Volumes of Revolution: Disk vs. Shell Method Explained

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Homework Help Overview

The discussion revolves around calculating the volume of a solid obtained by rotating a region under the curve y = 1/x from x = 1 to x = 4 about the x-axis. Participants are exploring the differences between the disk method and the shell method for this problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply both the disk method and the shell method, expressing confusion over differing results. They question the validity of their shell method setup.
  • Another participant suggests that the shell integral should be divided into two parts, indicating a misunderstanding in the original setup of the shell method.
  • Further clarification is provided regarding the correct length of the shell for certain intervals.

Discussion Status

The discussion is ongoing, with participants providing insights into the setup of the shell method. There is recognition of a mistake in the original shell method, and some guidance has been offered to clarify the approach.

Contextual Notes

Participants are working within the constraints of a textbook problem that primarily covers the disk method, leading to exploration of the shell method as an alternative. There is an emphasis on understanding the geometric interpretations involved in both methods.

LHC
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I've encountered a weird problem in my text...somewhat by accident =P

My text only covers volumes of revolution through the disk method, and one of the questions was:

Find the volume of the solid obtained when the given region is rotated about the x-axis.

c) Under y = 1/x from 1 to 4

Using the disk method, I got the answer \frac{3\pi}{4}...

Ok, so I wonder...what happens if I try the shell/rings method?

So this is what I do:

I thought that the radius of such shells would be the height of the function, so it would be "y". And, the length of such shells would be the distance from the function to the line x = 1, ...so (1/y - 1)...

Because of that, I ended up trying this:

V = \int_{0}^{1} 2\pi \ y\ (\frac{1}{y} - 1) dy

This turns out to yield \pi

I'm so confused right now haha...could someone please tell me what I did wrong? Either my shell method was wrong, or the disk method was...or...both =S...
 
Last edited:
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I just found out that I get the answer if I do this:

<br /> V = \int_{\frac{1}{4}}^{1} 2\pi \ y\ (\frac{1}{y} - 1) dy + \pi \times (\frac{1}{4})^2 \times 3

And that's basically taking shells from y = 1/4 to y = 1, then adding that cylinder that's left behind (from x = 1 to x = 1, and from y = 0 to y = 1/4).

So...that made sense. But can anyone tell me why my shell method previously described in the original post was wrong?
 
Your shell integral needs to be broken into two parts. For y<1/4 the length of the shell isn't (1/y-1), it's just 3.
 
Ohhh...*LED above head suddenly flickers*...I get it. I had the wrong length of the shell! Thanks for explaining that to me. =D
 

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