Volumes of Revolution: Disk vs. Shell Method Explained

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SUMMARY

The discussion focuses on the calculation of volumes of revolution using both the disk method and the shell method, specifically for the region under the curve y = 1/x from x = 1 to x = 4. The disk method yields a volume of 3π/4, while the shell method, when correctly applied, results in a volume of π. The confusion arose from an incorrect interpretation of the shell's dimensions, particularly the length of the shell, which needed to be adjusted for values of y less than 1/4. Ultimately, the correct approach involves breaking the shell integral into two parts to accurately account for the geometry of the solid.

PREREQUISITES
  • Understanding of calculus concepts, particularly integration
  • Familiarity with the disk method for calculating volumes of revolution
  • Knowledge of the shell method for volumes of revolution
  • Ability to interpret and manipulate definite integrals
NEXT STEPS
  • Study the application of the disk method in different scenarios
  • Learn how to derive volumes using the shell method with varying functions
  • Explore the concept of breaking integrals into segments for complex shapes
  • Practice problems involving volumes of revolution for various curves
USEFUL FOR

Students and educators in calculus, particularly those focusing on volumes of revolution, as well as anyone seeking to clarify the differences and applications of the disk and shell methods in integration.

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...
 
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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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