= Solution
The <product rule> for the <divergence> gives
$$
\nabla\cdot(u\nabla v)=\nabla u\cdot\nabla v+u\nabla^2v,\qquad\nabla\cdot(v\nabla u)=\nabla v\cdot\nabla u+v\nabla^2u.
$$
Subtracting cancels the two <gradient> products. \b[Therefore]
$$
\boxed{\nabla\cdot(u\nabla v-v\nabla u)=u\nabla^2v-v\nabla^2u.}
$$
To obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the <divergence theorem> on the spherical shell. On the outer sphere the shell's outward <unit normal> is $\widehat{\mathbf r}$; on the inner sphere it is $-\widehat{\mathbf r}$. Write $\partial_r=\widehat{\mathbf r}\cdot\nabla$ on both spheres. Then the resulting <Green second identity> is
$$
\boxed{\int_{\rho\leq|\mathbf x|\leq r}(v\nabla^2u-u\nabla^2v)\,dV=\int_{|\mathbf x|=r}(v\partial_ru-u\partial_rv)\,dS-\int_{|\mathbf x|=\rho}(v\partial_ru-u\partial_rv)\,dS.}
$$
The minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.
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