Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 12C (Vector Calculus)

The curl is constant:Orient the spanning planar disk by the unit normal $\mathbf{n}=(a,b,c)$ and orient its boundary by the right-hand rule, equivalently counterclockwise when viewed from the side towards which $\mathbf{n}$ points. Stokes theorem converts the line integral to a constant flux integral through a disk of area $\pi R^2$. With this orientationReversing the orientation changes the sign. The centre of the circle does not enter because the curl is constant.

Symmetry of the derivative matrix means ${\partial }_i{F}_j={\partial }_j{F}_i$. Contracting this symmetric pair with the antisymmetric Levi-Civita symbol givesEvery circle bounds its planar disk, and the vector field is smooth on all of ${\mathbb{R}}^3$, including that disk. Stokes theorem therefore gives for either orientationThe global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve line integral vanish.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

Set $s=x^2+y^2>0$. The first two components of the curl vanish since the horizontal components are independent of $z$ and the last component depends only on $z$. For the third,so this is a curl-free vector field everywhere in its domain:Choose the orientation in which the projection onto the $xy$-plane runs counterclockwise, and parametrizeOn this curve the horizontal contribution is $(-\sin t,\cos t)\cdot (-\sin t,\cos t)dt=dt$, while the vertical contribution is $zdz=d(z^2/2)$. Its integral over a closed curve is zero. Thereforewith $-2\pi$ for the opposite orientation. This does not contradict Stokes theorem: the planar disk spanning this curve meets the excluded $z$-axis at $(0,0,200)$, where the vector field is undefined. The horizontal one-form is the angular differential and records one winding about that axis; thus this curl-free vector field has no single-valued global potential of a conservative vector field on its domain.