Goat Grazing Problem: Radial Cutoff Integral
A direct polar-coordinate derivation centered at the tether point. Field radius is \(1\).
The key idea
This is the same classical goat-grazing solution and the same numerical answer, but the coordinate system makes the geometry especially direct.
Result
\[ r \approx 1.1587284730181215 \]For a circular field of radius \(R\), the answer scales linearly:
\[ r \approx 1.1587284730181215\,R. \]Equivalent transcendental form
\[ \sin a-a\cos a=\frac{\pi}{2}, \qquad r=2\cos\left(\frac a2\right). \]The diagram samples many radial directions to draw the grazed shape. The displayed area uses the closed-form expression from the derivation below.
Derivation
Put the goat's tether point at the origin, and put the center of the circular field at \((1,0)\). Since the field has radius \(1\), its boundary is
\[ (x-1)^2+y^2=1. \]Use polar coordinates centered at the tether point:
\[ x=\rho\cos\theta, \qquad y=\rho\sin\theta. \]Substitute into the circle equation:
\[ (\rho\cos\theta-1)^2+\rho^2\sin^2\theta=1. \]Expanding gives
\[ \rho^2\cos^2\theta-2\rho\cos\theta+1+\rho^2\sin^2\theta=1. \]Since \(\cos^2\theta+\sin^2\theta=1\), this simplifies to
\[ \rho^2-2\rho\cos\theta=0. \]Therefore
\[ \rho(\rho-2\cos\theta)=0. \]The nonzero distance from the tether point to the fence is
\[ \rho=2\cos\theta. \]This is meaningful for
\[ -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}. \]If the rope length is \(r\), the radial grazing limit at angle \(\theta\) is
\[ \min(r,2\cos\theta). \]Using the polar area element \(dA=\rho\,d\rho\,d\theta\), the grazed area is
\[ A(r)= \int_{-\pi/2}^{\pi/2} \int_0^{\min(r,2\cos\theta)} \rho\,d\rho\,d\theta. \]Evaluating the inner integral gives
\[ A(r)= \frac12 \int_{-\pi/2}^{\pi/2} \min(r,2\cos\theta)^2\,d\theta. \]Now split where the rope length equals the distance to the fence:
\[ r=2\cos\theta. \]Define
\[ \alpha=\arccos\left(\frac r2\right). \]For \(|\theta|\leq\alpha\), the rope stops the goat first. For \(\alpha\leq|\theta|\leq\pi/2\), the fence stops the goat first. So
\[ A(r)= \frac12 \left[ \int_{-\alpha}^{\alpha}r^2\,d\theta + 2\int_{\alpha}^{\pi/2}4\cos^2\theta\,d\theta \right]. \]The first integral is \(r^2\alpha\), so
\[ A(r)= r^2\alpha+ 4\int_{\alpha}^{\pi/2}\cos^2\theta\,d\theta. \]Use
\[ \int\cos^2\theta\,d\theta= \frac{\theta}{2}+\frac{\sin(2\theta)}{4}. \]Then
\[ 4\int_{\alpha}^{\pi/2}\cos^2\theta\,d\theta = \pi-2\alpha-\sin(2\alpha). \]Therefore
\[ A(r)= r^2\alpha+\pi-2\alpha-\sin(2\alpha). \]Since
\[ \alpha=\arccos\left(\frac r2\right) \]and
\[ \sin(2\alpha)=\frac r2\sqrt{4-r^2}, \]the area can be written entirely in terms of \(r\):
\[ A(r)= r^2\arccos\left(\frac r2\right) +\pi -2\arccos\left(\frac r2\right) -\frac r2\sqrt{4-r^2}. \]The goat must graze exactly half the field, so \(A(r)=\pi/2\):
\[ r^2\arccos\left(\frac r2\right) +\pi -2\arccos\left(\frac r2\right) -\frac r2\sqrt{4-r^2} = \frac{\pi}{2}. \]Solving this equation numerically gives
\[ r\approx 1.1587284730181215. \]Reduction to the usual transcendental equation
Let \(a=2\alpha\). Then
\[ r=2\cos\left(\frac a2\right). \]Starting from
\[ A(r)=r^2\alpha+\pi-2\alpha-\sin(2\alpha), \]substitute \(\alpha=a/2\) and \(r=2\cos(a/2)\):
\[ A= 2a\cos^2\left(\frac a2\right) +\pi-a-\sin a. \]Since
\[ \cos^2\left(\frac a2\right)=\frac{1+\cos a}{2}, \]this becomes
\[ A=\pi+a\cos a-\sin a. \]Set \(A=\pi/2\) and rearrange:
\[ \sin a-a\cos a=\frac{\pi}{2}. \]Thus the answer can also be written as
\[ r=2\cos\left(\frac a2\right), \qquad \sin a-a\cos a=\frac{\pi}{2}. \]