Since the radius of the lighthouse is 1 the length of the straight part of the chain is the same as x. Through a little trigonomtry we can get the coordinates of the dog given x as (cos(x)+x*sin(x),sin(x)-x*cos(x)). The distance from (0,0) to the dog, given x, conveniently works out to (1+x2)1/2, using the length of a line segment formula.
Lets call y the angle fromed by (1,0), (0,0), and the location of the dog. Through some more trigonomty (again left up to you) we can solve for y in terms of x as y=x-tan-1. Now we are ready to integrate.
The area in terms of polar coordinates is 1/2 * the integral over the range of the angle of the radius squared. We can not simply take the integral from 0 to Π of 1+x2 because this is not the radius formed by the angle x, but the radius formed by y. We could take the integral over the range of the angle y but finding the coordinates of the dog in terms of y is very hard (I couldn't do it). However we can take the ingregal as x goes from 0 to Π of 1+x2 multiplied by the change in y given a change a change in x. Since y=x-tan-1(x). dy/dx = 1-1/(1+x2)!
So now the area becomes the integral as x goes from 0 to Π of 1/2 the product of 1+x2 and 1-1/(1+x2) which equals 1/2 the integral from 0 to Π of x2 which equals Π3/6.
This integral, however, does not cover the area bounded by the triangle (0,0), (-1,0) and (-1,Π). So we must add this area which is Π/2. Yet we must subtract the area inside the lighthouse which the dog may not go in, which is also Π/2. So these two modifications conventiently cancel each other out.
The area of the quarter circle to the left of x=-1 and above y=0 is Π3/4, since the length of the chain is Π and the dog covers a semicircle to the left of the lighthouse. So the total area above y=0 is 5/12 * Π3. Finally double this for the area under y=0 and the total area is 5/6 * Π3
Here is another solution I received from Patrick.
I would like to thank Steven Lutz for sending me this problem.
Michael Shackleford, A.S.A.