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Simple divergence/Green's theorem question

  1. Nov 25, 2014 #1
    I'm exploring the divergence theorem and Green's theorem, but I seem to be lacking some understanding. I have tried this problem several times, and I am wondering where my mistake is in this method.

    The problem:
    For one example, I am trying to find the divergence of some vector field from a hemisphere. Let the hemisphere be given by $$x^2 + y^2 + z^2 = 9.$$ Also, the vector field in question is given by $$ \textbf{V} = \bigg(y,\hspace{2mm} xz,\hspace{2mm} 2z-1\bigg) $$
    Now, I want to evaluate the integral over the surface:
    $$\iint\textbf{V}\cdot\textbf{n}\hspace{2mm}d\sigma$$

    Attempt at a solution:
    Here is how I try to solve it. I instead use (by Green's theorem, where tau is a volume element) $$\iiint\nabla\cdot\textbf{V}\hspace{2mm}d\tau.$$
    Taking the gradient of the the vector field, I get 2 (only the z-hat component of the field will contribute). And since it is a simple hemisphere, I can integrate over the volume in spherical coordinates with the following limits:

    $$r \hspace{1mm}\epsilon\hspace{1mm}[0,3]$$
    $$\phi \hspace{1mm}\epsilon\hspace{1mm}[0,2\pi]$$
    $$\theta \hspace{1mm}\epsilon\hspace{1mm}[0,\pi/2]$$
    The Jacobian is standard for going from Cartesian to spherical coordinates: $$r^2 \hspace{1mm}sin(\theta)$$
    Lastly, evaluating this integral (and not forgetting to include the gradient of the vector field in the integral), I get $$36\pi$$
    The answer given in the text is $$27\pi$$ This is not a hard problem, and I am most certain that my integration and arithmetic is correct. There must be some fundamental step that I am missing.
     
  2. jcsd
  3. Nov 25, 2014 #2
    I think I have figured it out. If you consider the outward flux from the bottom of the hemisphere (ie. a disk), then that is equal to 9 Pi. The total flux through the curved surface is then (36 - 9) Pi, or 27 Pi
     
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