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* Round Silo
Then
I asked: What if the goat is tethered to a round silo with a radius of 10 feet?
It is not quite so straightforward, is
it? No longer do you have definite lengths of rope at
succeeding
corners. Now you will find that after a half circle (yellow area) the rope will
begin to shorten continually as it wraps around the silo. This curve
is the involute of a circle. The area under this curve is
obtained by adding up an infinite number of infinitesmally small
sectors of a circle, with the radius decreasing a bit for each one,
as shown in the diagram. There are a couple of other ways to look at
it but all my sources, after doing the calculus to show how it is
done, bring it down to one equation: Ai=L^3/6R. This is for
just the one side involute. To get the whole area we double this and
add the free semicircle. The resultant
But what if the rope or tether is longer than piR? This makes it even more complicated. Just like the square barns, if the rope is longer than 1/2 the perimeter there will be an overlap. Now the nice neat equation will not work. We will have to actually perform the integration. But we will have to stop the integration at the point of intersection to avoid the overlap, and do other calculations before we can obtain the total area the goat can graze. The problem then is to find the point of intersection mathematically. It was a matter of trial and error so I turned to BASIC, which I have not used for many years.
As I have previously demonstrated the rope can describe a 1/2 circle
freely , but as it begins to wrap around the silo it shortens. At
each of several steps we can calculate the area of a portion of a
circle by using the remaining rope at that point as the radius. The
formula for a sector is: sector area = (pi * remaining^2)*step/360.
Once I found the rotation angle, I plugged that into the regular calculus formula and came up with the same answer as the professional calculator Wolfram's Alpha, for another proof that my system works. My BASIC program and comparison of results can be seen below 

LET starttime=time 
Results for default rope
at radius*pi compared to Wolfram Alpha My program radius of silo = 10 length of rope = 31.415927 rotation angle is 179.6697 Angle A = .33030236 Angle A + rotation = 180. involute to 179.6697 = 516.7713 Wolfram Alpha double this = 1033.5426 Add semicircle + 1550.3139 Total Area = 2583.8565 My special program with breakout radius of silo = 10 length of rope = 50 rotation angle is 107.777 Angle A = 72.223001 Angle A + rotation = 180. involute to 107.777 = 1577.662 add final triangle area + 155.94683 subtract sector area  63.026458 equals final involute = 1670.5824 double this = 3341.1647 Add semicircle + 3926.9908 Total Area = 7268.1555 When I submitted this program to Dr. Peterson, of Drexel University's Dr. Math program, he replied with the following equation to find the stopping point: thetatan(theta)=L/R. This returns many values, but only one will fit since it must be between pi/2 and pi radians. Dr. Peterson suggested that my BASIC routine is unnecessary, but some sort of trial and error method must still be used with this equation. The conclusion is that if the tether is longer than 1/2 the circumference of the circle the problem cannot be solved by just one equation. 
References
Scholarly
http://www.math.harvard.edu/~knill/hatsumon/goat.pdf General
http://euclidsmuse.com/members/polymnia/apps/app/133/
http://jwilson.coe.uga.edu/EMAT6680Fa09/Thrasher/FINAL/Grazing.html 
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