The Cow and the Lighthouse – Solution

This is a detailed mathematical solution to the problem posed here.

Overview

A cow is tethered to the south side of a cylindrical lighthouse of radius $ R $ with a chain of length $ L $, where $ L \leq \pi R $. We are asked to compute the area $ A $ in which the cow can graze, assuming the ground is flat and grassy.

Solution Breakdown

The accessible area consists of two parts:

A semicircular region south of the lighthouse where the cow can swing freely, with area $ \frac{1}{2} \pi L^2 $
Two side lobes formed by the taut chain wrapping around the east and west flanks of the lighthouse. These are bounded by involutes of the circle and each have area $ \frac{L^3}{6R} $, for a total of $ \frac{L^3}{3R} $.

Final Result

The total accessible grazing area is therefore:

$$ A = \frac{\pi L^2}{2} + \frac{L^3}{3R} $$

Coordinate Transformation

To compute the involute area cleanly, we define a coordinate transformation $ (\varphi, e) $, where:

$ \varphi $: the angle from the x-axis to the tangent point on the lighthouse
$ e $: the extension length of the chain from the tangent point to the cow

The mapping from these coordinates to Cartesian coordinates is given by:

$$ \begin{aligned} x &= R \cos \varphi - e \sin \varphi \\ y &= R \sin \varphi + e \cos \varphi \end{aligned} $$

The Jacobian determinant of this transformation is $ J = e $, which simplifies the area integral considerably.

Area of One Involute Region

The area of a single lobe can be computed as:

$$ \displaystyle \int_0^{\frac{L}{R}} \int_0^{L - R\varphi} e \, de \, d\varphi = \frac{L^3}{6R} $$

Hence the total involute area is twice that:

$$ \frac{L^3}{3R} $$

Conclusion

Combining both grazing regions gives the final formula:

$$ \boxed{A = \frac{\pi L^2}{2} + \frac{L^3}{3R}} $$

This solution demonstrates how geometry, coordinate transformations, and a well-chosen Jacobian can elegantly resolve a problem that initially appears intractable.