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A sealed cylindrical drum of radius r is filled with 9% of water. Now if the drum is tilted to rest on its side, show that the fraction of the curved surface area (not counting the flat sides) that will be under is water is less than 1/12 .

What I know so far:

$$w = \textrm{water_volume} = \pi r^2 \cdot 9x$$

Total curved surface area is:

$$s = 2 \pi r \cdot 100x$$

What I need to know is the relationship between $s$ and $w$.

I'm not sure where to start.

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The height and radius of the cylinder are irrelevant. Think of a unit circle in the plane. You have a chord, parallel to the y axis, that cuts your circle. The cut piece accounts for 9% of the area of the circle, and you're interested in the percentage of circumference between the two endpoints of the chord. –  user641 Oct 21 '12 at 8:56

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