Dipstick problem (volume of a cylinder)

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SUMMARY

The discussion revolves around calculating the height of fluid in a cylinder using a dipstick, specifically focusing on a circle with a radius of 5 bisected by the y-axis. The objective is to determine how far to move the circle down the y-axis to achieve specific percentages of the circle's area below the x-axis, such as 10% and 20%. A proposed solution involves creating a triangle connecting the center of the circle to the intersection points with the x-axis, allowing for the calculation of the area beneath the x-axis by subtracting the triangle's area from the wedge formed by the circle. The user ultimately finds the answer but seeks further assistance with the trigonometric calculations involved.

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dipique
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There is a problem I'm doing that I understand except for one part. There is a cylinder lying flat, and I want to measure the percentage of fluid in the cylinder by lowering a dipstick inside. In order to do this, I need to find out the height of the fluid when the cylinder is 10% full, 20% full, etc. Here is the question for the part of the problem I can't do:

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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I just got an idea, but I don't think I know enough to make it work. One solution (I think) is to create a triangle connecting the center of the circle with the two points where the x-axis intersects the circle. That angle forms both a wedge of the circle and a triangle; subtract the triangle from the wedge, and there is the area beneath the x-axis. Since we knew the radius of the circle, finding the percentage would be simple. Unfortunately, I'm not good enough with trig to push this through. Help?
 
All right, I got the answer. If anybody finds this and can't figure it out, feel free to shoot me an e-mail.

Dan
 

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