anonymous
  • anonymous
Show that the area under the curve y=1/x from 1 to Infinity is not finite. Now rotate this curve about the x-axis and find the volume using the same interval. Please help! Im lost at the 2nd part...:(
Mathematics
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anonymous
  • anonymous
Show that the area under the curve y=1/x from 1 to Infinity is not finite. Now rotate this curve about the x-axis and find the volume using the same interval. Please help! Im lost at the 2nd part...:(
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
you only need the part about volume?
anonymous
  • anonymous
|dw:1361170984139:dw|
anonymous
  • anonymous
so: \(\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx \)

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anonymous
  • anonymous
Okay but do I just keep on taking the integral of that? Or do I just leave it as is.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Okay how do I evaluate it at infinity?
anonymous
  • anonymous
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....
anonymous
  • anonymous
Okay then I just get -4(pi)/b^3
TuringTest
  • TuringTest
try that integration again\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx\]
anonymous
  • anonymous
Is the answer -b^-2(pi)
whpalmer4
  • whpalmer4
\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx = \pi[ \lim_{x\rightarrow\infty} [ - x^{-1}] - (-(1)^{-1})] =\pi( 0+1) = \pi\]

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