The Dipstick Problem - area of a Cyclinder

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SUMMARY

The discussion focuses on calculating the percentage of a circle's area above and below the x-axis as it is moved along the y-axis. Specifically, a circle with a radius of 5 is analyzed, with three key cases identified: when the circle is centered at the x-axis (50% above and below), when it is fully above (a >= 5), and when it is fully below (a <= -5). The integral calculations for the area percentages are provided for the cases where the circle is partially above or below the x-axis, specifically using the formulas involving the integral of the circle's equation.

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dipique
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There is a circle with a radius of 5 bisected by the y-axis. If we move it up and down the y-axis, different percentages of the circle will be above and below the x-axis. What I need to find out is how to relate the percentage below/above the x-axis (either would work) to the distance between the bottom of the circle and the x-axis.

So: we'll start with the circle sitting right on top of the x-axis. 100% of the circle area is above the x-axis. If we want to move it down so that 10% of the circle's area is below the x-axis, how many units would the circle have to be moved down?

Dan
 
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There are three special cases in this problem that I'll note first.

1) When a = 0, 50% of the area is above and 50% is below the x-axis
2) When a >= 5, 100% of the area is above the x-axis
3) When a <= -5, 100% of the area is below the x-axis

The next two cases are between these bounds of course.

When 0 < a < 5 the percentage below the x-axis is the integral from -sqrt(25 - a^2) to sqrt(25 - a^2) of -sqrt(25 - x^2)*dx divided by 25*Pi.

When -5 < a < 0 the percentage above the x-axis is the integral from -sqrt(25 - a^2) to sqrt(25 - a^2) of sqrt(25 - x^2)*dx divided by 25*Pi.
 

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