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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katieb
  • katieb
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anonymous
  • anonymous
\[\pi \int\limits_{1}^{e^2} 4 - \ln(x)^2\]
anonymous
  • anonymous
okay so in know that, this would be the volume if it was spun around the x- axis just not sure what to do when its revolved around y = -2
anonymous
  • anonymous
@ganeshie8 @Hero @jim_thompson5910 any idea ?

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jim_thompson5910
  • jim_thompson5910
|dw:1438914029970:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438914089545:dw|
anonymous
  • anonymous
yep yep i know what it looks like, i've graphed it
jim_thompson5910
  • jim_thompson5910
|dw:1438914110494:dw|
jim_thompson5910
  • jim_thompson5910
you agree it's this region? |dw:1438914141386:dw|
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
we're going to rotate that region around the line y = -2 |dw:1438914218489:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438914231869:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438914302702:dw|
jim_thompson5910
  • jim_thompson5910
does that help?
anonymous
  • anonymous
it does tell me if this is right
jim_thompson5910
  • jim_thompson5910
in a sense, you'll have 2 cylinders cylinder A with radius 4 cylinder B with radius ln(x)+2
jim_thompson5910
  • jim_thompson5910
volume of washer = cylinder A - cylinder B
anonymous
  • anonymous
\[\pi \int\limits_{1}^{e^2}16 - (\ln(x) + 2)\]
anonymous
  • anonymous
is that right @jim_thompson5910
anonymous
  • anonymous
oh woops forgot to square the (ln(x) +2)
jim_thompson5910
  • jim_thompson5910
\[\Large \pi \int\limits_{1}^{e^2}16 - (\ln(x) + 2)^2\] is correct
anonymous
  • anonymous
thank's !!
jim_thompson5910
  • jim_thompson5910
np

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