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Unformatted text preview: 0 centered at , and by V , the region between S ± and S , as shown in Fig. 1. Figure 1: A ﬁgure showing the domain of integration. 1. Show g ( x ) is harmonic. 2. Apply the formula Z V f ∇ 2 gg ∇ 2 f d V = Z ∂v ± f ∂g ∂ng ∂f ∂n ² d A to obtain Z S ± f ∂g ∂ng ∂f ∂n ² d A = Z S ± ± f ∂g ∂ng ∂f ∂n ² d A ? 3. Observe that Z S ± f ∂g ∂ng ∂f ∂n ² d A =1 a 2 Z S f d A1 a Z S ∂f ∂n d A and similarly for the right hand side of ? with a replaced by ± because ∂g ∂n =1 r 2 . 2 Winter 2010 Math 255 4. Use the divergence theorem to show that Z S ∂f ∂n d A = Z S ± ∂f ∂n d A = 0 so equation ? reduces to1 a 2 Z S f d A =1 ± 2 Z S ± f d A ? ?. 5. Now obtain the average value property for f by taking the limit in ?? as ± → 0. 3...
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This note was uploaded on 12/31/2011 for the course MATH 255 taught by Professor Jackwaddell during the Fall '08 term at University of Michigan.
 Fall '08
 JackWaddell
 Math, Vector Calculus

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