Here's the question you clicked on:
zerosniper123
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...:(
you only need the part about volume?
so: \(\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx \)
Okay but do I just keep on taking the integral of that? Or do I just leave it as is.
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.
Okay how do I evaluate it at infinity?
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....
Okay then I just get -4(pi)/b^3
try that integration again\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx\]
Is the answer -b^-2(pi)
\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx = \pi[ \lim_{x\rightarrow\infty} [ - x^{-1}] - (-(1)^{-1})] =\pi( 0+1) = \pi\]