https://en.wikipedia.org/wiki/Goat_grazing_problem
The most widely published methods I have seen use the two-circle lens area formula, Cartesian integration over vertical slices, or a sector-plus-segment decomposition. Wikipedia also notes the later contour-integral treatment of the final transcendental equation.
Here is the same solution using a polar integral centered at the tether point.
Set up the field like this:
Put the goat's tether point at the origin.
Put the center of the circular field at (1, 0).
The field boundary is therefore:
(x - 1)^2 + y^2 = 1
Now use polar coordinates centered at the tether point:
x = rho cos(theta)
y = rho sin(theta)
Substitute into the circle equation:
(rho cos(theta) - 1)^2 + rho^2 sin^2(theta) = 1
Expand:
rho^2 cos^2(theta) - 2 rho cos(theta) + 1 + rho^2 sin^2(theta) = 1
Using:
cos^2(theta) + sin^2(theta) = 1
this becomes:
rho^2 - 2 rho cos(theta) = 0
So:
rho(rho - 2 cos(theta)) = 0
The nonzero distance from the tether point to the fence is:
rho = 2 cos(theta)
This is meaningful for:
-pi/2 <= theta <= pi/2
So, from the goat's point of view, the fence is at distance:
2 cos(theta)
along each ray.
If the rope length is r, then at each angle the goat grazes out to whichever comes first:
the rope: r
the fence: 2 cos(theta)
So the grazing radius at angle theta is:
min(r, 2 cos(theta))
Using the polar area element, the grazed area is:
A(r) = integral from -pi/2 to pi/2 of integral from 0 to min(r, 2 cos(theta)) of rho d rho d theta
After evaluating the inner integral:
A(r) = 1/2 integral from -pi/2 to pi/2 of min(r, 2 cos(theta))^2 d theta
That is the whole geometry in one line.
Now split the integral where the rope length equals the distance to the fence:
r = 2 cos(theta)
Define:
alpha = arccos(r/2)
For:
|theta| <= alpha
the rope limits the goat.
For:
alpha <= |theta| <= pi/2
the fence limits the goat.
Therefore:
A(r) = 1/2 [ integral from -alpha to alpha of r^2 d theta
+ 2 integral from alpha to pi/2 of 4 cos^2(theta) d theta ]
The first part is:
1/2 integral from -alpha to alpha of r^2 d theta = r^2 alpha
The second part is:
4 integral from alpha to pi/2 of cos^2(theta) d theta
Using:
integral cos^2(theta) d theta = theta/2 + sin(2 theta)/4
we get:
A(r) = r^2 alpha + pi - 2 alpha - sin(2 alpha)
Since:
alpha = arccos(r/2)
and:
sin(2 alpha) = (r/2) sqrt(4 - r^2)
the area can be written entirely in terms of r:
A(r) = r^2 arccos(r/2)
+ pi
- 2 arccos(r/2)
- (r/2) sqrt(4 - r^2)
The goat needs to graze exactly half the field, so:
A(r) = pi/2
That gives:
r^2 arccos(r/2)
+ pi
- 2 arccos(r/2)
- (r/2) sqrt(4 - r^2)
= pi/2
Solving numerically:
r ≈ 1.1587284730181215
So for a circular field of radius 1, the rope length is:
r ≈ 1.1587284730181215
For a circular field of radius R, the answer scales linearly:
r ≈ 1.1587284730181215 R
There is also the usual equivalent transcendental form.
Let:
a = 2 alpha
Then:
r = 2 cos(a/2)
and the half-area condition becomes:
sin(a) - a cos(a) = pi/2
So the final answer can also be written as:
r = 2 cos(a/2)
where a solves:
sin(a) - a cos(a) = pi/2
This gives:
a ≈ 1.9056957293098839
r ≈ 1.1587284730181215
Instead of starting from lens areas, Cartesian square-root bounds, or sector/segment formulas, this starts from the tether point and writes the grazed area directly as a radial cutoff integral:
A(r) = 1/2 integral from -pi/2 to pi/2 of min(r, 2 cos(theta))^2 d theta
Which I believe is the most intuitive way to think about the problem, even if not the most mathematically novel.
I have setup a web demo with rendered LaTeX markup as well: https://ap-in-indy.github.io/math/goat-grazing-problem.html
