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I would like to integrate over the following surface: $\Omega=\{(v_1,\dots,v_n):\sum_{i=1}^N\phi(|v_i|^2)=N, \sum_{i=1}^N v_i=p, v_i\in R^3,p \in R^3\}$.

If $\phi(|v|^2)=|v|^2$, it is easy to see that $\Omega$ is a sphere, but for a general $\phi$ the surface measure on $\Omega$ is a complicated expression.

Is there a suitable change of variables to transform $\Omega$ to a sphere?

user177197
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1 Answers1

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perhaps this could work: $v_i=\frac{t_i}{||t_i||}\sqrt{\phi^{-1}{(||t_i||^2)}}$ , not sure though since it's only a week i've started m.v.c. Proof: $$ \Omega=\{(v_1,\dots,v_n):\sum_{i=1}^N\phi(||v_i||^2)=N, \sum_{i=1}^N v_i=p, v_i\in R^3\}=\\ =\{(t_1,\dots,t_n):\sum_{i=1}^N\phi(||\frac{t_i}{||t_i||}\sqrt{\phi^{-1}{(||t_i||^2)}}||^2)=N, \sum_{i=1}^N \frac{t_i}{||t_i||}\sqrt{\phi^{-1}{(||t_i||^2)}}=p, t_i\in R^3\}=\\ =\{(t_1,\dots,t_n):\sum_{i=1}^N\phi(\phi^{-1}{(||t_i||^2)})=N, \sum_{i=1}^N \frac{t_i}{||t_i||}\sqrt{\phi^{-1}{(||t_i||^2)}}=p, t_i\in R^3\}=\\ =\{(t_1,\dots,t_n):\sum_{i=1}^N||t_i||^2=N, \sum_{i=1}^N \frac{t_i}{||t_i||}\sqrt{\phi^{-1}{(||t_i||^2)}}=p, t_i\in R^3\} $$

Frank
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