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Apply Stokes theorem to $\mathbf{F}=\psi \mathbf{B}$, with $\mathbf{B}$ an arbitrary constant vector. The curl product rule for a scalar field reduces to $\nabla \times (\psi \mathbf{B})=\nabla \psi \times \mathbf{B}$. Using the scalar triple product,Since these scalar products agree for every $\mathbf{B}$, the vectors agree. This proves the vector-valued Stokes identityIn particular, the ordering of the cross product is normal crossed with gradient; reversing it would change the sign.

For constant $b$, the curl product rule for a scalar field givesUsing part (b) and the order of the cross product,Here $dS$ inside a vector product denotes the outward vector area element $ndS$. Closedness is the condition that permits replacing the entire surface by the boundary integral of a volume.

Away from the origin, write $\mathbf{F}=r^{-\beta -1}\mathbf{r}$. The curl product rule for a scalar field and the gradient of a radial function giveConsequently the surface integral of curl over the upper hemisphere is zero. Its outward unit normal vector induces a counterclockwise equator when viewed from above. Parametrize that boundary byThe radial vector field is perpendicular to this tangent, soThis verifies Stokes theorem directly without passing through the singular origin.

Apply the product rule inside the Cartesian expression for the curl:The first sum is the gradient of the scalar field and the second is the curl of the vector field. Thus the curl product rule for a scalar field is

On the unit circle, the positively oriented unit tangent vector is $\mathbf{u}=(-y,x,0)$, and $|\mathbf{u}|=1$. Hence $d\mathbf{r}=\mathbf{u}ds$ andSince the scalar field $\psi$ has no $z$ dependence, the curl product rule for a scalar field givesOn the unit upper hemisphere $S_1$, the outward unit normal vector is $(x,y,z)$, so its vertical component is $z=\cos \theta$. Applying the Stokes theorem to the vector field $\psi \mathbf{u}$ therefore provesThe smoothness needed here is that $\psi$ extend as a continuously differentiable scalar field to a neighbourhood of the projected unit disk. The left side is a scalar line integral with respect to arc length; the positive boundary orientation is used when identifying it with the vector line integral.