# A cow is tied to a silo with radius r by a rope just long enough to reach the opposite side of the silo. Find the grazing area available for the cow?

## Edit: I found this, would that work in solving the problem?

May 22, 2018

$\frac{5}{6} {\pi}^{3} {r}^{2}$

#### Explanation:

First, we need to know the length of rope, and we know the rope can reach the opposite side of the silo.

By letting the length of the rope be $l$,

$l = \frac{1}{2} \cdot 2 \pi r$
$\textcolor{w h i t e}{l} = \pi r$

This picture depicts the area the cow can graze and ignore the $10$ and replace it with $r$.

Now to find the area we shall split it into 2 portions,

1. The semicircle on the left of the diagram

2. The circle involute on the right of the diagram

1. To find the semicircle,

Find area of the semicircle

${A}_{1} = \frac{1}{2} \cdot \pi {\left(\pi r\right)}^{2}$
$\textcolor{w h i t e}{a} = \frac{1}{2} {\pi}^{3} {r}^{2}$

2. To find the circle involute,

Find the length of wrapped rope $s$,

$s = r \theta$

Find the length of unwrapped rope $r$,

$r = l - s$
$\textcolor{w h i t e}{r} = \pi r - r \theta$
$\textcolor{w h i t e}{r} = r \left(\pi - \theta\right)$

Find the area of the circle involute,

${A}_{2} = | \frac{1}{2} {\int}_{0}^{\pi} {r}^{2} d \theta |$
$\textcolor{w h i t e}{{A}_{2}} = | \frac{1}{2} {\int}_{0}^{\pi} {r}^{2} {\left(\pi - \theta\right)}^{2} d \theta |$
$\textcolor{w h i t e}{{A}_{2}} = | {r}^{2} / 2 {\left[{\left(\pi - \theta\right)}^{3} / 3\right]}_{0}^{\pi} |$
$\textcolor{w h i t e}{{A}_{2}} = | {r}^{2} / 6 \left[{\left(\pi - \pi\right)}^{3} - {\left(\pi - 0\right)}^{3}\right] |$
$\textcolor{w h i t e}{{A}_{2}} = | - \frac{{\pi}^{3} {r}^{2}}{6} |$
$\textcolor{w h i t e}{{A}_{2}} = \frac{{\pi}^{3} {r}^{2}}{6}$

Now, we can find the total area,

$A = {A}_{1} + 2 {A}_{2}$
$\textcolor{w h i t e}{A} = \frac{1}{2} {\pi}^{3} {r}^{2} + 2 \left(\frac{{\pi}^{3} {r}^{2}}{6}\right)$
$\textcolor{w h i t e}{A} = \frac{1}{2} {\pi}^{3} {r}^{2} + \frac{1}{3} {\pi}^{3} {r}^{2}$
$\textcolor{w h i t e}{A} = \frac{5}{6} {\pi}^{3} {r}^{2}$

Therefore, the grazing area for the cow is $\frac{5}{6} {\pi}^{3} {r}^{2}$.