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Thread: triple integral

  1. #1
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    triple integral

    Hey.

    What's the answer for this:

    $\displaystyle \int_{0}^{r}dr \int_{0}^{2\pi}rd\phi \int_{0}^{l}dl \rho (1-r/R)$
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  2. #2
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    $\displaystyle \rho l2\pi r^{2}[\frac{1}{2}-\frac{r}{3R}]$
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  3. #3
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    Quote Originally Posted by jacek View Post
    Hey.

    What's the answer for this:

    $\displaystyle \int_{0}^{r}dr \int_{0}^{2\pi}rd\phi \int_{0}^{l}dl \rho (1-r/R)$
    Actually, that is a very badly posed integral. It is never a good idea to use the same letter to represent the variable of integration and a limit of integration and that is done twice here. Other than that, it's pretty close to trivial. We can "separate" those integrals and do them separately:
    $\displaystyle \left(\int_0^r (1- r/R)r dr\right)\left(\int_0^{2\pi} d\phi\right)\left(\int_0^l dl\right)$.

    And, of course, $\displaystyle \int_0^{2\pi} d\phi$ and $\displaystyle \int_0^l dl$ are trivial!

    $\displaystyle \int_0^r (1- r/R)r dr= \int_0^r rdr- \frac{1}{R}\int_0^r r dr$ should also be easy though not "trivial".
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