I am trying to find a function of the form $y=f(x)$ such that the volume of the solid generated by the function between any two points around the $x$ axis is numerically equal to its length bewteen the same $2$ points. Using integrals, we can express this as an equation: $$\pi\int y^2dx=\int\sqrt{\left(\frac{dy}{dx}\right)^2+1}\quad dx$$ ie $$\pi y^2=\sqrt{\left(\frac{dy}{dx}\right)^2+1}$$ Rearranging, we get the differential equation $$\frac{dy}{dx}=\sqrt{\pi^2y^4-1}$$ I tried to solve this first by separating the variables: $$\int\frac{1}{\sqrt{\pi^2y^4-1}}~dy=x+c$$ So the next step is obviously to find $$I=\int\frac{1}{\sqrt{\pi^2y^4-1}}~dy$$ Now here lies the problem. I tried to go about this using $2$ methods, each using different substitutions:
1.
Let $\pi y^2=\cosh u$: $$\sinh u\frac{du}{dy}=2\pi y\implies dy=\frac{1}{2\pi y}\sinh{u}~ du$$ $$\implies I=\frac{1}{2\pi}\int\frac{1}{y}du=\frac{1}{2\sqrt{\pi^3}}\int\cosh^\frac{1}{2}u~du$$ How can I find the following? $$\int\cosh^\frac{1}{2}u~du$$ 2.
Using the substitutions $\pi y^2=p$ followed by $p=\cosh \theta$ gives us the result $$I=\frac{1}{2\sqrt{\pi}}\int\frac{1}{\cosh^\frac{1}{2}\theta}~d\theta$$ Again, how I can I find the following? $$\int\frac{1}{\cosh^\frac{1}{2}\theta}~d\theta$$
Thanks for your help.
As mentioned in comments, there is no solution with elementary functions. However, if you have Byrd and Friedman – the great big book of elliptic integrals and functions – this problem falls very easily.
The second integral $\int\cosh^{-1/2}\theta\,d\theta$ is the slightly easier one to tackle, for B&F 296.00 gives $$\int\frac1{\sqrt{\cosh\theta}}\,d\theta=\sqrt2F\left(\cos^{-1}\frac1{\sqrt{\cosh\theta}},m=\frac12\right)+K$$ Substituting $\cosh\theta=\pi y^2$ gives $$I=\frac1{\sqrt{2\pi}}F\left(\cos^{-1}\frac1{y\sqrt\pi},\frac12\right)+K$$