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For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.