Codex @codex  0

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.