Past exam of the mathematical methods II course of the University of Oxford / 2018 / 9

Apply the divergence theorem to $A(r)=(C\cdot r)(C^{\prime }\cdot r)J(r)$. Since $C,C^{\prime }$ are constant and $J$ is a solenoidal vector field, the product rule givesIts boundary flux vanishes when $J$ has compact support, or when its normal component vanishes on the boundary of the integration volume. Integrating therefore provesFor a discontinuous field, the divergence-free hypothesis must include its boundary jump, in the sense of a weak derivative. A field that merely has zero divergence at interior points can still have a nonzero surface source; that weaker hypothesis would not justify discarding the flux. The result is the antisymmetric first moment of a localized solenoidal field.

Use the vector triple product identity:For the $i$th component of the second integral, the antisymmetric first moment of a localized solenoidal field with $C^{\prime }=e_i$ gives ${\int }_V{r}_i(C\cdot J)dV=-{\int }_V(C\cdot r)J_{i}dV$. Thus the two terms on the right have opposite integrals. Consequently

Inside the unit ball, $\nabla \cdot J=0$. On its sphere, $J\cdot r=xy-xy=0$, so the jump to zero outside adds no surface divergence. The field therefore satisfies the required weak derivative condition.

By reflection symmetry, integrals of $xy,xz,yz$ over the ball vanish. Rotational symmetry and spherical coordinates giveHence the left side is $\int x(y,-x,0)dV=(0,-4\pi /15,0)$. Meanwhile $J\times r=(-xz,-yz,x^2+y^2)$, so its integral is $(0,0,8\pi /15)$. Thereforeexactly the same vector.