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Can Someone Explain This Change of Variable?

  1. Sep 3, 2011 #1
    [tex] \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u(y)\,dS(y)=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz)\,dS(z) [/tex]

    Why does [tex]dS(y)\to dS(z)[/tex] and not [tex] dS(y)\to dS(x+rz)[/tex]?

    If you want more information, it comes from http://www.stanford.edu/class/math220b/handouts/laplace.pdf on page 8, it's used to prove the mean value formula for the laplacian.

    Is it because it doesn't matter what radius the surface we are integrating over is because we are taking the average?
     
    Last edited: Sep 3, 2011
  2. jcsd
  3. Sep 3, 2011 #2

    hunt_mat

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    The notation [itex]\partial B(x,r)[/itex] refers to the boundary of the ball of radius r with centre x, i.e. the sphere of radius r with centre x. All the change of variables is doing, is moving to the point x and rescaling so you're only integrating over the unit ball. Suppose that [itex]\mathbf{y}\mapsto x+r\mathbf{z}[/itex] with [itex]\| \mathbf{z} \| =1[/itex].

    So your question is just saying that the centre of the integration is just be been shifted.

    Good set of notes by the way.
     
  4. Sep 3, 2011 #3
    Does this only work because we are taking the average? Because if r>1, then dS(y) would be larger than dS(z), right? And so the left integral (without taking the average) would be larger than the right integral (without taking the average)...
     
  5. Sep 3, 2011 #4

    hunt_mat

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    You're integrating over the sphere, I think that as the radius if the sphere is constant, it can just be taken outside of the integral.
     
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