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Integration Problem

  1. Jan 15, 2008 #1
    y = 2pi Int dz int(1-r^2)*r dr and the limits are (0,1) (note int = Integral)

    I was trying to integrate it by parts for the second part and I think I am not doing it correctly. Which way can I apply for it.
     
  2. jcsd
  3. Jan 15, 2008 #2

    quasar987

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    Try a change of variable.
     
  4. Jan 15, 2008 #3

    HallsofIvy

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    Which limits are 0, 1? Both? If this is
    [tex]2\pi \int_0^1 dz \int_0^1 (1- r^2)r dr[/tex]
    I see no reason to integrate by parts. The first integral is
    [tex]\int_0^1 dz= \left[ z\right]_0^1= 1[/tex]
    and the second is
    [tex]\int_0^1 (1-r^2)r dr[/tex]
    Let [itex]u= 1-r^2[/itex]. Then du= -2r dr. When r= 0, u= 1 and when r= 1, u= 0.
    [tex]\int_1^0 u(-(1/2)du= \left{(-1/2)(u^2/2)\right]_1^0= (-1/2)(0- 1)= 1/2[/itex]
    [tex]2\pi\int_0^1 dz \int_0^1 (1-r^2)r dr= 2\pi (1)(1/2)= \pi[/tex]
     
  5. Jan 15, 2008 #4

    quasar987

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    (I assumed the OP meant r/(1-r^2) for the integrand. Otherwise, there is no need to use any "integration technique" for this integral.)
     
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