Cylindrical shells for calculating volume question.

[Dravin]

Okay, I get the basic premise, and I get that you need to do the integration magic by the other axis than you want to rotate around. So integrate by dx when you want to rotate your cylindrical shell around the y axis and integrate by dy when you want to rotate around the x axis. And you're integrating by 2*pi∫xf(x)dx (or yf(y)dy).

Now I have this 'curves' bounding my shape, x=0, y=1, y=3, x=1/y. Now you want to integrate from 1 to 3 (it being when 1/y=3 or 1 respectively). Now the set-up should be 2*pi∫(y^-2)dy, except it isn't for a reason that's eluding me, it's not, as far as I can tell a simply arithmetic error, evaluating the anti-derivative (-y^-1) from 1 to 3 gives me 2pi[(-1/3)-(-1)] = 4pi/3. I can get the right answer (4pi) using the washer method (I need to solve it both ways) but the cylindrical method is eluding me.

[Vort]

Okay, I get the basic premise, and I get that you need to do the integration magic by the other axis than you want to rotate around. So integrate by dx when you want to rotate your cylindrical shell around the y axis and integrate by dy when you want to rotate around the x axis. And you're integrating by 2*pi∫xf(x)dx (or yf(y)dy).

Now I have this 'curves' bounding my shape, x=0, y=1, y=3, x=1/y. Now you want to integrate from 1 to 3 (it being when 1/y=3 or 1 respectively). Now the set-up should be 2*pi∫(y^-2)dy, except it isn't for a reason that's eluding me, it's not, as far as I can tell a simply arithmetic error, evaluating the anti-derivative (-y^-1) from 1 to 3 gives me 2pi[(-1/3)-(-1)] = 4pi/3. I can get the right answer (4pi) using the washer method (I need to solve it both ways) but the cylindrical method is eluding me.

I'm trying to understand the shape you're integrating. I understand you to say that the quadrilateral bounded by y=1, y=3, x=0, and x=1/y is rotated about the y axis, and you're trying to find the volume by integration. Is this correct?

To integrate using cylindrical shells, first draw out the shape and see that it consists of two regions: The region between x=0 and x=1/3, where the height is 2, and the region between x=1/3 and x=1, where the height is 1/x - 1 (because when x=1/y, that means y=1/x).

So make your cylinders and integrate. Each cylinder has a curved surface area of 2πx (x is the radius at any given point) times the height (either 2 or 1/x - 1). Multiply by the differential thickness and integrate:

V(inner) = ∫(2πx)(2) dx [between x=0 and x=1/3]

= 4π ∫x dx

= 4πx²/2

= 2πx² [between x=0 and x=1/3]

= 2π[ (1/3)² - 0 ] = 2π/9

V(outer) = ∫(2πx)(1/x - 1) dx [between x=1/3 and x=2/3]

= 2π ∫(1 - x) dx

= 2π (x - x²/2)

= π (2x - x²) [between x=1/3 and x=2/3]

= π [4/3 - 4/9 - 2/3 + 1/9]

= π [12/9 - 4/9 - 6/9 + 1/9]

= π [3/9]

= π/3

V(total) = V(inner) + V(outer)

= 2π/9 + π/3

= 5π/9

Checking, the inner cylinder must have a volume of π(1/3)²(2) = 2π/9. The outer cylinder is twice as wide at the base but goes as 1/x in width, so it will have something less than twice the volume of the inner cylinder. So 5π/9 looks right. Is that what integrating with disks gave you?

[Dravin]

Disks gives me 4pi. I'm supposed to be integrating in respect to the x-axis, since cylinders are 'backwards' you need to integrate them with respect to dy not dx unless I'm mistaken. I just reread my post and noticed which axis I was working off of wasn't particularly clear.

I just realized I was over thinking the problem (or not thinking enough, I was, due to working with disks, squaring 1/y instead of multiplying it by the radius (in this case y)). Integrating and evaluating from 1 to 3:

2pi∫y(1/y)dy=2pi(y)| = 2pi(3-1) = 2pi(2) = 4pi.

Now if I can just get the hang of calculating r when when it doesn't just equal x or y (due to my rotational axis not being x=0 or y=0). So far my process is involves test points and I'm not sure if that's simply how the math is or if I'm missing an observation that would be making my life easier.

Edited November 20, 2011 by beefche

[pam]

I say it's just A-Z. :)

[Guest]

blink. blink.

[Vort]

Hmmm. I was thinking about this in the shower, and I did a disk integration that suggested 2π/3. I have a very full Sunday and don't want to take away from my family time later on, but I will try to get back to this today or tomorrow morning.

[MsMagnolia]

I thought the answer to "life, the universe and everything" was 42?

:)

[Vort]

2π/3. Actually writing the problem down helps...

[Dravin]

Vort, you're integrating around the wrong axis, the problem calls for integration around the x-axis not the y-axis. That is to say (picturing you're looking at the graph with the x axis running from left to right) that it needs to spin like it's on a horizontal lathe, not spin like a top. My apologies if I'm messing up the terminology and confusing you.

Edited November 22, 2011 by Dravin

[Vort]

Ah. That makes a difference. That's why I tried to describe the problem before attempting to solve it.

[PrinceofLight2000]

Hmmm. I was thinking about this in the shower, and I did a disk integration that suggested 2π/3. I have a very full Sunday and don't want to take away from my family time later on, but I will try to get back to this today or tomorrow morning.

