# Proving the area of circle and hyperbolic function is the same

I am an A-level student and due to the schools being closed down, I am reading ahead in Further Maths. I am stuck on a question regarding integrating an area due to a hyperbolic function. The question is below: I know how to find the area of the circle but I am stuck on the rectangular hyperbola. I have tried to integrate the function $$y=\sqrt{a^2+x^2}$$

by substituting in: $$\frac{x}{a}=\sinh{u}$$

However, that doesn't seem to prove it. I then tried to also tried to integrate it parametrically by using

$$x=\cosh{\theta}$$ $$y=\sinh{\theta}$$

But I still haven't been able to prove it. Could you help?

• Looks like a misprint in the second diagram: $P_2$ should be $(a\cosh\phi,a\sinh\phi)$. May 16, 2020 at 10:15
• Have a look at: nabla.hr/CL-DefiniteIntAppl2.htm. May 16, 2020 at 10:33

If you want to set up the integral, you need the area between the hyperboloid and the line. Right now you are getting the area below the hyperboloid. But notice that

$$A = \frac{a^2\cosh \phi \sinh \phi}{2} - \int_a^{a\cosh\phi}\sqrt{x^2-a^2}\:dx$$

where the first term is the area of the whole triangle with the side lengths of the point $$P_2$$. Then using $$x = a\cosh t$$ we have that

$$A = \frac{a^2\cosh \phi \sinh \phi}{2} - a^2 \int_0^\phi \sinh^2 t\:dt$$

Some hyperbolic identity magic gives us that

$$\sinh^2 t = \frac{1}{2}\cosh 2t - \frac{1}{2}$$

$$\cosh \phi \sinh\phi = \frac{1}{2}\sinh2\phi$$

simplifying the formula to

$$A = \frac{1}{2}a^2\phi$$

In the second diagram, regard the blue region as made up of a bunch of "infinitesimal" triangles with vertices $$(0,0)$$, $$(a\cosh t,a\sinh t)$$ and $$(a\cosh(t+\delta t) ,a\sinh(t+\delta t))$$ for $$t$$ between $$0$$ and $$\phi$$. The area of this triangle is $$\frac{a^2}2\left(\cosh t\sinh(t+\delta t)-\sinh t\cosh(t+\delta t)\right) =a^2\frac{\sinh(\delta t)}2=a^2\frac{\delta t}2+\text{higher order infinitesimals}.$$ Therefore the blue area is $$a^2\int_0^\phi\frac{dt}{2}$$ etc.