# Really hard Volume Question.

• DANIELLYMA
In summary: It looks like the integral can be solved without integrating. The area under the curve from -r to h is given by $$A(z) = \int_{-R}^{h} r^2 dz$$
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

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I can suggest:
http://img101.imageshack.us/img101/1570/49965064rf9.jpg And Volume = area*length

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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.

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.

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

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.

This kind of integral is simple...
What you actually need is the length 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]

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

hokie1 said:
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 see.
I failed to consider the other case.

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

would that work?

## 1. What is the formula for calculating volume?

The formula for calculating volume depends on the shape of the object. For a cube or rectangular prism, the formula is V = length x width x height. For a cylinder, the formula is V = π x radius2 x height. For a sphere, the formula is V = 4/3 x π x radius3.

## 2. How is volume different from capacity?

Volume refers to the amount of space occupied by an object, while capacity refers to the maximum amount of substance that an object can hold. For example, a cup may have a volume of 8 ounces, but its capacity may be 10 ounces if it is filled to the brim.

## 3. Can volume be negative?

No, volume cannot be negative. Volume is a measure of physical space, and negative values do not make sense in this context.

## 4. How does changing the dimensions of an object affect its volume?

Changing the dimensions of an object can greatly impact its volume. For example, doubling the length, width, and height of a cube will result in a volume that is 8 times larger. However, for a sphere, doubling the radius will result in a volume that is 8 times larger.

## 5. Can the volume of an object change?

Yes, the volume of an object can change. For example, if an ice cube melts, its volume will decrease as it turns into liquid. Additionally, if an object is compressed, its volume will decrease.

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