= Solution
For an oriented <smooth surface> $S$ with piecewise smooth, consistently oriented boundary, <Stokes theorem> gives
$$
\boxed{\int_S(\nabla\times B)\cdot n\,dS=\oint_{\partial S}B\cdot dx}
$$
for 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 by
$$
r(r,\theta)=(r\cos\theta,r\sin\theta,r^2),\qquad 1/3\leq r\leq1,\quad0\leq\theta<2\pi.
$$
Using $\partial_r r\times\partial_\theta r$ as the oriented <vector area element> gives
$$
\boxed{d\mathbf S=(-2r^2\cos\theta,-2r^2\sin\theta,r)\,dr\,d\theta,\qquad
dS=r\sqrt{1+4r^2}\,dr\,d\theta.}
$$
The sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-3-paraboloid.png]
{title=Annular paraboloid with upward normal and opposite induced orientations on the outer and inner boundary circles}
Differentiating the given <vector field> yields
$$
\nabla\times B=(0,0,3x^2+3y^2)=(0,0,3r^2).
$$
The <surface integral> is therefore
$$
I=\int_0^{2\pi}\int_{1/3}^1 3r^3\,dr\,d\theta
=\frac{3\pi}{2}(1-3^{-4})
=\boxed{\frac{40\pi}{27}}.
$$
For 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 and
$$
B\cdot dx=R^4(\sin^4\theta+\cos^4\theta)\,d\theta.
$$
Since each fourth power integrates to $3\pi/4$, the two-circle <line integral> is
$$
\oint_{\partial S}B\cdot dx=\frac{3\pi}{2}\left(1-\frac1{81}\right)=\frac{40\pi}{27},
$$
confirming <Stokes theorem> and the <Stokes flux through an annular paraboloid>. Reversing the chosen <orientation> changes both integrals to $-40\pi/27$.
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