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Solids of revolution

  1. Mar 10, 2010 #1
    Hi,

    I have the area [tex]D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1[/tex]

    That is rotated about the x axis, and i need to calculate the area

    [tex]\pi \int_0^1 3^2-y^2 = \pi \int_0^1 9-xe^{2x^2}[/tex]

    [tex]\frac{-9\pi}{4}\cdot (e^{2x^2}-1)\bigg|_0^1[/tex]

    But this is all wrong, why?
     
    Last edited: Mar 10, 2010
  2. jcsd
  3. Mar 10, 2010 #2

    tiny-tim

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    Hi James889! :smile:

    First, are you trying to find the area of a surface of revolution, or the volume of a solid of revolution? :confused:

    Anyway, how did you get 9/4 out of that?

    Do the two parts separately (you seem to be suffering from a sort of human-fly syndrome :rolleyes:).
     
  4. Mar 10, 2010 #3
    Hi Tim,

    Im trying to find the area of a solid.
    I just factored out the 9 from the integral, the [tex]1/4[/tex] is from integrating [tex]xe^{2x^2}[/tex]
     
  5. Mar 10, 2010 #4

    tiny-tim

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    Yes, but how did they get toegther? :confused:

    Anyway
    You mean the surface area?

    But you're using ∫πy2dx, which is a volume.

    For the correct formula, see http://en.wikipedia.org/wiki/Surface_of_revolution.
     
  6. Mar 10, 2010 #5
    I am so bad at this :tongue:
     
  7. Mar 10, 2010 #6

    tiny-tim

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    Have you got it now?

    If not, show us what you have so far. :smile:
     
  8. Mar 10, 2010 #7
    Turned out i had misread the question, they did ask for the volume of the solid
    [tex]
    D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1
    [/tex]

    Same as before, baby steps.
    [tex]\pi\int_0^1 9-(\sqrt{x}e^{x^2})^2[/tex]

    [tex]9x-\frac{e^{2x^2}}{4}\bigg|_0^1[/tex]

    [tex]\pi\cdot\frac{36-e^2}{4} - (0-\frac{e}{4})[/tex]
     
  9. Mar 10, 2010 #8

    tiny-tim

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    Fine, except the last term should be e0/4, = 1/4. :wink:

    (and use more brackets, to show you have the π in the right place)
     
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