A cylinder with base radius 5 and height 10 is sitting on a table. A spotlight with a cylindrical beam of radius 5 is shining directly at the cylinder. The beam is parallel to the table and the bottom of the beam just touches the table. The cylinder just barely blocks the entire light beam.

What is the illuminated area?

(In reply to

re: Solution by Jer)

Either that or integrate just the bottom half of the circle {x=0 to 5} and double the result.

Thats where I tripped up; I got too hasty to take the integral over the entire range.

2*integral {x=0 to 5} 10*arccos((5-x/5) dx

Substituting y=(5-x)/5 simplifies a bit:

integral {y=0 to 1} 100*arccos(y) dy

The antiderivative is y*arccos(y)-sqrt(1-y^2), then:

100*[y*arccos(y)-sqrt(1-y^2)] {y=-1 to 1}

= 100*[1*arccos(1) - sqrt(1-1^2) - 0*arccos(0) + sqrt(1-0^2)]

= 100*[0 - 0 - 0 + 1] = **100**