Jason W.

asked • 08/05/15# A thin walled circular cylinder is filled 91 percent with water, closed up and then tipped over on a table so it would

A thin walled circular cylinder is filled 91 percent with water, closed up and then tipped over on a table so it would roll if pushed. What is the height of the water level over the tabletop divided by the cylinder diameter? I'm trying to find the area of the part of the circle base, now the location of the height, but can't seem to find the answer.

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## 1 Expert Answer

Gregg O. answered • 08/05/15

Engineering Valedictorian Available for Math Tutoring

I don't think there's an analytic solution. Let k=.91. We know the volume of the water is k*(volume of cylinder)

=k(pi*r

^{2}*h). Now, we tilt the cylinder on its side. Since the cylinder is more than half-full, we know the water line is going to be above the diameter parallel with the surface of the table (you're going to want to sketch along on a piece of paper, because it starts getting a bit complicated). Let the distance from the center of the circle to the water line be a. Then, the height of the water line above the table is r + a.Good so far? Because now it gets a bit tricky. Start with your drawing of the circle, centered at the origin. Draw your axes, but label the vertical axis x (instead of y). Don't label the horizontal axis. Draw your water line, which intersects the x-axis. Label the point of intersection a. As you continue vertically along the x-axis, you will intersect the circle. Label this point of intersection r.

We know that the volume of a cylinder equals the base area times the height(h). Except now, the base area is the area of the circle below the waterline. In other words, it's equal to the area of the circle, minus the area of the circle above the waterline. To find the volume, we multiply this area by the height. To solve for a, we must set both volumes of water equal (for the upright cylinder, and the cylinder on its side).

How to calculate the area above the waterline? Turn the drawing sideways, so the x-axis is facing to the right. The circle is formed by 2 curves, y = ±√(r

^{2}-a^{2}). To find the area of the circle above the waterline, we find the area in between both curves, for a≤x≤r. , and subtract it from pi*r^{2}. This means we integrate:A=pi*r

^{2}-∫^{r}_{a}[√(r^{2}-x^{2})-(-√(r^{2}-x^{2}))]dx=pi*r

^{2 }- 2∫^{r}_{a}√(r^{2}-x^{2})dx.To integrate, we need to use trigonometric substitution. The details are thorny, so I'm going to skip the derivation. I'd be happy to provide it if you'd like. So, here's the area:

(pi*r

^{2})/2 +a√(r^{2}-a^{2}) + pi*r^{2}*sin^{-1}(a/r).Since this is the base area, the volume is the above expression times h.

Setting volumes equal (since we have the same amount of water),

k*pi*r

^{2}*h = (All of the above craziness)*h. Divide by h:k*pi*r

You can't solve for a, since it's impossible to separate out. But, if you could,

^{2}= (pi*r^{2})/2 + a√(r^{2}-a^{2}) + pi*r^{2}*sin^{-1}(a/r).You can't solve for a, since it's impossible to separate out. But, if you could,

The height of the waterline above the table divided by the diameter of the cylinder would be (r+a)/(2r).

Given a particular value for r, you could use a numeric solver to approximate a. I'm trying it right now on my TI-89, with r=2. It'd been gridlocked for about 5 minutes now.

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Michael J.

08/05/15