Chapter 6, Problem 12P

To determine

To find : The area available for grazing by the cow.

Answer to Problem 12P

The required area is $\frac{5}{6}{\pi }^3{r}^2$ .

Explanation of Solution

Given information :

A cow is tied to a silo with radius $r$ with length of its half perimeter.

  

Formula used :

  $\text{Area of involute }=\left|\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\left(\text{length of unwrapped rope}\right)}^{2}d\theta }\right|$

Calculation :

  

Required area is the shaded region $\left(A\right)$

  $A$ is divided into $3$ parts (the above figure shows) $A_1$ , $A_2$ and $A_3$

Clearly $A_2=A_3$

So, $A=A_1+2A_2$

Now, length of the rope $=\frac{1}{2}\left(2\pi r\right)=\pi r$

So, area of semicircle $A_1=\frac{1}{2}\pi {\left(\pi r\right)}^2=\frac{{\pi }^3{r}^2}{2}$

Calculation of $A_2$

  

Now, length of wrapped rope $=S$

  $S=r\theta$

And length of the unwrapped rope $=l$

  $\begin{array}{c}l=R-S\\ =\pi r-r\theta \\ =r\left(\pi -\theta \right)\end{array}$

So, area of the circle involutes,

  $\begin{array}{c}{A}_2=\left|\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}{l}^{2}d\theta }\right|\\ =\left|\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}{r}^2{\left(\pi -\theta \right)}^{2}d\theta }\right|\\ =\left|\frac{r^2}{2}{{\displaystyle \left[\frac{{\left(\pi -\theta \right)}^3}{3}\right]}}_0^{\pi }\right|\\ =\left|\frac{-r^2}{6}{\pi }^3\right|=\frac{{\pi }^3{r}^2}{6}\end{array}$

So, the required area is

  $\begin{array}{c}A=A_1+2A_2\\ =\frac{1}{2}{\pi }^3{r}^2+2\frac{{\pi }^3{r}^2}{6}\\ =\frac{5}{6}{\pi }^3{r}^2\end{array}$