TITLE: Grazing area for a goat around a circle.
QUESTION [5 upvotes]: I am doing this math question and i am really confused on how to approach it.
This is the question:
A retired mathematics professor has decided to raise a goat. He owns a silo and a barn. The barns front wall is tangent to the silo at the corner. The silo has a circular base with a radius of 10 feet. The professor has decided to tether the goat to a chain that is anchored at the corner of the barn, the point of tangency. He has also cut the chain so that it is long enough to wrap around the silo exactly once- that is, the length of the chain equals the circumference of the silo. The barn length is longer than the chain.
This is an image of the barn (square), the silo (circle) and the goat's grazing area.

The answer I got is -4762.48876, but area is positive, so i made it 4762.48876.
These are the steps i took to getting the answer I have:


$$x(\phi)= -R\sin\phi + \phi R\cos\phi, \qquad y(\phi) = R-R\cos\phi - \phi R\sin\phi$$

REPLY [3 votes]: I have found another, much simpler derivation of the result. Let us divide the area of grazing into infinitesimaly thin triangles, as below:

They all approximately isoceles. The triangle with the apex at point $(R\sin\phi,R-R\cos\phi)$ has the legs of the length  $L(\phi) = (2\pi-\phi)R$ (the length of unwinded chain). they are supposed to be very thin, with the angle at the apex being $\Delta\phi$. Therefore, the area of each triangle is
$$\Delta S \approx \frac12 L(\phi)^2 \sin( \Delta\phi) \approx \frac12 (2\pi-\phi)^2R^2 \Delta\phi$$
The full area is therefore equal to
$$ S = \int_0^{2\pi} \frac12 (2\pi-\phi)^2 R^2d\phi = -\frac{(2\pi-\phi)^3R^2}{6}\Big|_{\phi=0}^{2\pi} = \frac43\pi^3R^2$$
The previous method I've given is more general: it can be used for any area with a boundery with known parametrization. This answer is specific to this problem.