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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

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  1. ByteMe Group Title
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    you only need the part about volume?

    • one year ago
  2. ByteMe Group Title
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    |dw:1361170984139:dw|

    • one year ago
  3. ByteMe Group Title
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    so: \(\large V=\pi \int_1^{\infty}r^2dx=\pi \int_1^{\infty}(\frac{1}{x})^2dx \)

    • one year ago
  4. zerosniper123 Group Title
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    Okay but do I just keep on taking the integral of that? Or do I just leave it as is.

    • one year ago
  5. ByteMe Group Title
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    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
  6. zerosniper123 Group Title
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    Okay how do I evaluate it at infinity?

    • one year ago
  7. ByteMe Group Title
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    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
  8. zerosniper123 Group Title
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    Okay then I just get -4(pi)/b^3

    • one year ago
  9. TuringTest Group Title
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    try that integration again\[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx\]

    • one year ago
  10. zerosniper123 Group Title
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    Is the answer -b^-2(pi)

    • one year ago
  11. whpalmer4 Group Title
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    \[\pi\int_1^\infty\frac1{x^2}dx=\pi\int_1^ \infty x^{-2}dx = \pi[ \lim_{x\rightarrow\infty} [ - x^{-1}] - (-(1)^{-1})] =\pi( 0+1) = \pi\]

    • one year ago
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