Cylinder Tilted Volume 2 (filled half the base)

[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 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

 

[Deleted member 4993]

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 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

View attachment 1391

Cylinder Tilted Volume

"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.

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."

Can you please provide a solution to this problem and ? and if possible a diagram of your cross sections and computations ? Thanks

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Aren't these same problem - with small modification??

Show some effort - the other one was done for you completly!!

 

[galactus]

Max, have you made progress?. If so, let me see what you've come up with.

I got to nosing around wolfram and found there is actually a name for this. It is called a "cylindrical hoof".

If using rectangles, you can use similar triangles to solve for h in terms of L, x, and r.

\(\displaystyle \frac{x}{h}=\frac{r}{L}\)

The volume of each slice is then \(\displaystyle dV=width \cdot height\cdot dy\).

where h is the height of each slice, w is the width, and dy is the thickness.

If using circular segments, remember the area of a circular segment is \(\displaystyle \frac{1}{2}r^{2}(\theta-sin\theta)\)

The circular segments are perpendicular to the water surface. So, the water surface remains level when the cylinder is tilted and had different length than L. Looking from the side, the water surface can be thought of as the hypoteneuse of a right triangle.

 

[daon2]

Pretty sure I did the triangle one correctly in your first thread as an example.

 

[maxhk]

Solution by rectangular cross sections

Solution by rectangular cross sections :
A rectangular cross section at a depth \(\displaystyle y \) has width \(\displaystyle 2 \sqrt{r^2-y^2} \) and length \(\displaystyle y\(\dfrac{L}{r}\) \)

So the volume is \(\displaystyle V = \int_{0}^{r} 2 \sqrt{r^2-y^2} * y *\(\dfrac{L}{r}\) dy = \dfrac{2}{3} r^2 L \)

What do you think ?

 

[maxhk]

Solution by triangular cross sections using your suggestions :

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

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

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

The volume \(\displaystyle V = \int_{-r}^r \dfrac{1}{2} * \sqrt{r^2-x^2} * \dfrac{L}{r}\sqrt{r^2-x^2} dx = \dfrac{2}{3} r^2 L\)
which the same as in my rectangular cross sections before.

 

[daon2]

I said replace it because I made a mistake. L is the width of half the ellipse that intersects the horizontal diameter of the base of the cylinder. I explained this somewhat in the other thread: The equation of an ellipse at the origin is (x/a)^2+(y/b)^2=1. The values "+/-a" give the x-intercepts, the values "+/-b" give the y-intercepts. Think of your r as being a, and the distance from the center of the base to the outer edge of the glass is your b. Then H(x) will be y.

If you're not getting the same answer, perhaps I am wrong, I dunno

 

[galactus]

Solution by rectangular cross sections :
A rectangular cross section at a depth \(\displaystyle y \) has width \(\displaystyle 2 \sqrt{r^2-y^2} \) and length \(\displaystyle y\(\dfrac{L}{r}\) \)

So the volume is \(\displaystyle V = \int_{0}^{r} 2 \sqrt{r^2-y^2} * y *\(\dfrac{L}{r}\) dy = \dfrac{2}{3} r^2 L \)

What do you think ?

Yes, that is the correct solution.

You can make it a little easier by using x instead.

\(\displaystyle \frac{L}{r}\int_{-r}^{r}x^{2}dx=\frac{2}{3}Lr^{2}\)

The circular segments are a little trickier because the volume of a circular segment is

\(\displaystyle \frac{1}{2}r^{2}(\theta-sin\theta)\)

The cross sections will be perp. to the water surface, not the cylinder.

The plane that slices the cylinder to form the 'cylindrical hoof' has equation.

\(\displaystyle z=\frac{Lx}{r}\).

It looks like a tongue.

The length of the water surface is \(\displaystyle \sqrt{r^{2}+L^{2}}\)

There are various ways to set it up.

See here:

http://mathworld.wolfram.com/CylindricalWedge.html

 

[galactus]

as a last challenge I am looking for solution by circular cross sections using single integrals and elementary geometry if possible (no advanced mathematics or double or triple integrals)!

Hence, I refer to my last post.

 

[galactus]

The distance from the z-axis down to the top of the water at some point can be found by similar triangles.

From the side, the water surface forms a line with coordinates (0,0) at one end and (L,-r) at the other.

Let y be the distance from the z axis through the center of the cylinder down to some point on the water surface.

\(\displaystyle \frac{r}{L}=\frac{-y}{z}\)

\(\displaystyle y=\frac{-rz}{L}\)

If you look at the cylinder from the end, you can imagine a line from the radius down to the chord forming the cross section of the water surface. This is the same as above \(\displaystyle \frac{r}{L}z\)

But, this distance is also the radius minus the middle ordinate. The middle ordinate is the distance from the chord to the circle.

\(\displaystyle r-r(1-cos\frac{\theta}{2})=rcos\frac{\theta}{2}\)

So, the triangle formed has angle \(\displaystyle \frac{rcos\frac{\theta}{2}}{r}=cos\frac{\theta}{2}\)

Therefore, \(\displaystyle cos\frac{\theta}{2}=\frac{\frac{rz}{L}}{r}=\frac{z}{L}\)

\(\displaystyle \theta=2cos^{-1}(\frac{z}{L})\)

Sub this into the area for a circular segment, then integrate w.r.t z from 0 to L.

I had a rough time trying to describe this. I am afraid if I try to draw a picture, it will not turn out so good.

I hope my diagram relates what I am trying to impart.

 

[maxhk]

Thanks a lot "Galactus" for your interesting work!

I actually was thinking along the same line but I thought I was wrong since I get an integral of an inverse trigonometric functions and don't know how to integrate such things yet.

I use same context but to avoid confusion and trouble (like \(\displaystyle cos^{-1}(\dfrac{-z}{L})\) at \(\displaystyle L\) is \(\displaystyle \pi\) not \(\displaystyle 0\) which would not cancel out), I put the y-axis pointing downward to have everything positive.

\(\displaystyle y=\frac{rz}{L} = r cos(\frac{\theta}{2})\)

==> Therefore, \(\displaystyle cos(\frac{\theta}{2})=\frac{z}{L}\)

and \(\displaystyle \theta=2*cos^{-1}(\frac{z}{L})\)

\(\displaystyle Area = \dfrac{r^2}{2} *(2*cos^{-1}(\frac{z}{L}) - 2 * \dfrac{z}{L}*\sqrt{1 -\dfrac{z^2}{L^2}})\)

Integrating this from \(\displaystyle 0\) to\(\displaystyle L \) gives \(\displaystyle volume = \dfrac{2}{3} r^2 L\) which is the same as before.