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 August 19th, 2010, 03:39 PM #1 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 surface integral Hi; Let S be a sphere of radius r and let p be a fixed point inside or outside the sphere but not on it. Show that $\int \!\!\! \int_D \frac{1}{||x - p||} \,\,dxdy$ gives $4 \pi r$ if p is inside S $4 \pi r^2/d$ if p is outside S Thanks...
 August 21st, 2010, 10:55 AM #2 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: surface integral Hi, I am sorry, I would like to help you but you should learn first to exactly formulate your problem. For example it is completely nonsense that p is a point ($p\in \mathbb{R}^n$) and then you write something like $|x-p|$ where you use $dx$ in the integrand. You are talking about a "sphere" which can be defined in any dimension, definition $R^2=\sum_{i=1}^d (x_i)^2$ where $d\in \mathbb{N}$ ,$x_i\in\mathbb{R},R\in V(\mathbb{R}^n)$. What is D? I believe I know what you want but please help with a correct definition. A hint in the forefront: Gauß. Regards, Jens

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# surface integral 4pir, 4pir^2/d

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