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.

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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
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\[\pi \int\limits_{1}^{e^2} 4 - \ln(x)^2\]
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

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Other answers:

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

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