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Question Number 215535 by MrGaster last updated on 10/Jan/25

$${\int }_{\sum _{1\le i\le n}{x}_i^2\le R^2}\sum _{1\le i\le n}{x}_i\frac{\partial f}{\partial x_i}\prod _{1\le i\le n}{dx}_i=?$$

Answered by MrGaster last updated on 10/Jan/25

$${\int }_0^{R}rdr{\int }_{\sum _{1\le i\le n}{x}_i^2-r^2}\sum _{1\le i\le n}\frac{x_i}{r}\frac{\partial f}{\partial x_i}dS$$$$={\int }_0^{R}rdr{\int }_{\sum _{1\le i\le n}{x}_i^2=r^2}⟨n,▽f⟩dS$$$$={\int }_0^{R}rdr{\int }_{\sum _{1\le i\le n}{x}_i^2\le r^2}{▽}^{2}f\prod _{1\le i\le n}{dx}_i={\int }_0^{R}rdr{\int }_{\sum _{1\le i\le n}{x}_i^2\le r^2}g\left(\sum _{1\le i\le n}{x}_i^2\right)\prod _{1\le i\le n}{dx}_i$$$$\mathrm{Let}\sum _{1\le i\le n}{x}_i^2=u^2\Rightarrow \prod _{1\le i\le n}{dx}_i=d{\int }_{\sum _{1\le i\le n}{x}_i^2=u^2}\prod _{1\le i\le n}{dx}_i=\frac{2{\pi }^{\frac{n}{2}}{u}^{n-1}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}du$$$${\int }_{\sum _{1\le i\le n}{x}_i^2\le R^2}\sum _{1\le i\le n}{x}_i\frac{\partial f}{\partial x_i}\prod _{1\le i\le n}{dx}_i=\frac{2{\pi }^{\frac{n}{2}}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{R}rdr{\int }_0^{r}g\left(u^2\right)u^{n-1}du$$

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