Getting two difference results when calculating volume of cylinder?

[Nat3]

This actually isn't a homework problem -- I'm just trying to understand an example in my textbook. The example shows how to calculate the volume of a cylinder (maybe it's actually a shell, I'm not sure) using an integral, but it occurred to me that I should be able to simply "unwrap" the cylinder so that it's just a rectangle and then multiple the dimensions instead of integrating.

For some reason the answers are not the same, and I don't understand why. Here's my work:

What did I do wrong?

 

[gopher_p]

1) You have an antiderivative error in your first calculation.

2) While the rectangular sheet is a good approximation to the volume of the cylindrical shell, especially for very thin sheets, it is not exact. The edges where the shell is cut would actually be beveled once the sheet was flattened.

Edit: If you want a geometric/algebraic/not calculus way to do the problem, consider that a cylindrical shell is just a cylinder with a cylindrical hole.

 

[SammyS]

This actually isn't a homework problem -- I'm just trying to understand an example in my textbook. The example shows how to calculate the volume of a cylinder (maybe it's actually a shell, I'm not sure) using an integral, but it occurred to me that I should be able to simply "unwrap" the cylinder so that it's just a rectangle and then multiple the dimensions instead of integrating.

For some reason the answers are not the same, and I don't understand why. Here's my work:

What did I do wrong?

The radius is not the same for any of the cylindrical shells.

You used the outer radius.

Using the inner radius of 3, you get (2π)(3)(6)(1) = 36π .

By the way $\ \displaystyle \int {\rho\ }d\rho = \frac{\rho ^2}{2}+C\ .$

 

[Simon Bridge]

...it occurred to me that I should be able to simply "unwrap" the cylinder so that it's just a rectangle and then multiple the dimensions instead of integrating.

That can give you a good approximation when the thickness is small, but the volume calculated this way will always be too big. You can see why by actually making a fat cylindrical shell out of plasticene (or something), cutting it, and carefully unrolling. Imagine the inner radius is zero - what shape do you get when you unroll?

Consider - which dimensions did you use when you unrolled the cylindrical shell? Did you use the inner or the outer dimensions? The height is the same but the circumference isn't. So which one is right?

The easiest non-integrating way to find the volume is to use the equation for the volume of a cylinder and subtract the volume of the hole.

If we take: height h, inner radius r, outer radius R (R>r), we can work out the general formula by different methods and compare them.

"Subtracting the hole" gets you:
$$V=\pi(R^2-r^2)h$$ You can see this one has to be correct right?

Using the unrolling approach, you one of these: $$V=2\pi R(R-r)h\qquad V=2\pi r(R-r)h$$... depending on if you use the inner or outer dimensions. Either way, it's not the same.

By integration:$$V=2\pi h\int_r^R \rho \; d\rho = \pi h(R^2-r^2)$$... here the volume element is the cylindrical shell between $\rho$ and $\rho+d\rho$
... which is $dV = 2\pi h \rho \;d\rho$

I got the volume element by the "unrolling" method - which I can get away with this time because this shell is infinitesimally thin.

When you unroll a fat cylinder, you end up with a slab with trapezium-shaped ends.
What happens when you work out the volume of that?

 

[Nat3]

Edit: If you want a geometric/algebraic/not calculus way to do the problem, consider that a cylindrical shell is just a cylinder with a cylindrical hole.

Good point!

By the way $\ \displaystyle \int {\rho\ }d\rho = \frac{\rho ^2}{2}+C\ .$

Oh snap! Thanks.

That can give you a good approximation when the thickness is small, but the volume calculated this way will always be too big. You can see why by actually making a fat cylindrical shell out of plasticene (or something), cutting it, and carefully unrolling. Imagine the inner radius is zero - what shape do you get when you unroll?

I was having trouble visualizing this until you said "Imagine the inner radius is zero".. So the edges of the "rectangle" on the ends would be angled?

Consider - which dimensions did you use when you unrolled the cylindrical shell? Did you use the inner or the outer dimensions? The height is the same but the circumference isn't. So which one is right?

I used the outer radius. Could I have used the average of the radii?

The easiest non-integrating way to find the volume is to use the equation for the volume of a cylinder and subtract the volume of the hole.

If we take: height h, inner radius r, outer radius R (R>r), we can work out the general formula by different methods and compare them.

"Subtracting the hole" gets you:
$$V=\pi(R^2-r^2)h$$ You can see this one has to be correct right?

Using the unrolling approach, you one of these: $$V=2\pi R(R-r)h\qquad V=2\pi r(R-r)h$$... depending on if you use the inner or outer dimensions. Either way, it's not the same.

By integration:$$V=2\pi h\int_r^R \rho \; d\rho = \pi h(R^2-r^2)$$... here the volume element is the cylindrical shell between $\rho$ and $\rho+d\rho$
... which is $dV = 2\pi h \rho \;d\rho$

I got the volume element by the "unrolling" method - which I can get away with this time because this shell is infinitesimally thin.

When you unroll a fat cylinder, you end up with a slab with trapezium-shaped ends.

Wow, thanks for such a detailed explanation! That helps so much!

What happens when you work out the volume of that?

The calculated volume will be too much, since it's including the non-solid part of the trapezium-shaped ends? So with my incorrect method, I got $48\pi$ when the correct answer is $42\pi$.

 

[Simon Bridge]

I used the outer radius. Could I have used the average of the radii?

You could try it and see ;)

The calculated volume will be too much, since it's including the non-solid part of the trapezium-shaped ends? So with my incorrect method, I got $48\pi$ when the correct answer is $42\pi$.

... area of the end times the height will work. Replace the rectangular end with the area of the trapezium.