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+x^{2})^{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+x^{2} 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+x^{2} multiplied by the change in y given a change a change
in x. Since y=x-tan^{-1}(x). dy/dx = 1-1/(1+x^{2})!

So now the area becomes the integral as x goes from 0 to Π of 1/2 the product
of 1+x^{2} and 1-1/(1+x^{2}) which equals 1/2 the integral from
0 to Π of x^{2} which equals Π^{3}/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 Π^{3}/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 * Π^{3}. Finally double
this for the area under y=0 and the total area is 5/6 * Π^{3}

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.