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Flux through a surface

  1. Jun 18, 2006 #1
    Hi all,

    Let [itex]D[/itex] be the top half of a ball of radius [itex]a>0[/itex] and let

    [itex]\mathbf{f} = xz \mathbf{i} + y \mathbf{j} + x \mathbf{k}[/itex]

    Calculate the outward flux across S using the definition of the surface integral.

    [itex]\int_{S} \mathbf{f} \cdot \mathbf{n} dS = \int_{cap} \mathbf{f} \cdot \mathbf{n} dS + \int_{disc} \mathbf{f} \cdot \mathbf{n} dS[/itex]

    [itex]\frac{1}{a}\int_{cap} (xz,y,x) \cdot (x,y,z) d S[/itex]
    [itex]\frac{1}{a}\int_{cap} x^2z + y^2 + xz d S[/itex]
    [itex]\frac{1}{a}\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^2\theta\cos\theta + r^2\sin^2\varphi\sin^2\theta + r^2\cos\varphi\sin\theta\cos\theta) (a^2 \sin\theta) d\theta d\varphi[/itex]
    [itex]a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\sin^2\varphi\sin^3\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi[/itex]

    [itex]a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi[/itex]

    I assume that one would then evaluate the integral by parts. But is there an easier method?
     
    Last edited: Jun 18, 2006
  2. jcsd
  3. Jun 18, 2006 #2
    I figured it out using integration by substitution.
     
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