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Thread: Divergence Theorem

  1. #1
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    Divergence Theorem

    Use the Divergence Theorem to find the flux of F across the surface sigma with outward orientation.

    F(x,y,z) = 2xzi + yzj + z^2k, where sigma is the surface of the solid bounded above by z = sqrt(a^2-x^2-y^2) and below by the xy-plane.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by noles2188 View Post
    Use the Divergence Theorem to find the flux of F across the surface sigma with outward orientation.

    F(x,y,z) = 2xzi + yzj + z^2k, where sigma is the surface of the solid bounded above by $\displaystyle z = \sqrt{a^2-x^2-y^2}$ and below by the xy-plane.
    $\displaystyle \text{div}(F) = 2z+z+2z=5z$

    So you want to evaluate $\displaystyle 5\int\int\int z\,dV$ bounded above by a sphere of radius $\displaystyle a$ and bounded below by the plane $\displaystyle z=0$.

    Use spherical coordinates (where $\displaystyle \rho$ is the radius, $\displaystyle \phi$ is the azimuth angle, and $\displaystyle \theta$ is the angle around the circle.
    $\displaystyle z=\rho\cos\phi$
    $\displaystyle dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$

    The bounds:
    $\displaystyle \rho=0..a$
    $\displaystyle \phi=0..\pi/2$
    $\displaystyle \theta=0..2\pi$

    So you want to evaluate $\displaystyle 5\int_0^{2\pi}\int_0^{\pi/2}\int_0^a \rho^3\sin(\phi)\cos(\phi)\,d\rho\,d\phi\,d\theta = \frac{5}{2}\int_0^{2\pi}\int_0^{\pi/2}\int_0^a \rho^3\sin(2\phi)\,d\rho\,d\phi\,d\theta$

    Evaluating this integral should be straightforward. I have included the final answer below.
    Spoiler:
    $\displaystyle \frac{5\pi a^4}{4}$
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