anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
drawing it is always a good idea, especially if you are a visual thinker; but it is not obligatory.....
anonymous
  • anonymous
@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.
anonymous
  • anonymous
Please check for me.

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anonymous
  • anonymous
@bibby are you aware of the topic?
anonymous
  • anonymous
;(
anonymous
  • anonymous
Saw you browsing the calc section so I thought you knew.
anonymous
  • anonymous
@timo86m would you happen to know?
anonymous
  • anonymous
@IrishBoy123
anonymous
  • anonymous
@SolomonZelman
anonymous
  • anonymous
Can you draw it?
anonymous
  • anonymous
Do you not see the attachment?
anonymous
  • anonymous
Okay... no need to get so edgy :(
anonymous
  • anonymous
Didn't mean it like that, I thought you weren't able to see it, I was gonna reupload for you :)
anonymous
  • anonymous
No wonder you aren't asked to evaluate... Well, anyway, you got the curves right, you just shaded the wrong part ^^
anonymous
  • anonymous
That's what I wanted to know, which part I am looking for
anonymous
  • anonymous
So I am basically looking for the part on top of it correct?
anonymous
  • anonymous
If so then it would be \[2-\ln(x)\]
anonymous
  • anonymous
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:
1 Attachment
anonymous
  • anonymous
Yes, exactly ^^
anonymous
  • anonymous
Just shaded that.
anonymous
  • anonymous
So, no more problems, then? No problem setting up the bloody integral? :D
anonymous
  • anonymous
So it should be \[\int\limits_{0}^{2} \pi (2-\ln(x))^2\]
anonymous
  • anonymous
Right?
anonymous
  • anonymous
Since y=2 is at the top
anonymous
  • anonymous
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
anonymous
  • anonymous
Oops, misread that. I thought x was equal to 2
anonymous
  • anonymous
:) And don't forget the dx at the end of the integral.
anonymous
  • anonymous
So integral setup right (aside from no dx)?
anonymous
  • anonymous
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:
anonymous
  • anonymous
Yeah, I noticed that too but since I'm not solving it, I ignored it.
anonymous
  • anonymous
@PeterPan can you look over some of my other problems?
anonymous
  • anonymous
Sure ^^
anonymous
  • anonymous
Tagged you. Thanks

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