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Show that the area under the curve y=1/x from 1 to Infinity is not finite. Now rotate this curve about the xaxis and find the volume using the same interval.
Please help! Im lost at the 2nd part...:(
 one year ago
 one year ago
Show that the area under the curve y=1/x from 1 to Infinity is not finite. Now rotate this curve about the xaxis and find the volume using the same interval. Please help! Im lost at the 2nd part...:(
 one year ago
 one year ago

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ByteMeBest ResponseYou've already chosen the best response.2
you only need the part about volume?
 one year ago

ByteMeBest ResponseYou've already chosen the best response.2
so: \(\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx \)
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay but do I just keep on taking the integral of that? Or do I just leave it as is.
 one year ago

ByteMeBest ResponseYou've already chosen the best response.2
What do you mean "just keep on taking the integral..." ??? Evaluate \(\large \pi \int_1^{\infty}(\frac{1}{x})^2 dx \) will give you the volume.. a finite volume.
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay how do I evaluate it at infinity?
 one year ago

ByteMeBest ResponseYou've already chosen the best response.2
But that's what you did in the first part of this question when you proved that the area is not finite... Replace the infinity with b and take the limit of the integral as b approaches infinity....
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Okay then I just get 4(pi)/b^3
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
try that integration again\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{2}dx\]
 one year ago

zerosniper123Best ResponseYou've already chosen the best response.0
Is the answer b^2(pi)
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.0
\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{2}dx = \pi[ \lim_{x\rightarrow\infty} [  x^{1}]  ((1)^{1})] =\pi( 0+1) = \pi\]
 one year ago
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