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Math Help - Quick Integration Check please!

  1. #1
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    Quick Integration Check please!

    Hi can you just check that I have this integration correct?

    (1 - r^2/R)

    Solution

    ( r - r^3/3)

    Cheers!
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  2. #2
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    e^(i*pi)'s Avatar
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    What is R and how is it related to r? Convention dictates that variables are case sensitive

    \displaystyle \int 1- \dfrac{r^2}{r} dr = \int (1-r) dr

    If R is some constant:

    \displaystyle \int 1- \dfrac{r^2}{R} dr = \int 1 - ar^2 where  a = \dfrac{1}{R}


    edit: you can check integration by differentiating the result and since \dfrac{d}{dr} \left(r - \dfrac{r^3}{3}\right) = 1-r^2 it's not correct
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  3. #3
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    I would get a different answer to yours but I'm not certain what R is....
    My answer: r - (r^3)/3*R
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  4. #4
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    Sorry Ill explain myself better. The question relates to charge distribution that is spherically symmetrical. This charge distribution is over the range:

    A(1 - r/R) for r< R

    and

    0 for r > R

    Where A is a constant

    Making a few assumptions about the charge distribution we get:

    Q = A( 1 - r/R) 4πr^2* δr = 4πA(1 - r^3/R)δr

    π = pi

    So to find the total charge over the whole of sphere we take integral from R to 0:

    Q = 4πA(1 - r^3/R) = 4πA [This is the bit i am obviously getting wrong!] = ??????

    Sorry about using the correct symbols but i don't seem to be able to cut and paste from word into the forum
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