Really hard Volume Question. !

1. Jan 29, 2009

DANIELLYMA

Really hard Volume Question. URGENT!

Can someone help me with this question? I don't know where to begin. I don't want the answers, just hints in the right direction

*edit* can a mod please fix the topic title? It's supposed to read 'Really hard Volume Question. URGENT!' << Done. >>

Thanks

Last edited by a moderator: Jan 29, 2009
2. Jan 29, 2009

rootX

Re: ally hard Volume Question. URGENT!

I can suggest:
http://img101.imageshack.us/img101/1570/49965064rf9.jpg [Broken]

And Volume = area*length

Last edited by a moderator: Apr 24, 2017 at 10:44 AM
3. Jan 29, 2009

alle.fabbri

Re: ally hard Volume Question. URGENT!

Call z the axis which h belongs to. You have to find an expression for the area of the rectangle obtained by intersecting the cylinder with an horizontal plane as a function of z and integrate this area from 0 to h.

4. Jan 29, 2009

hokie1

Re: ally hard Volume Question. URGENT!

The area of the disc (what's inside the circle) from -r to h, where h is between -r and r can be figured out by doing half a circle at a time. The function is sqrt (r^2-x^2). Since a disc is symmetrical just find the area above the x-axis and multiply by 2. So the fun part here is finding out what the integral is.

5. Jan 29, 2009

rootX

Re: ally hard Volume Question. URGENT!

think simple ...
I think it can be solved without integration using bunch of trigonometry and sector, triangle area formulas

6. Jan 29, 2009

hokie1

Re: ally hard Volume Question. URGENT!

Sure area 2 is a fraction of pi r^2. I bet the fraction formula as a function of h is not simple. Area 1 is + when h is above the center of the circle and - when h is below the center. I think I'll stick with my integral since I've already solve a related problem: The probability of intersection between a tossed square and a target disc.

7. Jan 29, 2009

alle.fabbri

Re: ally hard Volume Question. URGENT!

This kind of integral is simple...
What you actually need is the lenght of the segment obtained by intersecting the circle [tex] x^2 + z^2 = R^2 [\tex] with a straight line z = const. And this is given by

[tex]
b(z) = 2 \sqrt{R^2 - z^2}
[\tex]

Thus, let L be the height of the cylinder, the area of the rectangle is given by

[tex]
A(z) = L b(z) = L 2 \sqrt{R^2 - z^2}
[\tex]

And then the volume of the water

[tex]
V(h) = \int_{-R}^{h} A(z) dz
[\tex]

8. Jan 29, 2009

alle.fabbri

Re: Really hard Volume Question. URGENT!

I dont know why but latex command doesn't work....

9. Jan 29, 2009

rootX

Re: ally hard Volume Question. URGENT!

I see.
I failed to consider the other case.

10. Jan 29, 2009

miller8605

Re: Really hard Volume Question. URGENT!

(1/2(pi r^2) + 1/2(pi (r-h)^2 ) * l

would that work?

11. Jan 29, 2009