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?
Area = 42.131292... sq. units
I solved the problem two ways.
First, by simulation:
The slant height of the cone is sqrt(h^2 + r^2) = sqrt(8^2 + 6^2) = 10.
The cone can be unrolled as a circle of radius 10 that is missing a sector.
Theta of this incomplete circle runs from 0 to (r/h) 360 or 0 to = 216 deg.
If we pepper the plane of the circle with random dots and see how many lie within
the circle vs. circle and radius of the cylinder (in its intersection with the cone),
we can get the area of the illuminated patch. The program is here
. An example run is here pictured here
, with the lighter gray points lying within the patch. the unrolled cylinder showing the intersection (to scale) looks like this
. (Being without desmos,
I cut the shapes out of paper to make sure they fit :-)
simulation runs (10^8 points):
lord@rabbit 12475 % cone2
Intersection area: 42.109028
Intersection area: 42.274119
Intersection area: 42.181443
Intersection area: 42.150056
The average is Area = 42.15 +/- 0.15
Second, doing it analytically proved difficult in cartesian coordinates: I broke the region of intersection into two parts: region A and region B+C as show in Figure 1
in the math given here
The answers agree. Area = 42.131292... sq units
Edited on November 13, 2021, 2:59 pm