# A definite integral problem related to sliding down frictionless QUARTER CIRCLE

SOLVE FOR $$\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$$

THEORY BELOW

Let there be a block of unit mass sliding down a circular curve of radius $$R$$ such that it started with $$0$$ initial speed from top most point as shown.

I attempted to find the time which it should take in sliding down the friction less curve, a given.

Now, from the principle of conservation of energy, the velocity $$V$$ of the block at a distance $$y$$ below the starting point should be that $$V^{2} = 2*g*y$$, where $$g$$ is gravity

Now, considering only the vertical motion, we have $$\frac{dy}{dt} = V \sin \theta = V \sin A$$ and that $$\sin A = \sqrt{1 - (y/R)^2}$$

or Time to slide = $$T = \int_0^{T} dt = \int_0^{R} \frac{1}{V \sin A} dy = \int_0^{R} \frac{1}{\sqrt{2*g*y} \sqrt{1 - (y/R)^2}} dy = \frac{R}{\sqrt(2*g)} * \int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$$

i.e. it essentially breaks down to solving $$\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy$$, which seems unintegrable!

KINDLY HELP/GUIDE!

• have you tried converting to polar coordinates to integrate? Commented Dec 23, 2021 at 17:58
• $\int_0^{R} \frac{1}{\sqrt{y(R^2 - y^2)}} dy=\,(y=Rx)\,\frac{1}{\sqrt R}\int_0^{1} \frac{1}{\sqrt{x(1 - x^2)}} dx=\,(x=\sqrt t)\,\,\frac{1}{2\sqrt R}\int_0^{1} \frac{t^{-\frac{3}{4}}}{\sqrt{(1 - t)}} dt=\frac{1}{2\sqrt R}B\Big(\frac{1}{4};\frac{1}{2}\Big)$ Commented Dec 23, 2021 at 18:59
• Thanks, I will study this! Commented Dec 24, 2021 at 3:20

Render $$y =R\cos\theta$$, then the integral becomes
$$R^{-1/2}\int_0^{\pi/2}\dfrac{d\theta}{\sqrt{\cos\theta}}$$
Next put in $$\cos\theta=1-2\sin^2(\theta/2)$$ and compare with the complete elliptic integral of the first kind described in Wikipedia.