What people seem to be missing is that the original field is circular, and that now the goat is being staked out at the edge. So he doesn't get to eat within a circular region, but within the intersection of his rope-circle with the original field-circle.I'm afraid I don't have the drawing tools to explain this very well, but I can do the calculation for you:a) Goat's rope is tied at the leftmost end of the circular field. If he restricts himself to the region between radius r and radius (r + dr), he can eat within a partial annulus with radial depth dr, and with angular extension +/- theta, where theta is the angle at which he runs out of field. Through some simple geometry, which I cannot illustrate for you:r/2 = Rsin(A/2)A = 2*Arcsin(r/(2R))2*theta + A = pitheta = (pi - A)/2 , sotheta = pi/2 - Arcsin(r/(2R))(Note that when r = 0, theta = pi/2; and when r = 2R, theta = 0. That's right!)So the area the goat can eat in is, for rope length L:integral [r*(2 theta) dr] {from r = 0 to r = L}= integral [pi*r dr] - integral [Arcsin(r/(2R)*rdr]= pi*r^2/2 - integral [r * Arcsin(r/(2R) dr]= pi*L^2/2 - integral [r * Arcsin(r/(2R) dr]To do the integral:Let z = r/2R , so thenr = 2RzLet y = Arcsin(r/(2R)) = Arcsin(z) , so thensin(y) = z and thenr = 2R sin(y) anddr = 2R cos(y) dySo: integral [r * Arcsin(r/(2R) dr] = integral [2Rsin(y)*y*2Rcos(y)dy]= 4R^2 integral [y sin(y)cos(y)dy] {y = 0 to Arcsin(L/(2R))}= 2R^2 integral [y * 2 sin(y) cos(y) dy]= R^2 integral [y * sin(2y) d(2y)]= R^2 integral [u * sin(u) du]/2 {u = 0 to u = 2*Arcsin(L/(2R))}= (R^2/2) integral [u sin(u) du]But we can integrate this by parts:integral [u sin(u) du] = integral [- u*d(cos(u))] = -u*cos(u) + integral [cos(u) du] = - u*cos(u) + sin(u)At u = 0, this is 0At u = 2*Arcsin(L/(2R))cos(u) = cos(2*Arcsin(L/(2R)) = (cos(Arcsin(L/(2R))^2 - sin(Arcsin(L/(2R))^2 = 1 - 2*(sin(Arcsin(L/2(R))^2 = 1 - 2* (L/(2R))^2sin(u) = 1 - (cos(u))^2 = 1 - (1 - 2*(L/(2R))^2)^2) = 1 - (1 -4(L/(2R))^2 + 4(L/(2R))^4) = 4(L/(2R))^2 - 4 (L/(2R))^4 = 4(L/(2R))^2(1 - (L/(2R))^2)So this integral is:- 2*Arcsin(L/(2R) )(1 - 2((L/(2R))^2) + 4(L/(2R))^2 (1 - (L/(2R))^2)This is pretty awful:Multiply by R^2/2:- R^2(Arcsin(L/(2R)) (1 - 2 (L/(2R))^2) + 2R^2(L/(2R))^2(1-(L/(2R))^2)Subtract from pi*L^2/2

i*L^2/2 + R^2 (Arcsin(L/(2R))(1 - 2(L/(2R))^2) - 2R^2 (L/(2R)^2(1 - (L/(2R))^2)Sanity check: If L = 2R, this gives

i(2R)^2/2 + R^2(Arcsin(1))(1-1) - 2R^2(1)(1-1)= 2piR^2Nope, the answer should be pi*R^2, the original area of the field. I can't do long calculations using this keyboard, and I can't draw pictures. So I've made a mistake somewhere.I'll describe the calculation in words, and maybe you can do it on your own:a) On a very tight rope, the goat can get a little bit of the far-western bit of the original circular lawn. The angle he can traverse is from south to north: +/- 90degrees.b) On a looser rope, he can get farther in towards the center. But his angle is restricted, because he runs out of lawn on north and south.c) When his rope has length sqrt(2) R, he can just reach the northmost and southmost bits of the lawn, so his angle of range is +/- 45 degrees.d) When his rope gets even longer, he can cover more and more of the lawn, but his angle of range starts to decrease, because he's running out of lawn.e) Finally, when his rope has length = 2R, he's eaten the entire lawn, and his range is just the easternmost tip of the lawn. So, the solution to the problem is when the total integrated range = pi*R^2/2.