Volume vs. Area of a Surface of Revolution

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Discussion Overview

The discussion centers on the differences between calculating the volume and surface area of a solid of revolution formed by revolving a continuous function around the x-axis. Participants explore the mathematical reasoning behind the formulas used for each calculation and question the apparent inconsistencies in the approaches taken.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes the process of finding volume using cylinders and surface area using frustums, questioning why the arc length is necessary for surface area but not for volume.
  • Another participant suggests that the choice of approximating surface does not significantly affect the answer, while also acknowledging that it does change the outcome.
  • A different participant emphasizes that the formula for surface area includes the term sqrt[1+(dy/dx)^2], which is not present in the volume calculation, and seeks clarification on this difference.
  • One participant explains that to find surface area, one must consider the arc length, which involves the distance formula, leading to the inclusion of the sqrt[1+(dy/dx)^2] term.
  • Another participant argues that using horizontal segments to approximate a sloped line will not yield the correct length, drawing a parallel to why cylinders cannot be used to find surface area.
  • Links to online applets for calculating volume and surface area are shared, although their relevance to the discussion is questioned by other participants.
  • A participant expresses regret for posting links and acknowledges the feedback received regarding their contributions.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of arc length in surface area calculations versus volume calculations. There is no consensus on the implications of using different approximating surfaces, and the discussion remains unresolved regarding the perceived inconsistencies.

Contextual Notes

Some participants note that the differences in approach may stem from the geometric properties of the shapes involved, but these observations do not lead to a definitive resolution of the questions raised.

WillJ
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f(x) is a continuous function of x, whose domain is [a, b]. Revolve the graph around the x axis. In doing so you will create a solid.

Apparently, to find the volume of this solid, partition the solid into n cylinders along the x-axis from a to b, each partition of the x-axis containing some x*. A cylinder's radius is f(x*), so its volume is pi(f(x*)^2). Add the cylinders' volumes together. Take the limit as n approaches infinity; that is, take the integral of pi(f(x)^2)dx from x=a to x=b.

So that's volume. To find the area of the surface of revolution, instead of using cylinders, partition the solid into n frustums of cones along the x-axis from a to b, each frustum having two different circular sides, one with radius f(xi-1) and the other with radius f(xi). The surface area of a frustum is 2pi times the average of the radii times the arc length of the curve cut by the two circular sides of the frustum, this arc length being sqrt[1+(dy/dx)^2]. Add the frustums' surface areas together. Take the limit as n approaches infinity; that is, take the integral of 2pif(x)sqrt[1+(dy/dx)^2]dx from x=a to x=b.

My question: Why the inconsistency? Why can't we take the area of the surface of revolution just by taking cylinders, like we did with volume (except of course taking the surface area of the cylinders, not the volume), and not have to worry about the arc length (that is, not have to worry about the fact that, for a given partition, one circular side will be different from the other, since you'd think it wouldn't matter when the partition becomes infinitely thin)? And if we're not allowed to do that, why are we allowed to use cylinders instead of frustums and ignore the arc length when taking volume?

In my calculus II class we're using Stewart's Calculus as our textbook. The book doesn't really explain the inconsistency. I also own Spivak's Calculus, which doesn't seem to explain it either (although Spivak at least admits there's some "fudging" in what he's doing in that chapter, although I'm not sure if this is what he was referring to or not).

I might not have explained my question well in the post. If you're confused about what I'm asking, I can try clearing it up.
 
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Off the top of my head (that is, take this with a huge grain of salt), I think the approximating surface doesn't really matter in the answer, but the cone/cylinder might give the simplest (or maybe easiest to visualize) way to get to the answer.
 
Mystic998 said:
Off the top of my head (that is, take this with a huge grain of salt), I think the approximating surface doesn't really matter in the answer, but the cone/cylinder might give the simplest (or maybe easiest to visualize) way to get to the answer.
No, it definitely changes the answer.

The formula for the surface area of revolution is the integral of 2pif(x)sqrt[1+(dy/dx)^2]dx from x=a to x=b.

What I'm asking is why the sqrt[1+(dy/dx)^2]dx is there for surface area, but for volume it's just dx. It certainly affects the answer.
 
Here's my shot at helping you:

To find surface area you are essentially rotating an (arc length*circumference) around a given axis, in this case the x-axis. Now for some continuous function f(x) from x=a to x=b in order to approximate an arc length you will utilize the distance formula such that

Arc Length ~ \sqrt{[(\Delta x)^2] + [(\Delta y)^2]}

You can then multiply by:
\sqrt{(\Delta x)^2}/\sqrt{(\Delta x)^2}

Which gives:
[\sqrt{1+[(\Delta y)^2/(\Delta x)^2]}]\Delta x

Then take the lim as n approaches infinity, aka the integral to end up with your

\int \sqrt{1+(dy/dx)^2} dx

And that would be where it comes from, I hope it helped some
 
As Will J said, cylinders will not give you the surface area for the same reason that horizontal lines will not approximate a line at an angle to the x-axis.

That is, you cannot approximate the the length of, say, the line from (0,0) to (1, 1) by using segments parallel to the x-axis. Even if you take a large number of short horizontal segments, their sum will be 1, not \sqrt{2}. Even in the limit, you get 1, not \sqrt{2}.
 
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That's enough. DexteronLine, if you do not stop posting links to your own applets, I will start deleting your posts that contain links.

I am very impressed with them but they do not help the posters understand their specific problems and do not belong here.
 
HallsofIvy

I apologize if the links I added were not suitable for the forum. I will not add links to my applets anymore.

Thanks for the remarks about the program.
I will learn a lot from the postings on this forum

Asad
 

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