What sort of man thinks about math in the shower?

[Dravin]

What sort of man thinks about math in the shower?

A productive one. While I tend not to think math in the shower I know I've thought about papers in the shower.

[Vort]

What sort of man thinks about math in the shower?

Sister Vort would roll her eyes and sigh at the question.

Actually, I expect that most people with a mathematical bent or who just enjoy math do so. It's too much fun not to play around with. Kind of like composing poetry; once you start, you don't want to stop.

[Dravin]

Sister Vort would roll her eyes and sigh at the question.

Actually, I expect that most people with a mathematical bent or who just enjoy math do so. It's too much fun not to play around with. Kind of like composing poetry; once you start, you don't want to stop.

I'm that way with geology, when I'm on a road trip my eyes are glued to the road cuts. And when flying over mountains I'm looking for glacial features. Reminds me of a quip one of my Geology professors made, "Always wear a seat-belt when a geologist is driving, he won't be looking at the road."

[Vort]

Vort, you're integrating around the wrong axis, the problem calls for integration around the x-axis not the y-axis. That is to say (picturing you're looking at the graph with the x axis running from left to right) that it needs to spin like it's on a horizontal lathe, not spin like a top. My apologies if I'm messing up the terminology and confusing you.

Okay, so I don't have a scanner available to scan my answer. Let's see if I can give it succinctly by typing it:

In this case, cylinders are actually a lot easier to do than disks (washers). As you note, integration has to go along the y direction because the differential thicknesses are measured along y, not x.

The height of each differential cylinder (distance between the y axis and the curve) is just 1/y.

The circumference of each differential cylinder is 2π times the radius, or 2πy.

So the surface area of each cylinder is 2πy (1/y) = 2π (!!)

Therefore, the volume is

V = ∫2π dy [between 1 and 3]

= 2πy [between 1 and 3]

= 2π (3 - 1)

= 4π

Turns out to be very simple -- much simpler than integrating along the x axis using washers, though of course the answer is the same.

[MarginOfError]

What sort of man thinks about math in the shower?

I think there is nothing more appropriate for him to be thinking about in the shower. Unless, of course.....nevermind.

[Vort]

I'm that way with geology, when I'm on a road trip my eyes are glued to the road cuts. And when flying over mountains I'm looking for glacial features. Reminds me of a quip one of my Geology professors made, "Always wear a seat-belt when a geologist is driving, he won't be looking at the road."

A long time ago, maybe fifteen years or more, my wife bought Roadside Geology of Washington and read it to me as we drove along. Now my kids (or at least one or two of them) love to hear it as we drive across the mountains. Really amazing stuff.

[Dravin]

Turns out to be very simple -- much simpler than integrating along the x axis using washers, though of course the answer is the same.

Yep, I did a face palm when I spotted my mistake and realized how simple it was back in my second post. It is a lot easier than doing disk integration.

[Vort]

Yep, I did a face palm when I spotted my mistake and realized how simple it was back in my second post. It is a lot easier than doing disk integration.

Oh. Uh, yeah. Your second post. Of course.

Why, no, I didn't miss that at all. Why do you ask?

[Dravin]

Sister Vort would roll her eyes and sigh at the question.

Actually, I expect that most people with a mathematical bent or who just enjoy math do so. It's too much fun not to play around with. Kind of like composing poetry; once you start, you don't want to stop.

I just caught myself realizing as I poured myself a cup of soda that my cup was a frustum of a cone. As soon as I realized that it occured to me that using similar triangles one could figure out what it would be as a complete cone and set things up to calculate how much work it would take to pump all the soda (given a density) out of the cup ala one of my textbook problems.

[Vort]

I would laugh, if there were but a laugh button...

[Dravin]

I would laugh, if there were but a laugh button...

Calculus has corrupted me more than algebra ever did. I think the only similar experience I had with algebra was after doing solution dilution problems realizing I could use the same technique to get my hydrochloric acid to the desired concentration.

[Vort]

Calculus has corrupted me more than algebra ever did. I think the only similar experience I had with algebra was after doing solution dilution problems realizing I could use the same technique to get my hydrochloric acid to the desired concentration.

When I tell people that calculus is easier than algebra, they tend to look at me funny. But I believe it's true. Certainly was for me. Algebra required a real shift in my paradigm of understanding. I had such difficulty figuring out the purpose of "the unknown value", and it took me a while to learn to view an equation as a balance. In contrast, calculus required gaining an understanding of infinite series and sums (which wasn't that hard), then learning some tricks for derivatives and integrals. There's also something very satisfying about decomposing a seemingly difficult integral problem and finding the answer, step by step.

[Dravin]

There's also something very satisfying about decomposing a seemingly difficult integral problem and finding the answer, step by step.

Yep. Even more fun though is when you realize the short way (when applicable). Like when I was asked to calculate the volume of a solid of revolution. Looking at the formula for a little bit I realize it was the formula for a circle situated on the origin.

No washers, no shells, just a nice simple volume of a sphere calculation to do (and no, the point wasn't to prove the formula for a sphere).

It is nice when you are beating your head against a problem to have what you need to do come to you. There is a certain feeling of victory, almost a thought of, "Bow before your betters puny expression/equation." Though maybe I'm just on a power trip.