MHB Calculating the Area Between Two Curves: A Shorter Method?

[Saitama]

Problem:

Calculate the area of region defined by the inequalities:

$$-1<xy<1$$
$$-1<x^2-y^2<1$$

Attempt:

Although I have solved the problem but I am not very satisfied with the method I used. The graph of region is symmetrical in all the four quadrants so I calculated the area in the first quadrant and multiplied by 4. I found the points of intersection and got the following integrals:

Wolfram Alpha

But I feel that given the shape of region, there exists an easier method. Has someone got any idea about a shorter method?

Here's an image depicting the region:

Any help is appreciated. Thanks!

 

[MarkFL]

I would use polar coordinates and divide the region into octants.

The curve:

$$x^2-y^2=1$$

becomes:

$$r^2=\sec(2\theta)$$

and the curve:

$$xy=1$$

becomes:

$$r^2=2\csc(2\theta)$$

we find they intersect at:

$$\theta=\frac{1}{2}\tan^{-1}(2)$$

and so we may express the area as:

$$A=8\left(\frac{1}{2}\int_0^{\frac{1}{2}\tan^{-1}(2)} \sec(2\theta)\,d\theta+\int_{\frac{1}{2}\tan^{-1}(2)}^{\frac{\pi}{4}} \csc(2\theta)\,d\theta \right)$$

You should then be able to show that:

$$A=2\sinh^{-1}(2)-4\ln\left(\frac{2}{1+\sqrt{5}} \right)\approx4.81211825059603$$

When you multiply the result you give from W|A by 4, you find they are equivalent.

 

[Saitama]

I would use polar coordinates and divide the region into octants.

The curve:

$$x^2-y^2=1$$

becomes:

$$r^2=\sec(2\theta)$$

and the curve:

$$xy=1$$

becomes:

$$r^2=2\csc(\theta)$$

we find they intersect at:

$$\theta=\frac{1}{2}\tan^{-1}(2)$$

and so we may express the area as:

$$A=8\left(\frac{1}{2}\int_0^{\frac{1}{2}\tan^{-1}(2)} \sec(2\theta)\,d\theta+\int_{\frac{1}{2}\tan^{-1}(2)}^{\frac{\pi}{4}} \csc(2\theta)\,d\theta \right)$$

You should then be able to show that:

$$A=2\sinh^{-1}(2)-4\ln\left(\frac{1}{1+\sqrt{5}} \right)\approx4.81211825059603$$

When you multiply the result you give from W|A by 4, you find they are equivalent.

That's beautiful MarkFL! Thanks a bunch. :) (Clapping) (Clapping)

You seem to have made a very small error in the end. I get $4\ln\left(\frac{2}{1+\sqrt{5}}\right) $ instead of $4\ln\left(\frac{1}{1+\sqrt{5}}\right)$, rest everything is great. Thanks once again.

 

[MarkFL]

That's beautiful MarkFL! Thanks a bunch. :) (Clapping) (Clapping)

You seem to have made a very small error in the end. I get $4\ln\left(\frac{2}{1+\sqrt{5}}\right) $ instead of $4\ln\left(\frac{1}{1+\sqrt{5}}\right)$, rest everything is great. Thanks once again.

Yes, that was a typo, when I entered the expression into the calculator, I used the correct number. I will fix my post. :D

I also had a typo when I stated the polar equivalent of $xy=1$, but I used the correct form in the integrand. :D I have now fixed that as well.