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anonymous

  • 5 years ago

what is surface of revolution of y=(x^2+1)^(1/2), [0<x<3]

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  1. anonymous
    • 5 years ago
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    I think they are trying to say find the integral over 0 to 3.

  2. anonymous
    • 5 years ago
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    your ksaa

  3. anonymous
    • 5 years ago
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    there is no such thing as the integral of (2x^2+1)^(1/2)

  4. anonymous
    • 5 years ago
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    am i right?

  5. anonymous
    • 5 years ago
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    I am not sure what you are working on in class. Usually when I see questions like this they something like what is the volume of the solid revolved around a certain x or y. But you can integrate this, let u=2x^2+1 then when you get dx you can solve for u and substitute for x.

  6. anonymous
    • 5 years ago
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    yes that is ture but, the du doesnt cancel out the x that is there cause of the equation

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