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Homework Help: Surface Integral over a Hemisphere (Work check please! I end up with zero!)

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Seawater has density 1025 kg/m^3 and flows in a velocity field v=yi+xj, where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the hemisphere x^2+y^2+z^2=9, z≥0

    2. Relevant equations
    Surface integral of F over S is ∫∫ F • dS
    In this case,

    p * ∫∫s F • n dS

    Where n = the cross product between rtheta and rphi.

    3. The attempt at a solution
    First, I parameterized the surface:

    [tex]\vec r(\theta,\phi) = \langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi \rangle[/tex]
    Where 0 < theta < 2pi and 0 < phi < pi/2.

    Partial with respect to theta:
    [tex]\vec r_{\theta}(\theta,\phi) = \langle -3\sin\phi\sin\theta, 3\sin\phi\cos\theta, 0 \rangle[/tex]

    Partial with respect to phi:
    [tex]\vec r_{\phi}(\theta,\phi) = \langle 3\cos\phi\cos\theta, 3\cos\phi\sin\theta, -3\sin\phi \rangle[/tex]

    [tex]\vec r_{\phi}(\theta,\phi) \times r_{\theta}(\theta,\phi) = \langle 9\sin^{2}\phi\cos\theta, 9\sin^{2}\phi\sin\theta, 9\cos\phi\sin\phi \rangle[/tex]

    Next, I look at the velocity field and grab the velocity vector:
    [tex]\vec v = \langle 3\sin\phi\sin\theta,3\sin\phi\cos\theta,0 \rangle[/tex]

    I am now set to integrate:
    [tex]\int\int_S \delta\vec v \cdot d\vec S = \int\int_{(\phi,\theta)} \delta\vec v \cdot \vec r_\phi \times \vec r_\theta\ d\phi d\theta[/tex]

    [tex]\int^{2\pi}_{0}\int^{\pi/2}_{0} (1025)* \langle 3\sin\phi\sin\theta,3\sin\phi\cos\theta,0 \rangle \cdot \langle 9\sin^{2}\phi\cos\theta, 9\sin^{2}\phi\sin\theta, 9\cos\phi\sin\phi \rangle d\phi d\theta[/tex]

    After the dot product I end up with....
    27sin^3(phi)cos(theta)sin(theta) + 27sin^3(phi)cos(theta)sin(theta)
    which, I simpify to:

    Split the integral in two.
    [tex](1025)*54*(\int^{2\pi}_{0} \cos\theta\sin\theta d\theta) (\int^{pi/2}_{0} \sin_^{3}\phi d\phi[/tex]

    Trig Identity Substitution:
    [tex](1025)*27*(\int^{2\pi}_{0} -1/2\sin2\theta d\theta) (\int^{pi/2}_{0} (1-\cos\phi^{2})\sin\phi d\phi[/tex]

    So, I end up with...

    [tex]1025*27*((-1/2\cos\theta)^{2\pi}_{0}) ( \cos^{3}\theta/3-\cos\theta)^{\pi/2}_{0})[/tex]

    Giving me...
    [tex]1025*27*(0) ( \cos^{3}\theta/3-\cos\theta)^{\pi/2}_{0}) = 0[/tex]

    Last edited: Apr 28, 2010
  2. jcsd
  3. Apr 29, 2010 #2


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    Science Advisor

    Yes, that is correct (and nice work!), the total flow through the hemisphere is 0. The flow is "anti- symmetric" about the z- axis. The flow through a given point, (x, y, z), on the hemisphere is canceled by the flow through (-x, -y, z).
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