I came across a question:

Find $f(r)$ and prove the centre of mass formula:
$\vec{r_{cm}} = \frac{1}{V} \int f(r) \vec{dS} $
Where V is the total volume and our surface integral is over a body with uniform density.

I'm not even quite sure where to start. I spent a while fiddling around with the divergence theorem but to no avail. I think $f(r) =\frac{r^2}{2} $ but this is only a guess. Any hints would he great to get me started along the right track. Thanks


The intended formula suggests to use Gauss' divergence theorem. This theorem is about the flux of some vector field ${\bf v}$ through the boundary surface $\partial B$ of a body $B$. But your formula integrates a scalar function over $\partial B$, albeit in a "vectorial" way. The trick is to choose a fixed "test vector" ${\bf e}$, and to consider the vector field $${\bf v}({\bf x}):={|{\bf x}|^2\over2}{\bf e}={(x_1^2+x_2^2+x_3)^2\over2}(e_1,e_2,e_3)\ .$$ One computes $${\rm div}({\bf v})=x_1e_1+x_2e_2+x_3e_3={\bf x}\cdot{\bf e}\ .$$ The centroid ${\bf c}$ of $B$ is defined by the moment equation $${\bf c}\>V=\int_B{\bf x}\>{\rm d}({\bf x})\ ,\tag{1}$$ where $V$ denotes the volume of $B$. We now take the scalar product of $(1)$ with the vector ${\bf e}$ and apply Gauss' theorem: $$\eqalign{{\bf c}\>V\cdot{\bf e}&=\int_B{\bf x}\cdot{\bf e}\ {\rm d}({\bf x})=\int_B{\rm div}({\bf v})\>{\rm d}({\bf x})\cr &=\int_{\partial B}{\bf v}\cdot{\bf n}\>{\rm d}\omega=\int_{\partial B}{|{\bf x}|^2\over2}\>{\bf e}\cdot{\bf n}\>{\rm d}\omega\cr &=\left(\int_{\partial B}{|{\bf x}|^2\over2}\>{\bf n}\>{\rm d}\omega\right)\cdot{\bf e}\quad .\cr}$$ Since this is true for every "test vector" ${\bf e}$ it follows that the hoped for formula holds with $f({\bf x}):={1\over2}|{\bf x}|^2$.


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