Hard Problem (Cylinder)

[maxhk]

A cylindrical glass of radius r and height L is filled with water and then tilted until the water remaining in the glass exactly covers its base.

(a) Determine a way to "slice" the water into parallel rectangular cross-sections and then set up a definite integral for the volume of the water in the glass. Find the volume.

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(b) determine a way to "slice" the water into parallel cross-sections that are trapezoids and then set up a definite integral for the volume of the water. Find the volume.

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(c) Find the volume of water in the glass from purely geometric considerations.

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(d) Suppose the glass is tilted until the water exactly covers half the base. In what direction can you "slice" the water into triangular cross-sections? Rectangular cross-sections? Cross-sections that are segments of circles? Find the volume of water in the glass.

I have done correctly part (a) and (c). Any help with (b) and (d) ? Thanks
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[daon2]

Draw parallel chords along the circular base of the cylinder in the obvious way, and I believe cutting along them perpendicular to the circular base will produce rectangles. The trapezoids will come from cutting orthogonally to the chords from the rectangles. Though cutting along the axis in this case gives you a triangle. To get triangles when it is half, covered, it is the same as performing the cuts to get trapezoids.

The integral for the triangles would be \(\displaystyle \int_{-r}^r \frac{1}{2}B(x)H(x)dx\)

Where the base B(x) is: \(\displaystyle B(x)=\sqrt{r^2-x^2}\)

And the height is H(x) is a weird one. You get half an ellipse with equation (x/r)^2+(H(x)/L)^2=1. So

\(\displaystyle H(x)=\frac{L}{r}\sqrt{r^2-x^2}\)

 

[galactus]

Here are the respective cross sections:

http://www.ima.umn.edu/~arnold/calculus/crosssecs/all.gif


Here is a nice site on tilted cylinders:

http://www.lmnoeng.com/Volume/InclinedCyl.htm

 

[daon2]

Galactus' link shows what I was talking about with the cuts. And my integral is wrong as well. I assumed the ellipse has width 2L which is not true. The width is actually \(\displaystyle 2\sqrt{L^2+r^2}\), assuming the liquid reaches the rim. So replace \(\displaystyle L\) by \(\displaystyle \sqrt{L^2+r^2}\)