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Thread: Yet another triple integral..

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face Yet another triple integral..

    This is certainly not my favorite calculus topic.

    $\displaystyle \int_{-7}^7\int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}}\int_{-\sqrt{49-x^2-y^2}}^{\sqrt{49-x^2-y^2}} (x^2z+y^2z+z^3)dzdxdy$

    I am to solve this by converting to spherical coordinates.
    I guessed the region of integration was the volume of the sphere $\displaystyle x^2+y^2+z^2=49$.

    So I had:

    $\displaystyle \int_0^{2\pi}\int_0^{\pi}\int_0^7{\rho}^5cos{\phi} sin{\phi}d{\rho}d{\theta}d{\phi}$

    Went about solving, and..ended up with 0.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    This is certainly not my favorite calculus topic.

    $\displaystyle \int_{-7}^7\int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}}\int_{-\sqrt{49-x^2-y^2}}^{\sqrt{49-x^2-y^2}} (x^2z+y^2z+z^3)dzdxdy$

    I am to solve this by converting to spherical coordinates.
    I guessed the region of integration was the volume of the sphere $\displaystyle x^2+y^2+z^2=49$.

    So I had:

    $\displaystyle \int_0^{2\pi}\int_0^{\pi}\int_0^7{\rho}^5cos{\phi} sin{\phi}d{\rho}d{\theta}d{\phi}$

    Went about solving, and..ended up with 0.
    The bounds are correct and the answer can be zero.
    The integrand is

    $\displaystyle x^2z+y^2z+z^3=z(x^2+y^2+z^2)=(\rho\cos\phi)(\rho^2 )$

    Multipling that by the jacobian and then breaking this into 3 integrals we have

    $\displaystyle \int_0^{2\pi}d{\theta}\int_0^{\pi}\cos{\phi}\sin{\ phi}d{\phi}\int_0^7{\rho}^5 d{\rho}=0 $

    since $\displaystyle \int_0^{\pi}\cos{\phi}\sin{\phi}d{\phi}=0 $
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by matheagle View Post
    The bounds are correct and the answer can be zero.
    The integrand is

    $\displaystyle x^2z+y^2z+z^3=z(x^2+y^2+z^2)=(\rho\cos\phi)(\rho^2 )$

    Multipling that by the jacobian and then breaking this into 3 integrals we have

    $\displaystyle \int_0^{2\pi}d{\theta}\int_0^{\pi}\cos{\phi}\sin{\ phi}d{\phi}\int_0^7{\rho}^5 d{\rho}=0 $

    since $\displaystyle \int_0^{\pi}\cos{\phi}\sin{\phi}d{\phi}=0 $
    Oh good! I was at least on the right track for this one, an improvement..
    Would it be wrong to take:
    $\displaystyle 2\int_0^{2\pi}\int_0^{\pi/2}\int_0^7{\rho}^5cos{\phi}sin{\phi}d{\rho}d{\thet a}d{\phi}$
    To obtain a positive volume for this?

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  4. #4
    MHF Contributor matheagle's Avatar
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    NO, that's not correct. You need symmetry in BOTH the region and the integrand
    This isn't volume, that's the integral of ONE.

    $\displaystyle 1> 0$ hence $\displaystyle \int\int\int 1dV \ge \int\int\int 0dV =0$

    BUT your integrand isn't strictly positive $\displaystyle z\rho^2$ can be both positive and negative and it's balanced in the sense that the positive and negative parts of z cancel each other off to give you the answer of ZERO.
    Last edited by matheagle; Jun 5th 2010 at 02:25 PM.
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