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amistre64

  • 5 years ago

What is the volume of rotation of; 1/x; from [1,inf) when spun around the x axis ?

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  1. amistre64
    • 5 years ago
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    \[\pi \int\limits_{1}^{\infty} [1/x]^2 dx\]

  2. anonymous
    • 5 years ago
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    Since the curve is being rotated about the x axis, we can see that the radius of the inverted funnel so formed will be in the direction of the y axis. Now, since the funnel is never ending, it will not terminate at any one point and so we can take the integral of the area of the base of the funnel from 0 to infinity. This will give us the required volume. \[\int\limits_{1}^{\infty} \pi*y ^{2} dx\]\[= \pi \int\limits_{0}^{\infty} (1/x ^{2}) dx\]\[= \pi\] , when we solve the integral.

  3. anonymous
    • 5 years ago
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    1 to infinity in the 2nd integral*

  4. amistre64
    • 5 years ago
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    it is pi :) but that looks copied and pasted :) can you expand it out?

  5. anonymous
    • 5 years ago
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    what looks copied and pasted? we solve questions similar to this in my calculus class.. integral (1/x^2) = [-1/x] now in the limits 1 to infinity we get [0 - (-1)] * pi = pi.

  6. amistre64
    • 5 years ago
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    the breaks between the sentences; and the exact wording from the website I saw this on is a dead give away :)

  7. anonymous
    • 5 years ago
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    :O

  8. amistre64
    • 5 years ago
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    the solution is the same regardless of where the info is retrieved tho so kudos ;)

  9. anonymous
    • 5 years ago
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    lol alright

  10. amistre64
    • 5 years ago
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    \[F(x) = \pi \int\limits_{0}^{\infty} x^{-2} dx \rightarrow F(x)=\pi. \frac{-1}{x}\]

  11. amistre64
    • 5 years ago
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    \[F(\infty) = \pi.0 = 0\]\[F(1) = \pi.-1 = -\pi\]

  12. amistre64
    • 5 years ago
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    \[F(\infty) - F(1) = 0 -(-\pi) = \pi\]

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