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Partially illuminated cylinder (Posted on 2021-11-03) Difficulty: 3 of 5
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?

See The Solution Submitted by Jer    
Rating: 5.0000 (1 votes)

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Solution Solution | Comment 5 of 9 |
Let x represent how far up the height of the cylinder from the base we are at.

No matter how far up we are the radius of the cylinder is 5.

Draw a circle representing the beam, call its center O.
Draw the chord for the beam length, with endpoints AB.  
Draw radius OA and draw the radius that is the perpendicular bisector of AB.  Call the intersection C.
Then OAC is a right triangle with right angle C and has OA=5 and OC=5-x.
At this point we can calculate the chord length as sqrt(10x-x^2)

Because the cylinder and the light beam have the same radius the same diagram can be reused to find the arc length that the beam covers on the cylinder for a cross section x units up from the base.

Then the angle subtended is 2*arccos((5-x)/x).  Which makes the arc length 10*arccos((5-x)/x).

This arc length is a cross section of the illuminated area.  We can express the area as an integral:
integral {x=0 to 10} 10*arccos((5-x/5) dx

Substituting y=(5-x)/5 simplifies a bit:
integral {y=-1 to 1} 50*arccos(y) dy

The antiderivative is y*arccos(y)-sqrt(1-y^2), then:
50*[y*arccos(y)-sqrt(1-y^2)] {y=-1 to 1}
= 50*[1*arccos(1) - sqrt(1-1^2) - (-1)*arccos(-1) + sqrt(1-(-1)^2)]
= 50*[0 - 0 + pi + 0] = 50pi = 157.080

  Posted by Brian Smith on 2021-11-04 11:58:43
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