Showing $\iiint_R \mathbf{r} \times \nabla\phi \,dV = \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n} \,dS$ using divergence theorem

Question

Question: Let $\phi$ be a smooth scalar field defined on a region $R \subseteq \mathbb{R}^{3}$ with a smooth boundary $\partial R$. Then show that the following identity holds:

$$\iiint_R \mathbf{r} \times \nabla\phi \,dV = \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n} \,dS$$

In my course up until this point the divergence theorem has been covered, and I can see the similarities between the integrals above and the divergence theorem identity but am unsure how to go from $\nabla . F$ in the divergence theorem to $\mathbf{r} \times \nabla\phi$ above.

How should one begin to attempt showing an identity like this given they have use of the divergence theorem?

Answer 1

In Cartesian coordinates we have $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k},$ and, without showing every term,

$$\iiint_R\mathbf{r} \times \nabla \phi \, dV \\ = \left(\iiint_R y \frac{ \partial\phi}{\partial z} \, dV - \iiint_R z \frac{ \partial\phi}{\partial y} \, dV \right)\mathbf{i} + \cdot \cdot \cdot \\ = \left(\iiint_R \frac{ \partial }{\partial z}(y \phi) \, dV - \iiint_R \frac{ \partial}{\partial y}(z \phi) \, dV \right)\mathbf{i} + \cdot \cdot \cdot \\ = \left(\iiint_R \nabla \cdot (y \phi \mathbf{k}) \, dV - \iiint_R \nabla \cdot (z \phi \mathbf{j}) \, dV \right)\mathbf{i} + \cdot \cdot \cdot \\ = \left(\iint_{\partial R} (y \phi \mathbf{k}) \cdot \mathbf{n} \, dS - \iint_{\partial R} (z \phi \mathbf{j}) \cdot \mathbf{n} \, dS \right)\mathbf{i} + \cdot \cdot \cdot \\ = \left(\iint_{\partial R} y \phi n_z \, dS - \iint_{\partial R} z \phi n_y \, dS \right)\mathbf{i} + \cdot \cdot \cdot \\ = \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n} \, dS. $$

The divergence theorem familiar to you was applied in going from the third to fourth step.

Addendum

A proof with much more compact notation uses the component form of the divergence theorem

$$\int_R \partial_i f \, dV = \int_{\partial R} f \,n_i \, dS,$$

and the expression for curl and cross-product using the Levi-Cevita symbol

$$(\mathbf{r} \times \nabla \phi)_i = \varepsilon_{ijk} \,r_j \,\partial_k \phi, \\ (\mathbf{r} \times \phi \mathbf{n})_i = \varepsilon_{ijk}\,r_j\phi \,n_k.$$

Thus,

$$\begin{align} \int_R (\mathbf{r} \times \nabla \phi)_i \, dV &= \int_{\partial R} \varepsilon_{ijk} \,r_j \,\partial_k \phi \, dS \\ &= \varepsilon_{ijk}\int_{\partial R} \ \,\partial_k (r_j\phi) \, dS \\ &= \varepsilon_{ijk}\int_{\partial R} \, r_j\,\phi \,n_k \, dS \\ &= \int_{\partial R} \varepsilon_{ijk} \, r_j\,\phi \,n_k \, dS \\ &= \int_{\partial R} (\mathbf{r} \times \phi \mathbf{n})_i \, dS \end{align}$$

Answer 2

While the solution above is elegant and correct, I believe the following approach better demonstrates how the identity was obtained in the first place.

Consider setting $\mathbf{F} = \phi\mathbf{c} \times \mathbf{r}$ in the Divergence Theorem as you quoted above, where $\mathbf{c}$ is an arbitrary yet constant vector field. By the divergence product rule for a scalar field and a vector field:

$ \nabla \cdot (\phi\mathbf{c} \times \mathbf{r}) = \nabla \phi \cdot (\mathbf{c} \times \mathbf{r}) + \phi \nabla \cdot (\mathbf{c} \times \mathbf{r}) = \mathbf{c} \cdot (\mathbf{r} \times \nabla \phi)$, by the cyclic properties of the triple vector product and the fact that $ \nabla \cdot (\mathbf{c} \times \mathbf{r}) = \mathbf{0}$ (check this directly).

Now note from the Divergence Theorem for the vector field $\mathbf{F}$:

$ \iiint_R \mathbf{c} \cdot (\mathbf{r} \times \nabla \phi) \,dV = \iint_{\partial R} (\phi\mathbf{c} \times \mathbf{r}) \cdot \mathbf{n} \,dS $

$ \implies \mathbf{c} \cdot \iiint_R (\mathbf{r} \times \nabla \phi) \,dV = \mathbf{c} \cdot \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n}\,dS $

where the right hand side of the last expression follows again from the cyclic properties of the triple product. Recalling $\mathbf{c}$ is arbitrary, we're done.