1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook