Question

Using the washer/disk method. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid and a typical disk or washer. xy=1, y=0, x=1 about x=-1

Answer 1

Hello there, finally some Calculus.

Answer 2

Are you given a function to do this with? Or is it just points?

Answer 3

function seems to be y=1/x imo...

Answer 4

just xy=1, y=0, x=1 and x=2 and to rotate it about x=-1. I though of maybe looking at it as y=1/x

Answer 5

Oh, thank you for pointing that out, okay. Now we can go about this.

Answer 6

@kelberry :good luck! ;)

Answer 7

So, as I recall, the washer/disk method is \[\Pi \int\limits_{a}^{b}(f(x)-g(x))^2dx\]

Answer 8

Well, that's for washers, but for disks it's just \[f(x)^2\]

Answer 9

So \[(\frac{ 1 }{ x })^2 = \frac{ 1 }{ x^2 }\]

Answer 10

Oh, wait you mentioned an axis.

Answer 11

yes it has to rotate about x=-1

Answer 12

So, instead we do \[(\frac{ 1 }{ x } + 1)^2\]

Answer 13

Oh, wait how tricky, uhh, we're rotating about the x axis, so we need it in terms of y, so it should be 1/y not 1/x

Answer 14

Sorry, I haven't done this in a few months.

Answer 15

It is easier if you shift the graph 1 unit to the right, then you can rotate about the y-axis.

Answer 16

\[Pi*\int\limits_{0}^{1}(\frac{ 1 }{ y } + 1)^2dy\] Hm, let's see if this is right.

Answer 17

How does that look?

Answer 18

I think that would be right.

Answer 19

No, there is a problem with y=0...

Answer 20

Oh my, you're correct, haha.

Answer 21

Ah, I see, I misread the second portion. I understand now.

Answer 22

\[\int\limits_{a}^{b}((1-\frac{ 1 }{ y })+1)^2dy\]

Answer 23

Split up the area: one is the integral\[\pi \int\limits_{\frac{ 1 }{ 2 }}^{1}\left( \frac{ 1 }{ y }+1 \right)^2dy\]Anthe the bottom part is just a hollow cylinder, so you don't need an integral.