Area of described parametric region

Question

The problem is as follows:

A rope is tied to a cow and attached to the side of a circular silo with radius $r$. If the rope has length $\pi r$, what is the area of the land available for grazing by the cow?

Since this comes from the chapter on parametric equations, the first thing I did was to find the coordinates $(x,y)$ of the cow in terms of $\theta$:

The result was $$x=r(1+\cos\theta+\theta\sin\theta)$$ $$y=r(\sin\theta-\theta\cos\theta)$$

Now, to find the area of the region in the first quadrant, I set up the integral $$\int_a^b x\, dy$$ $$=r^2\int_0^\pi \theta\sin\theta\,(1+\cos\theta+\theta\sin\theta)\,d\theta$$

My question is, how do I find the anti-derivative of the integrand? Or is there an easier way to solve this problem (either with parametric equations or without)?

Answer 1

The differential form $\omega=x\,dy$ isn’t the only one that’ll give you the area swept out by the radius vector when integrated. Any other form $\tau =\omega+df$ will also work. A convenient choice can be $\frac12(x\,dy-y\,dx)$, which is the area of the triangle defined by $(x,y)$ and $(x+dx,y+dy)$. This yields, after simplification, $$\frac{r^2}2(\theta^2+\theta\sin\theta)\,d\theta$$ for the integrand, which looks a lot more manageable.

Answer 2

Polar coordinates choice is convenient.

A diagram of taut string around circular silo needs to be drawn to describe required part of the involute locus. From unwound length $ r\theta$ and radius $r$ by Pythagoras thm we compute radius to center of silo:

$$ \rho = r \sqrt{1+\theta^2} $$

Area $$ = \frac12 \int_0 ^{\pi-\tan^{-1} \pi } \rho^2 d \theta - \pi r^2/2 $$

You can take it further.