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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

R.Y
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  • If $a$ and $c$ are the same in the two equation the two hyperbolas has no common points. – Emilio Novati Sep 24 '16 at 13:43
  • Sorry, I forgot to metion that $a$ and $c$ are constants and are not equal to each other. – R.Y Sep 24 '16 at 14:34
  • If you know what a "Jacobian" is, then I suggest a change of variables, $u=a+cx^2$, $v=a+cy^2$. – B. Goddard Sep 24 '16 at 14:53
  • I already tried this, it however did not ease the problem. – R.Y Sep 24 '16 at 14:59
  • I understand that $a\ne c$ but if $a$ in the first equation is the same $a$ in the second equation and $c$ in the first equation is the same $c$ in the second equation than the two hyperbolas have the same asymptotes $y=\pm \sqrt{c}$ so they does not intersect. – Emilio Novati Sep 24 '16 at 15:09
  • Maybe a picture is more claryfing: see the link in the problem description. – R.Y Sep 24 '16 at 15:15

1 Answers1

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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.)

B. Goddard
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