 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Partially illuminated cylinder (Posted on 2021-11-03) 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) Comments: ( Back to comment list | You must be logged in to post comments.) re(3): Solution Comment 8 of 8 | (In reply to re(2): Solution by Brian Smith)

One more swing at this before I work in the sequel problem.

This time I'll use arcsin instead of arccos. So continuing from about halfway through my original solution:

This makes the angle subtended 2*arcsin(sqrt(10x-x^2)/5).  Which makes the arc length 10*arcsin(sqrt(10x-x^2)/5).

Then the area can be evaluated by an integral:
integral {x=0 to 10} 10*arcsin(sqrt(10x-x^2)/5) dx.

Substituting x/5 - 1 = cos(t) will reduce the integral to:
integral {t=0 to pi} 50*sin(t)*arcsin(sin(t)) dt

arcsin(sin(t)) simplifies to t on the interval [0,pi/2] and pi-t on the interval [pi/2, pi].
I'll note that just taking arcsin(sin(t))=t for all t will actually recreate the wrong answer I made with arccos in my first solution attempt.

Then the integral breaks into two parts:
integral {t=0 to pi/2} 50t*sin(t) dt + integral {t=pi/2 to pi} 50(pi-t)*sin(t) dt,

Each of these integrals evaluate to 50, so the final answer comes to 50+50=100.

 Posted by Brian Smith on 2021-11-13 11:10:58 Please log in:

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