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Integrating & finding Volumes!

  1. Aug 6, 2005 #1
    hi felles.

    I am trying to find what is the volume of the [tex]y=\frac{a}{x^2}+b[/tex] is when it is rotated in y-axis.

    The values of a is 1 and b is -1.

    max hight is 3 and min is 0.

    I was trying to integrade and ended up with [tex]V=\frac{-Pi}{y^2+2Y+1}[/tex] where y is 3.

    Is this right?

    I did a U substitution to integrade [tex]x^2=\frac{1}{y+1}[/tex]

    Plz help.
  2. jcsd
  3. Aug 6, 2005 #2


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    It seems like your integral is wrong.
    [tex]\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C \mbox{, } \alpha \neq -1[/tex]
    [tex]\int \frac{1}{x} dx = \ln{x} + C [/tex]
    Viet Dao,
  4. Aug 6, 2005 #3


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    Looks good, or you can say:
    [tex]V = \pi \int^3_0 \frac{1}{y + 1} dy = \pi(\ln{4} - \ln{1}) = \pi \ln{4}[/tex]
    Viet Dao,
  5. Aug 6, 2005 #4

    Is it possible for Mathematica to generate a 3D image from a function rotated around an axis using the integration method?

    Please demonstrate some source code and a graphic?

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