Our students were tasked, using the divergence theorem $ \int_V \nabla \cdot \vec{F} \mathrm{d}^3 r = \int_{\partial V} \vec{F} \cdot \mathrm{d}\vec{a} $, to show that for any bounded, connected subset V of $\mathbb{R}^3$ with surface $\partial V$ the following identity holds:

$$\int_V \vec{r} \mathrm{d}^3r = \frac{1}{4} \int_{\partial V} \vec{r} \left( \vec{r} \cdot \mathrm{d}\vec{a} \right)$$

where $$ \vec{r} = \begin{pmatrix} x_1\\x_2\\x_3\\ \end{pmatrix}$$

Now, the sample solution considered the i-th component of the right hand side (using Einstein sum convention):

$$ \left( \frac{1}{4} \int_{\partial V} \vec{r} \left( \vec{r} \cdot \mathrm{d}\vec{a} \right) \right)_i = \frac{1}{4} \int_{\partial V} r_i r_j \mathrm{d}a_j = \frac{1}{4} \int_{\partial V} (r_i r_j) \mathrm{d}a_j \stackrel{Div. theorem}{=} \frac{1}{4} \int_V \partial_j (r_i r_j) \mathrm{d}^3r = \frac{1}{4} \int_V (\delta_{ij} r_j + 3r_i) \mathrm{d}^3r = \left( \int_V \vec{r} \mathrm{d}^3r \right)_i $$

On the other hand, many students used a corollary of the divergence theorem for gradient fields $ \int_V \nabla \phi \mathrm{d}^3 r = \int_{\partial V} \phi\mathrm{d}\vec{a} $ by realizing that for $\phi = \frac{1}{2} \left( x_1^2 + x_2^2 +x_3^2 \right) = \frac{1}{2} r^2 = \frac{1}{2} \left( \vec{r} \cdot \vec{r} \right) $ we find

$$ \int_V \vec{r} \mathrm{d}^3 r = \int_V \nabla \phi \mathrm{d}^3 r = \int_{\partial V} \phi\mathrm{d}\vec{a} = \frac{1}{2} \int_{\partial V} \left( \vec{r} \cdot \vec{r} \right) \mathrm{d}\vec{a} $$

Now, if both identities are correct, then $$\frac{1}{4} \int_{\partial V} \vec{r} \left( \vec{r} \cdot \mathrm{d}\vec{a} \right) = \frac{1}{2} \int_{\partial V} \left( \vec{r} \cdot \vec{r} \right) \mathrm{d}\vec{a}$$

which looks temptingly similar. However, nobody was able to prove this relation and it is (to me) far from trivial to see why this relation should hold.

Can anybody either show where a mistake is hidden or prove that the last identity is correct?

I realize that the sample solution effectively applies the divergence theorem to the outer product of $\vec{r}$ with itself: $$\frac{1}{4} \int_{\partial V} \vec{r} \left( \vec{r} \cdot \mathrm{d}\vec{a} \right) = \frac{1}{4} \int_{\partial V} \vec{r} \otimes \vec{r} \cdot \mathrm{d}\vec{a} = \frac{1}{4} \int_{\partial V} \vec{r} \, \vec{r}^T \cdot \mathrm{d}\vec{a} \stackrel{Div. theorem}{=} \frac{1}{4} \int_V \nabla \cdot \left( \vec{r} \, \vec{r}^T \right) \mathrm{d}^3r $$

which I find very curious. I did not know you could apply the divergence theorem to matrix-like objects, however, for every example we constructed we always found that the identitiy of the problem was correct.


2 Answers 2


Notice for any constant vector $\vec{k}$, we have

$$\begin{align} \nabla \times ( r^2 (\vec{k} \times \vec{r}) ) &= \nabla r^2 \times (\vec{k} \times \vec{r} ) + r^2 \nabla \times( \vec{k} \times \vec{r} )\\ &= 2 \vec{r} \times (\vec{k} \times \vec{r} ) + r^2 ((\nabla \cdot \vec{r}) \vec{k} - \color{red}{(\vec{k} \cdot \nabla )\vec{r}})\\ &= 2 (r^2 \vec{k} - (\vec{k}\cdot \vec{r})\vec{r}) + (3-\color{red}{1})r^2 \vec{k}\\ &= 4 r^2 \vec{k} - 2(\vec{k}\cdot\vec{r})\vec{r} \end{align} $$ Divide by $8$, integrate over any surface $S$ and apply Stoke's theorem, we obtain

$$\begin{align} \vec{k} \cdot \left[ \frac12 \int_S r^2 d\vec{a} - \frac14 \int_S \vec{r} (\vec{r}\cdot d\vec{a})\right] = & \frac18 \int_S( 4r^2 \vec{k} - 2(\vec{k}\cdot\vec{r})\vec{r}) \cdot d\vec{a}\\ = & \frac18 \int_S (\nabla \times ( r^2(\vec{k} \times \vec{r}))\cdot d\vec{a}\\ = & \frac18 \int_{\partial S} r^2(\vec{k} \times \vec{r})\cdot d\vec{r} \end{align} $$ When $S = \partial V$ is the boundary of some region $V$, the boundary of $S$ will be empty. This means the leftmost integral over $\partial S = \emptyset$ vanishes. As a result, $$\vec{k} \cdot \left[ \frac12 \int_{\partial V} r^2 d\vec{a} - \frac14 \int_{\partial V} \vec{r} (\vec{r}\cdot d\vec{a})\right] = 0$$ Since this is true for all constant vector $\vec{k}$, what's inside the square bracket vanishes and

$$\frac12 \int_{\partial V} r^2 d\vec{a} = \frac14 \int_{\partial V} \vec{r} (\vec{r}\cdot d\vec{a})$$

  • $\begingroup$ Thank you very much for this nicely written proof! Your answer was in-depth and, as far as I can see, accurate and correct. I consider my question answered. $\endgroup$
    – Xeno
    Dec 9, 2019 at 4:42
  • $\begingroup$ I'm sorry, but I think you made a mistake: From line 3 to 4, you are summing $2r^2 \vec{k} + 3r^2 \vec{k}$ to be $4r^2 \vec{k}$ instead of $5r^2 \vec{k}$. $\endgroup$
    – Xeno
    Dec 22, 2019 at 11:48
  • $\begingroup$ @Xeno, you are right, there is a typo. The $(\nabla \cdot \vec{k}) \vec{r}$ should really be $(\vec{k}\cdot \nabla)\vec{r}$ which generate an extra copy of $-r^2\vec{k}$. the final sum is still $4r^2\vec{k}$. $\endgroup$ Dec 22, 2019 at 13:20

The last identity is in fact correct, and it follows from the Divergence Theorem. So this leads me to believe that it will not hold for arbitrary surfaces, i.e., those that are not boundaries of a region. Indeed, that is so. Let's try the easiest example, the unit disk $D$ in the $xy$-plane. Then $$\int_D \vec r(\vec r\cdot d\vec S) = \int_D \vec r (0) = \vec 0,$$ whereas $$\int_D \|\vec r\|^2\,d\vec S = \left(2\pi\int_0^1 r^3\,dr\right)\vec k = \dfrac{\pi}2\vec k.$$


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