 Solution

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.