From - Fri Dec  5 10:38:20 1997
From: kunkel@REMOVEnas.com (Paul Kunkel)
Newsgroups: sci.math
Subject: Re: Another problem with a goat...

In article <881073289.31648@dejanews.com>, jmccaul@iatcmail.ed.ray.com 
(Joe McCauley) says...
> A goat is tethered to the outside of a round building of radius r, on a
> tether whose length is half the circumference of the building, or pi*r.
> What is the area that can be reached by the goat?

     Let the rope be attached at the south point of the barn.  Suppose 
that the goat keeps the rope taught and walks the perimeter of the 
grazing area.  The grazing area will be the region crossed by the rope.  
On the south side there is a region where the goat can graze while 
keeping the rope straight.  It is a semicircle with radius (pi)*r.  Its 
area is (1/2)*(pi)^3*r^2.  Hold that thought.

     Now let the goat walk around the east side to the north point of the 
barn, which will take up all of the rope.  Still keeping the rope taught, 
he works his way back.  The part of the rope that is not wrapped around 
the barn will be a line segment, tangent to the barn wall.  Its length 
will be rt, the length of the arc from the north point of the barn to the 
tangent point, where t is the angle of the arc.

     When t changes by an arbitrarily small angle dt, the goat moves a 
distance of rtdt, and the rope sweeps across a triangular region with 
base rtdt and height rt.  The area of the triangle is (1/2)*r^2*t^2*dt, 
but the same thing would happen on the west side, so the combined areas 
are r^2*t^2*dt.  Integrate that from 0 to pi.  That area is 
(1/3)*(pi)^3*r^2.  Add this to the area of the semicircle.

(1/2)*(pi)^3*r^2 + (1/3)*(pi)^3*r^2 = (5/6)*(pi)^3*r^2

     Kunkel