# Proving the centre of mass formula with integral [closed]

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.

## closed as off-topic by sammy gerbil, John Rennie, Kyle Kanos, Jon Custer, M. EnnsDec 16 '17 at 4:16

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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 $f(r)$.

• I had never seen that integral identity before! Would you have any reference where I could read up on these/practice problems related to these identities ? Thanks – Matthew Dec 13 '17 at 12:19
• As I'm not quite sure why del dV = dS – Matthew Dec 13 '17 at 12:22
• You will find informations about such identity by seaching thinks like "vector analysis". Here, you can see the Gauss theorem for divergence: en.wikipedia.org/wiki/Divergence_theorem – kryomaxim Dec 13 '17 at 12:26
• video lectures are also more comprehensive than text khanacademy.org/math/multivariable-calculus/… – kryomaxim Dec 13 '17 at 12:27