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Math Help - Divergence Theorem limits

  1. #1
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    Divergence Theorem limits

    F = 2xz i + yz j + {z}^{ 2} k over the upper half of the sphere {x}^{2 } + {y}^{2 } + {z}^{2 } = {a}^{ 2}

    I thought it was dz d\theta z between a and 0 and \theta between \pi and 0 , but im wrong
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    F = 2xz i + yz j + {z}^{ 2} k over the upper half of the sphere {x}^{2 } + {y}^{2 } + {z}^{2 } = {a}^{ 2}

    I thought it was dz d\theta z between a and 0 and \theta between \pi and 0 , but im wrong
    I guess I am not sure what you are asking.

    Do you want to compute the flux out of the top half of the sphere using the divergence theorem?

    \iint_{S}\mathbf{F}\cdot d\mathbf{S}=\iiint_{V}\nabla \cdot \mathbf{F}dV

    If so you end up with the integral

    5\iiint_{V}zdV

    If you convert to spherical coordinates you get

    5 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{a} (\rho \cos(\theta))\rho^2\sin(\theta)d\rho d\theta d\phi

    Note that phi is the plane angle and theta is the angle measured from the positive z axis
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  3. #3
    Ant
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    Change to spherical polar coordintes would be my best bet. Then integrate wrt theta, phi and r with limits:

    theta [0, pi/2]
    phi [0, 2pi]
    r [0, a]

    with the Jacobian being r^2 sin(theta)

    that's just how I'd try it though!!
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
     (\rho \cos(\theta))\rho^2\sin(\theta)d\rho d\theta d\phi
    How do you get this, i have got the 5z?
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  5. #5
    Ant
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    Polar coords are:

     x = rsin(\theta)cos(\phi)
     y = rsin(\theta)sin(\phi)
     z = rcos(\theta)

    The Jacobian is

     r^2  sin(\theta)


    The Jacobian is used when changing from x, y, z into polar coordinates.

    So

     5x dx dy dz

    becomes

     5 rcos(\theta)     r^2  sin(\theta)
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  6. #6
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    Quote Originally Posted by Ant View Post
    The Jacobian is

     r^2  sin(\theta)
    Is the Jacobian always that?
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  7. #7
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    Quote Originally Posted by adam_leeds View Post
    Is the Jacobian always that?
    Can i use this for a cylinder?
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  8. #8
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    Quote Originally Posted by adam_leeds View Post
    Can i use this for a cylinder?
    The Jacobian in cylindrical coordinates

    r dr d\phi dz

    This can be calculated

    x = r\cos(\phi) \quad y = r\sin(\phi) \quad z= z

    This gives

    \begin{vmatrix} \cos(\phi) & \sin(\phi) & 0 \\ -r\sin(\phi) & r \cos(\phi) & 0 \\ 0 & 0 & 1 \end{vmatrix} = r\cos^2(\phi)+r\sin^2(\phi)=r
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