# divergence theorem; triple cross product

Let F=U $\times$ (U $\times$ x), where x is the position vector and U a uniform vector field. By using the divergence theorem, find the surface integral $\int_S$ F $\cdot$ dS, where S is the closed surface of the cube with verices $(\pm1,\pm1, \pm1)$.

By the divergence theorem $\int_S$ F $\cdot$ dS= $\int_V \nabla \cdot$ F dV.

But now,

$\nabla \cdot F = \nabla \cdot (U \times( U\times x))=(\nabla \times U)(U \times x)-U(\nabla(U \times x))=(\nabla \times U)(U \times x) -U((\nabla \times U)x-U(\nabla \times x))$

Because $U$ is a uniform vector field, I know $\nabla \times U = 0$ and $\nabla \cdot U = 0$, but I'm not sure if $\nabla \times x=0$, because that would imply $\nabla \cdot F = 0$ and the question wouldn't make sense.

Where am I wrong?

Also, how should I interpret the gradients, $\nabla F$ and $\nabla x$?

• Why would $\nabla \cdot F=0$ mean that the question wouldn't make sense? – Chappers Apr 12 '18 at 19:31

It's probably a better idea to expand the double cross product with Lagrange's identity, since we can then forget about cross products altogether: $$U \times (U \times x) = U(U \cdot x) - x(U \cdot U).$$ Now, $U$ is constant, so any derivative applied to it is zero. The divergence of the second term is simply $-3 \lvert U \rvert^2$. For the former term, it is easiest to use components: $$\nabla \cdot (U(U \cdot x)) = \partial_i (U_i U_j x_j ) = U_i U_j \delta_{ij} = U_i U_i = \lvert U \rvert^2,$$ so in fact $\nabla \cdot F = -2\lvert U \rvert^2$.
$\nabla F$, when $F$ is a vector field, is a two-index object with components $(\nabla F)_{ij} = \partial_i F_j$ (so $(\nabla x)_{ij} = \delta_{ij}$). It often called a (Cartesian) tensor.
Looking at your calculation, expanding using these identities gives $$\nabla \cdot (U \times (U\times x)) = (\nabla \times U) \cdot (U \times x) - U \cdot (\nabla \times (U \times x)) \\ = 0 - U \cdot ( U (\nabla \cdot x) - x (\nabla \cdot U) + (x \cdot \nabla) U - (U \cdot \nabla ) x) ) \\ = -3\lvert U \rvert^2 + 0 + 0 + U \cdot (U \cdot \nabla) x.$$ As we noted above, $(\nabla x)_{ij} = \delta_{ij}$, so $(U\cdot \nabla )x = U$, and we get the same answer as before.
• @harlem I've added another version of the calculation. I'm not sure what you mean by the $U(\nabla(U \times x))$ term. – Chappers Apr 12 '18 at 20:22