Involute of a circle by a sheep

It can be shown that the involute of a circle can be parametrized in $\theta$ as.

$x=r(\cos \theta+\theta\sin\theta)$ $y=r(\sin \theta-\theta\cos\theta)$

The equation of the circle

$x_{c}=rcos \theta$ $y_{c}=rsin \theta$

The area from 0 to $\pi$

${ A }_{ 1 }=\int _{ \pi }^{ 0 }{ (y-{ y }_{ c } } )dx=\int _{ \pi }^{ 0 }{ (r(\sin \theta -\theta \cos \theta )-rsin\theta } )r(sin\theta +\theta cos\theta )d\theta \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ r }^{ 2 }\int _{ 0 }^{ \pi }{ \theta cos\theta sin\theta d\theta } +{ r }^{ 2 }\int _{ 0 }^{ \pi }{ { \theta }^{ 2 }{ cos }^{ 2 }\theta } d\theta \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { \pi { r }^{ 2 } }{ 4 } +\frac { \pi { r }^{ 2 } }{ 4 } +\frac { 1 }{ 6 } { \pi }^{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 6 } { \pi }^{ 3 }{r}^{2}$ ,

If we continue on the blue curve(see figure), after going an angle of $\pi$ , it will move around a semicircle of radius $r\theta$ . thus

${ A }_{ 2 }=\frac { \pi { (r\pi ) }^{ 2 } }{ 2 } =\frac { { \pi }^{ 3 }{ r }^{ 2 } }{ 2 }$

By symmetry, Area covered from $0$ to $\pi$ is equal to area covered from $\pi$ to $2\pi$ . Thus $A_{1}$ = $A_{3}$ .

${ A }_{ tot }=\quad \quad { 2A }_{ 1 }+{ A }_{ 2 }\\ \quad \quad =\frac { { \pi }^{ 3 }{ r }^{ 2 } }{ 3 } +\frac { { \pi }^{ 3 }{ r }^{ 2 } }{ 2 } \\ \quad \quad =\quad \quad \frac { 5{ \pi }^{ 3 }{ r }^{ 2 } }{ 6 }$

As wanted,

Half the area of the bigger circle is obviously accessible to the sheep.

Let's calculate the silo side. sheep

When the rope touches the Silo till the angle $\theta$ as shown, the area element swept by sheep between $\theta$ and $d\theta$ would be $\frac{1}{2}(\pi - \theta)^{2}\times r^{2} d\theta$ . Integrate this over $theta$ from $0$ to $\pi$ and multiply by twice to get $\frac{1}{3} \pi^{3} r^{2}$ . Add them to get $\frac{5}{6} \pi^{3} r^{2}$ .

Hence the answer $\boxed{11}$