## zerosniper123 Group Title 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...:( one year ago one year ago

1. ByteMe Group Title

you only need the part about volume?

2. ByteMe Group Title

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3. ByteMe Group Title

so: $$\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx$$

4. zerosniper123 Group Title

Okay but do I just keep on taking the integral of that? Or do I just leave it as is.

5. ByteMe Group Title

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.

6. zerosniper123 Group Title

Okay how do I evaluate it at infinity?

7. ByteMe Group Title

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....

8. zerosniper123 Group Title

Okay then I just get -4(pi)/b^3

9. TuringTest Group Title

try that integration again$\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx$

10. zerosniper123 Group Title

$\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx = \pi[ \lim_{x\rightarrow\infty} [ - x^{-1}] - (-(1)^{-1})] =\pi( 0+1) = \pi$