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# Cylinder problem

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Assuming the cylinder is perfectly round, why wouldn't the liquid represent exactly 1/4 of the total volume?
Evnision the following: Split the cylinder vertically down the middle. Split the half-cylinder diagonally. Keeping the two pieces together, slice then horizontally into thin washers (the “empty” piece’s washers have 2 flat surfaces, the “water” piece’s, only 1).

If the volume of liquid is 1/4 of the original cylinders, it must be 1/2 of the half cylinder, so the volume of the empty piece and the water piece must each be 1/4. If you arrange from smallest to largest the washers for each piece, however, and compare them, you’ll find that the empty piece’s are slightly larger, having the same height and width, but always a slightly greater length.

The empty piece’s washers also have a “bevel” the water piece’s don’t, but if you make them thin enough, this becomes insignificant.

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One thing that puzzles me. If I mentally split the cylinder vertically down the middle, imagine the height of the cylinder approaching zero, it seems intuitively that the plane of water approaches a 50-50 split of the volume of that half cylinder. For a taller cylinder it is easy for me to visualize that the volume of water is something less than 25% of the cylinder's volume but at shorter and shorter heights my intuition seems to mislead me.

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I gave up trying to remember enough calculus to derive the necessary antiderivatives :shrug: and resorted to looking them up in sanctus’s post and a wikipedia list of integrals. So, not fully knowing what I’m doing, the solution to post #8:

$\int_0^1 \arccos x - x \sqrt{1-x^2} \, dx =$

$\int_0^1 \arccos x \, dx -\int_0^1 x \sqrt{1-x^2} \, dx =$

$x \, \arccos x - \sqrt{1-x^2} |_0^1 \, - \frac{-\left(\sqrt{1-x^2}\right)^3}3 |_0^1 = \, 1 - \frac13 = \frac23$

Dividing by the volume of a unit cylinder (radius and height 1), the volume of the water is $\frac2{3 \pi} \dot= 0.21220659$ of the total volume of the cylinder.

With 3 separate calculus approaches (once sanctus fixes his), a numeric approximation, and an actual physical measurement (vaguely) giving the same result, I’d say we’ve got a pretty high confidence answer to the original question :) :thumbs_up

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