Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 10A (Vector Calculus)

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field on a neighborhood of it, Stokes theorem statesThe oriented surface element and boundary direction must agree by the right-hand rule. Every boundary component is included; for an upward-oriented graph over an annulus the outer boundary is anticlockwise and the inner boundary clockwise when viewed from above.

Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt{2}\le z\le \sqrt{5}$, so its projection is the annulus $1\le r\le 2$. The surface is the corresponding band of the upper sheet of a two-sheeted hyperboloid, bounded by circles of radii $1$ and $2$ at heights $\sqrt{2}$ and $\sqrt{5}$.

For the parametrization $\mathbf{R}(x,y)=(x,y,g(x,y))$, its tangent vectors are $(1,0,x/z)$ and $(0,1,y/z)$. Their cross product gives the upward vector surface element of a graph:Its magnitude gives the scalar surface area of a graph:Both formulas apply on $1\le \sqrt{x^2+y^2}\le 2$. Reversing the orientation changes only the sign of the vector element.

Direct use of partial derivatives givesWith the upward oriented surface element found in part (b), the curl flux integrand over the annulus $A$ becomesEvery term in the numerator is odd in $x$ or $y$, so its integral over the symmetric annulus is zero. Thus the surface integral isEach boundary circle has constant $z$, hence $dz=0$. Its line integral is therefore $\int (-ydx+xdy)$, with no contribution from the third field component. An anticlockwise radius-$r$ circle gives $2\pi r^2$. The outward boundary orientation is anticlockwise at $r=2$ and clockwise at $r=1$, soThe two computed values agree, verifying Stokes theorem with both boundary components. A downward choice would give $-6\pi$ on both sides.