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.
If the density is uniform, the definition for the center of mass formula is:
$\vec{r}_{cm} = \frac{1}{V} \int \vec{r} dV = \frac{1}{V} \int 1 \vec{r} dV$.
Now you can use the rule for partial integration in three dimensions. By using
$\int \nabla(ab)dV = \int ab d \vec{S} = \int (b \nabla a + a \nabla b)dV$
and identifying $\nabla a = 1, b = \vec{r}, $ you get $a = b = \vec{r}$ and from above equation:
$\int \vec{r} dV = \int r^2 d \vec{S} - \int \vec{r} \nabla \vec{r}dV$.
The last term on the right hand side has the integrand $3 \vec{r}$ and you can put it on the left hand side. Then you will find $f(r)$.
