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For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

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 $-40\pi /27$.