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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 a smooth vector field near an oriented piecewise smooth surface $S$, Stokes theorem stateswhere the boundary orientation is induced by the surface unit normal through the right-hand rule. For a constant vector $\mathbf{k}$, the curl of $\varphi \mathbf{k}$ is $\nabla \varphi \times \mathbf{k}$. Apply Stokes theorem and use the scalar triple product:Since this holds for every constant $\mathbf{k}$, equality of all vector components proves the vector-valued Stokes identityFor the specified field, the two axis segments contribute zero because $x^2{y}^4$ vanishes on them. On the arc the direct line integral isIts two terms giveandThereforeOne can also check the orientation and sign by Stokes theorem. For the upward-oriented quarter disk, $(\nabla \times \mathbf{A}{)}_z=2x{y}^4-4x^2{y}^3$. Polar integration givesconsistent with the positively oriented boundary in the question.

Choose the scalar fieldThe allowed vector-valued Stokes identity givesEvery boundary point lies on the unit circle in the $xy$-plane, so $f=1/2$ throughout the boundary. Since the line integral of $d\mathbf{r}$ around a closed curve is the zero displacement,Only the boundary value of this scalar field mattered: the interior of the surface need not be part of a sphere. Either consistent choice of orientation gives the same zero answer.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.