Volumes of Revolution

1. Jul 31, 2008

LHC

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: Jul 31, 2008
2. Jul 31, 2008

LHC

I just found out that I get the answer if I do this:

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

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?

3. Jul 31, 2008

Dick

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.

4. Jul 31, 2008

LHC

Ohhh....*LED above head suddenly flickers*...I get it. I had the wrong length of the shell! Thanks for explaining that to me. =D