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Thread: Triple integral

  1. #1
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    Triple integral

    Please help with the following integral calculation:
    $\displaystyle \int\!\!\!\!\int\!\!\!\!\int_V e^{{(x^2+y^2+z^2)}^{\frac{3}{2}}}dxdydz$ where $\displaystyle V=\{(x,y,z)|x^2+y^2+z^2\leq1\}$
    I tried switching to spherical coordinates:
    $\displaystyle x=r\cos\varphi\sin\theta$
    $\displaystyle y=r\sin\varphi\sin\theta$
    $\displaystyle z=r\cos\theta$
    $\displaystyle dV=r^2 \sin \theta dr d \theta d\varphi$
    with $\displaystyle 0\leq r\leq1$ but I'm not sure about the range of $\displaystyle \varphi$ and $\displaystyle \theta$
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by sillyme View Post
    Please help with the following integral calculation:
    $\displaystyle \int\!\!\!\!\int\!\!\!\!\int_V e^{{(x^2+y^2+z^2)}^{\frac{3}{2}}}dxdydz$ where $\displaystyle V=\{(x,y,z)|x^2+y^2+z^2\leq1\}$
    I tried switching to spherical coordinates:
    $\displaystyle x=r\cos\varphi\sin\theta$
    $\displaystyle y=r\sin\varphi\sin\theta$
    $\displaystyle z=r\cos\theta$
    $\displaystyle dV=r^2 \sin \theta dr d \theta d\varphi$
    with $\displaystyle 0\leq r\leq1$ but I'm not sure about the range of $\displaystyle \varphi$ and $\displaystyle \theta$

    $\displaystyle 0\leq\varphi\leq2\pi , 0\leq\theta\leq \pi ,0\leq\rho\leq1$ now it is correct


    Edited:- Spec show me mistake that I did
    Last edited by Amer; Jun 10th 2009 at 12:08 PM.
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  3. #3
    Senior Member Spec's Avatar
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    Actually, with those variables the limits are $\displaystyle 0\leq \varphi \leq 2\pi,\ 0\leq \theta \leq \pi$

    It can be a bit confusing since a lot of sources use different variables for the angles.
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  4. #4
    MHF Contributor Amer's Avatar
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    Usually I use $\displaystyle \phi$ instead of $\displaystyle \theta$ so I didn't look to what he write thanks to Spec
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