Question

Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 2, and x = 1 is revolved around the line y = −2.

Answer 1

drawing it is always a good idea, especially if you are a visual thinker; but it is not obligatory.....

Answer 2

@IrishBoy123 I have graphed it but I am not sure which part I am trying to find the area of. I feel like he boundaries are mixed.

Answer 3

Please check for me.

Answer 4

@bibby are you aware of the topic?

Answer 5

;(

Answer 6

Saw you browsing the calc section so I thought you knew.

Answer 7

@timo86m would you happen to know?

Answer 8

@IrishBoy123

Answer 9

@SolomonZelman

Answer 10

Can you draw it?

Answer 11

Do you not see the attachment?

Answer 12

Okay... no need to get so edgy :(

Answer 13

Didn't mean it like that, I thought you weren't able to see it, I was gonna reupload for you :)

Answer 14

No wonder you aren't asked to evaluate...
Well, anyway, you got the curves right, you just shaded the wrong part ^^

Answer 15

That's what I wanted to know, which part I am looking for

Answer 16

So I am basically looking for the part on top of it correct?

Answer 17

If so then it would be
\[2-\ln(x)\]

Answer 18

Yeah... what you shaded was bounded by y=-2 and not y=2
See how the shaded part doesn't part the y=2 line? ^^

It should be something that looks like this:

Answer 19

Yes, exactly ^^

Answer 20

Just shaded that.

Answer 21

So, no more problems, then?
No problem setting up the bloody integral? :D

Answer 22

So it should be \[\int\limits_{0}^{2} \pi (2-\ln(x))^2\]

Answer 23

Right?

Answer 24

Since y=2 is at the top

Answer 25

No ^^
The limits on the integral, they should go from 0 to 1 only, as does indeed run from the y-axis (x=0) to the line x=1

Answer 26

Oops, misread that. I thought x was equal to 2

Answer 27

:)
And don't forget the dx at the end of the integral.

Answer 28

So integral setup right (aside from no dx)?

Answer 29

There's just one thing that bothers me... it may well be the reason you're not asked to evaluate the integral, but...

y=ln(x) is not DEFINED at x=0
as in... there is no ln(0)

So can we really have a limit from 0 to 1?

D:

Answer 30

Yeah, I noticed that too but since I'm not solving it, I ignored it.

Answer 31

@PeterPan can you look over some of my other problems?

Answer 32

Sure ^^

Answer 33

Tagged you. Thanks