= 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> give
$$
\int_{B_1}x^2\,dV=\frac13\int_{B_1}r^2\,dV=\frac{4\pi}{3}\int_0^1r^4\,dr=\frac{4\pi}{15}.
$$
Hence 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 $(0,0,8\pi/15)$. Therefore
$$
\frac12(1,0,0)\times(0,0,8\pi/15)=\boxed{(0,-4\pi/15,0),}
$$
exactly the same <vector>.
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