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anonymous
 3 years 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...:(
anonymous
 3 years 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...:(

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you only need the part about volume?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361170984139:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so: \(\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx \)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What 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
 3 years ago
Best ResponseYou've already chosen the best response.0Okay how do I evaluate it at infinity?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But 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
 3 years ago
Best ResponseYou've already chosen the best response.0Okay then I just get 4(pi)/b^3

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is the answer b^2(pi)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.0\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{2}dx = \pi[ \lim_{x\rightarrow\infty} [  x^{1}]  ((1)^{1})] =\pi( 0+1) = \pi\]
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