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?
Let x be how far up from the base we are.
Then the width of the beam is sqrt(6x-x^2)
And the radius of the cone is 6-0.75*x
Then the angle subtended is 2*arcsin(sqrt(6x-x^2)/(6-0.75*x))
So then the integral to calculate the area becomes:
integral {x=0 to 6} 2*(6-0.75*x)*arcsin(sqrt(6x-x^2)/(6-0.75*x)) dx
Rewrite sqrt(6x-x^2)/(6-0.75*x) into sqrt(1 - ((6-1.25x)/(6-0.75x))^2) for upcoming substitutions.
Note that the point x=24/5 is a special point where the behavior of the integrand changes. At that point is where the light beam is the same width as the cone. This is just like in the cylinder version of the problem where exactly halfway up the beam matched the cylinder.
First substitution to use is u = 6-0.75x. Then we get
integral {u=3/2 to 6} (8/3)*u*arcsin(sqrt(1-(5/3 - 4/u)^2) du
And the special point now occurs at u=12/5.
With the integrand cleaned up a bit then cos(A)=5/3 - 4/u becomes more apparent. This makes the integral into:
integral {A=0 to pi} (8/3) * 12/(5-3*cos(A)) * arcsin(sin(A)) * 36*sin(A)/(5-3*cos(A))^2 dA
= integral {A=0 to pi} 1152 * arcsin(sin(A)) * sin(A) / (5-3*cos(A))^3 dA
And the special point is at A=pi/2
Now eliminating arcsin will split the integral into two parts:
integral {A=0 to pi/2} 1152 * A * sin(A) / (5-3*cos(A))^3 dA
+ integral {A=pi/2 to pi} 1152 * (pi-A) * sin(A) / (5-3*cos(A))^3 dA
Use substitution A=B in the first and A=pi-B in the second to bring these definite integrals into the same range of integration. This yields:
integral {B=0 to pi/2} 1152 * B * sin(B) / (5-3*cos(B))^3 dB
+ integral {B=0 to pi/2} 1152 * B * sin(B) / (5+3*cos(B))^3 dB
These integrals differ by a single + vs - in their denominators.
At this point I headed over to wolfram for some integration and got 36/5 - 96pi/25 + 30arctan(2) = 28.351 for the first integral and 36/5 - 279pi/25 + 30arctan(2) = 5.354 for the second integral.
Then the total area is 72/5 - 15pi + 60arctan(2) = 33.705.
I really want to finish this analytically.
Added at edit:
And now I found how to finish the integral analytically
Apply integration by parts to each integral:
For integral {B=0 to pi/2} 1152 * B * sin(B) / (5-3*cos(B))^3 dB
use u=B and dv = sin(B) / (5-3*cos(B))^3 dB
Then du=dB and v=-192/(5-3cos(B))^2
then the integral becomes
eval {x=0 to pi/2}} -192B/(5-3cos(B))^2 + integral {B=0 to pi/2} 192/(5-3*cos(B))^2 dB
=-96pi + integral {B=0 to pi/2} 192/(5-3*cos(B))^2 dB
For integral {B=0 to pi/2} 1152 * B * sin(B) / (5+3*cos(B))^3 dB
use u=B and dv = sin(B) / (5-3*cos(B))^3 dB
Then du=dB and v=192/(5+3cos(B))^2
then the integral becomes
eval {x=0 to pi/2}} -192B/(5+3cos(B))^2 + integral {B=0 to pi/2} 192/(5+3*cos(B))^2 dB
=96pi - integral {B=0 to pi/2} 192/(5+3*cos(B))^2 dB
Now we can add these two back together. The constants cancel and the integrand simplifies to
integral {B=0 to pi/2} 192/(5-3*cos(B))^2 - 192/(5+3*cos(B))^2 dB
= integral {B=0 to pi/2} 11520*cos(B)/(9(sin(B)^2+16)^2
Then make the substitution (4/3)tan(C) = sin(B):
= integral {B=0 to arctan(3/4)} 11520*(4/3)*sec(C)^2/(16(tan(C)^2+16)^2 dC
= integral {B=0 to arctan(3/4)} 60*(cos(C)^2) dC
Then, finally, the antiderivative of cos(C)^2 is (C + sin(C)*cos(C))/2.
So then we can just evaluate 30*(C + sin(C)*cos(C)) at 0 to arctan(3/4). Which becomes
30*(arctan(3/4) + (3/5)*(4/5) - arctan(0) - 0*1)
= 30*arctan(3/4) + 72/5 = 33.705
This is the same as 72/5 - 15pi + 60arctan(2) by using the identity 2*arctan(2) - arctan(3/4) = pi/2.
And finally as Jer pointed out, I didnt consider the fact that the cone is slanted, our original integration assumed a vertical surface but the slant stretches it by 10:8 (slant height:vertical height).
Then the actual answer is (5/4)*(30*arctan(3/4) + 72/5) = (75/2)*arctan(3/4) + 18 = 42.131.
Edited on November 14, 2021, 3:18 pm
Some analytic progress
