Retake: Capacity Problem

[uberathlete]

Hi everyone. I've asked this question before but it was probably confusing and there was no picture to clear things up. So here's a retake of the problem:

I have a rectangular sheet of tin with length = 2 and width = Pi. I bend the the sheet to form a half-cylinder with semi-circle radius = 1 and length = 2. I then cover the bottom of the half-cylinder (ie. one of the semi-circle sides). I then fill this up with liquid, tilting it appropriately so that I can get as much liquid in it as possible. What is the capacity of the object?

Here's a picture of it: http://www.geocities.com/uberathlete/index.htm
The blue portion is the liquid that was poured into the object.

I apologize for the double post. Any help on this would be greatly appreciated. I've learned how to get the volume of objects using integration but I'm not sure how to apply it to this particular problem. It's driving me nuts!! :x

 

[stapel]

I apologize for the double post.

S'okay. Next time, though, post as a "reply" to the original thread, so people can remember what was said.

By the by, viewers, here is the image:
. . . . .

Nicely done! (And this confirms that I'd had the right picture in my head.) Thank you!

Eliz.

 

[Gene]

The key is picking what to use for dV.
Looking down from the top you see that the surface of the water at y above the bottom is a chord of length z, subtended by an angle 2t.
The distance from the center is x.
Cos(t)=x/r
sin(t)=(z/2)/r
Shifting to the side view
y/h = x/r
so we have
x=r*cos(t)
z=2r*sin(t)
y=hx/r=h*cos(t)
dx=-r*sin(t)dt
dV=y*z*dx =
(h*cos(t))(2r*sin(t))(-r(sin(t))*dt =
-2hr²sin²(t)cos(t)*dt
The integral is from t=pi/2 to 0
V=(-2hr²/3)(sin³(t))] =
(-2hr²/3)(0-1) =
2hr²/3

 

[uberathlete]

Whoa! Thanks for the reply Gene. I'll check it out and try to understand it ..

 

[uberathlete]

The key is picking what to use for dV.
Looking down from the top you see that the surface of the water at y above the bottom is a chord of length z, subtended by an angle 2t.
The distance from the center is x.
Cos(t)=x/r
sin(t)=(z/2)/r
Shifting to the side view
y/h = x/r
so we have
x=r*cos(t)
z=2r*sin(t)
y=hx/r=h*cos(t)
dx=-r*sin(t)dt
dV=y*z*dx =
(h*cos(t))(2r*sin(t))(-r(sin(t))*dt =
-2hr²sin²(t)cos(t)*dt
The integral is from t=pi/2 to 0
V=(-2hr²/3)(sin³(t))] =
(-2hr²/3)(0-1) =
2hr²/3

Hi Gene. I have a few questions if you don't mind ...

> When you say "looking down from the top" does this mean looking down at the liquid through the open "rectangular side" of the object?
> If I place the object onto x-y-z axes, how would it appear?
> What do you mean when you say "the distance from the center is x"? Which "center" do you mean?

Ok thanks, hope to read your reply soon.

 

[uberathlete]

Oh Gene, just as a follow up question ... what is r? Is that the semi-circle radius?

 

[galactus]

Couldn't we use 'washers' to find the volume?. Probably not, but it's a thought.

Using the equation of the line with coordinates P(0,-1) and P(2,0) and revolving it

halfway around the x-axis, we can set up an integral: \(\displaystyle \frac{{\pi}}{2}\int_{0}

^{2}(\frac{x}{2}-1)^{2}dx=\frac{{\pi}}{3}\)

Since the water peaks at the end which is tilted and forms a semicircle at the end

whichs holds in the water, we have a, sort of, 'half-cone' with volume \(\displaystyle \frac

{{\pi}}{3}.\)

I know, I'm probably off base, but this is an interesting problem to discuss.

Since this solution does not agree with Gene's I have my reservations about even

mentioning it.

 

[uberathlete]

Couldn't we use 'washers' to find the volume?. Probably not, but it's a thought.

Using the equation of the line with coordinates P(0,-1) and P(2,0) and revolving it

halfway around the x-axis, we can set up an integral: \(\displaystyle \frac{{\pi}}{2}\int_{0}

^{2}(\frac{x}{2}-1)^{2}dx=\frac{{\pi}}{3}\)

Since the water peaks at the end which is tilted and forms a semicircle at the end

whichs holds in the water, we have a, sort of, 'half-cone' with volume \(\displaystyle \frac

{{\pi}}{3}.\)

I know, I'm probably off base, but this is an interesting problem to discuss.

Since this solution does not agree with Gene's I have my reservations about even

mentioning it.

At first I was thinking the same thing. It's interesting cuz the curved side isn't the one slanting but the flat side (ie. the surface of the water) is.

 

[Gene]

I have reservations too, but I can't see anything wrong in my steps. Before I started, I would have bet a nickle (maybe even a quarter) that there would be a pi in the answer.

When you say "looking down from the top" does this mean looking down at the liquid through the open "rectangular side" of the object?

My top is the right hand semi-circle of the drawing with the water frozen.

> If I place the object onto x-y-z axes, how would it appear?

Pretty much like the drawing if you were lying down on your right side and looking at it.

> What do you mean when you say "the distance from the center is x"? Which "center" do you mean?

The center of the imaginary circle (of radius r) of the "pipe" to the chord.

 

[galactus]

I believe I may have a double integral which is viable and agrees with Gene's clever solution, 'taboot'.

Let's stand the half cylinder up with the bottom part holding back the water forming the semi-circle \(\displaystyle sqrt{1-x^{2}}\)

Now, the plane which cuts the half cylinder runs along the surface of our water. It has equation \(\displaystyle z=2y\)

In other words, it is bounded above by the plane \(\displaystyle z=2y\) and below by the region \(\displaystyle sqrt{1-x^{2}}\)

So, the integral is \(\displaystyle \int_{-1}^{1}\int_{0}^{sqrt{1-x^{2}}}2ydydx=\frac{4}{3}\)

This seems logical and I was pleased to see it agrees with Gene's solution.

 

[Gene]

I'm pleased too. I still have to fight the feeling that there would be a pi floating around.
-----------------
Gene

 

[galactus]

Yes, Gene, I thought the same thing. I reckon it's just one of those strange things.