Finding capacity/volume with integration

[uberathlete]

Hi all. I've got the following problem:

Say I have a rectangle with one side of length 2 and the other side of length Pi. Now I bend the rectangle to form a half cylinder of length 2 and semi-circle radius 1. I then cover the bottom of the half cylinder. To visualize the object, think of a cylinder with top and bottomsides closed. Then cut out the top of the cylinder, then slice the cylinder vertically in half.

Now, I fill it with some liquid, tilting it appropriately so that I get as much of the liquid in it as possible.

My question is, how do I find the capacity of the half-cylinder using integration?

Any help on this would be greatly appreciated. I've been wracking my brains for the past few days and its starting to hurt. Thanks!

 

[tkhunny]

How do you bend a rectangle (2D) into a cyllinder (3D)?

Why did you start with a closed object if you're just going to cut off the lid?

Why do you have to tilt it to achieve maximum volume?

I think it an unter-defined problem for the moment.

 

[stapel]

How do you bend a rectangle (2D) into a cyllinder (3D)?

I think the exercise means something like "take a rectangular sheet of tin, and roll and solder it to form a cylindrical tube", etc, etc. The resulting hollow cylinder will have an end circumference equal to one of the original sheet's dimensions, and a height equal to the other.

Eliz.

 

[uberathlete]

re

Maybe I should rephrase ...

Take a flat rectangular piece of tin with length 2 and width Pi. Bend it to form a half (or semi) cylinder with length 2 and semi-circle radius 1. Now, cover the bottom (ie. with a semi-circle). What is the capacity of this object?

 

[stapel]

Re: re

Take a flat rectangular piece of tin with length 2 and width Pi. Bend it to form a half (or semi) cylinder with length 2 and semi-circle radius 1. Now, cover the bottom (ie. with a semi-circle). What is the capacity of this object?

Oh... I think I see where the "tilting" is coming in:

You've got the classic "trough" with a semi-circular cross-section, but one of the end-caps is missing, so all the water will pour out unless you tip the thing a bit. Is that about right?

Thank you (and please forgive my earlier -- and possibly present -- confusion).

Eliz.

 

[galactus]

Is this something like you were descibing. Please forgive my hapohazard diagram.

I was trying to draw a cylinder cut in half lengthwise. That is what you meant by a

half cylinder, is it not?.

It is 2 feet long and has radius 1?.

\(\displaystyle 2\int_{-1}^{1}\sqrt{1-x^{2}}dx\)

The volume of this half cylinder could just be given by: \(\displaystyle (2)\frac{{\pi}r^{2}}{2}=\sout{2}\frac{{\pi}(1)^{2}}{\sout{2}}={\pi}\)

 

[uberathlete]

Re: re

Take a flat rectangular piece of tin with length 2 and width Pi. Bend it to form a half (or semi) cylinder with length 2 and semi-circle radius 1. Now, cover the bottom (ie. with a semi-circle). What is the capacity of this object?

Oh... I think I see where the "tilting" is coming in:

You've got the classic "trough" with a semi-circular cross-section, but one of the end-caps is missing, so all the water will pour out unless you tip the thing a bit. Is that about right?

Thank you (and please forgive my earlier -- and possibly present -- confusion).

Eliz.

Yes, that's right.

 

[uberathlete]

Is this something like you were descibing. Please forgive my hapohazard diagram.

I was trying to draw a cylinder cut in half lengthwise. That is what you meant by a

half cylinder, is it not?.

It is 2 feet long and has radius 1?.

\(\displaystyle 2\int_{-1}^{1}\sqrt{1-x^{2}}dx\)

The volume of this half cylinder could just be given by: \(\displaystyle (2)\frac{{\pi}r^{2}}{2}=\sout{2}\frac{{\pi}(1)^{2}}{\sout{2}}={\pi}\)

Hi Galactus. Your diagram is correct. The half cylinder however is closed at the top and open on the bottom and the side (ie. its rectangular side). So if I fill it with some liquid without tilting it, the liquid would just flow out of the bottom side. If I tilt it so that I can get as much liquid in it as possible, then that would be the maximum capacity of the object. So I sort of want to find the volume of the liquid, or in other words, the capacity of the object.