the area bounded by two hyperbolas(in the first quadrant) in a analytical way.
The functions are:
$y=\sqrt{a^2+cx^2}$
$x=\sqrt{a^2+cy^2}$
I've tried hyperbolic substitutions, but it did not lead to an elegant solution.
My proposition is:
$\int_0^b \int^\sqrt{a^2+cy^2}_\sqrt{\frac{cy^2-a^2}{c}}dxdy$
In which $b=\sqrt{\frac{a^2+ca^2}{1-c^2}}$
I hope someone can point me in the right direction!
Picture of the problem:
(http://i65.tinypic.com/155qbr9.jpg)
Cheers
PS. a and c are constants and not equal to each other
I think making it a polar integral does the job. The area in the first quadrant is symmetric, so it suffices to integrate from $0$ to $\pi/4$. The curve is $x^2-cy^2 = a^2$, so we have $r^2\cos^2\theta - cr^2\sin^2\theta = a^2$. The integral I get is
$${1\over 2}\int_0^{\pi/4} \frac{a^2}{\cos^2\theta -c \sin^2\theta} \; d\theta={a^2\over 2}\int_0^{\pi/4} \frac{1}{1 -(c+1)\sin^2\theta} \; d\theta.$$
Now it's time to use the $t = \tan(\theta/2)$ substitution. Multiply the final answer by 2 (or 8 if you want the area in your picture.)