A cone with radius 6 and height 8 is sitting on a table. A spotlight with a cylindrical beam of radius 3 is shining directly at the cone. The beam is parallel to the surface of the table and the bottom of the beam just touches the table. The cone just barely blocks the entire light beam.
What is the illuminated area?
(In reply to
Some analytic progress by Brian Smith)
Brian,
While the integral you set up does give the value you derived (I checked with numerical integration), I don't think it is the right integral. Since your terms are in cylindrical coordinates, I think the problem is done by integrating over the infinitesimal surface area element for cylindrical coordinates:
int int r(x) sqrt((dr/dx)^2 +1) d_theta dx,
where x is the cone axis and the derivatives are partials.
r(x) = 6 - (3/4) x, dr/dx = -3/4
Evaluating the square root gives:
sqrt( (-3/4)^2 + 1) = 5/4
The rest of the integral is the same as yours, so your result should be multiplied by (5/4)
area = 1.25 * 33.7050333 = 42.1312916 sq. units
I think of the 5/4 scaling as accounting for the fact that the dx part of the area elements is "leaning in" (like the slant of the cone) at a 10/8 pitch, and so gives more area.
Edited on November 14, 2021, 3:45 am
re: Some analytic progress
