Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 1

Let $D$ be a bounded planar region with a piecewise smooth, simple closed boundary $C$, oriented counterclockwise, and let $P,Q$ have continuous first partial derivatives on an open neighbourhood of its closure. Green's theorem states thatFor a region with holes, orient every boundary component with $D$ on its left: the outer component is counterclockwise and the inner components are clockwise.

Choose $P=-y/2$ and $Q=x/2$. Their indicated partial derivatives differ by one, so the area isThis is signed area as a boundary integral; reversing the direction of $C$ changes the sign of the integral.

In polar coordinates, put $x=r\cos \theta$ and $y=r\sin \theta$. Differentiating along the curve givesThe cardioid has $r\ge 0$ and is traversed once counterclockwise as $\theta$ increases from $0$ to $2\pi$. Its cusp at $\theta =\pi$ causes no difficulty: split the line integral there, or exclude a small arc and take its limit. Thus signed area as a boundary integral gives

For an oriented piecewise smooth parametric curve $\mathbf{r}:[a,b]\to {\mathbb{R}}^3$ and a continuous vector field $\mathbf{F}$ along it, the vector-field line integral isFor a piecewise smooth curve, sum these integrals over its smooth pieces. The dot product selects the component of the vector field in the direction of motion.

If $\mathbf{F}=\nabla \varphi$ for a smooth function $\varphi$, the multivariable chain rule givesThe fundamental theorem of calculus, with cancellation at intermediate endpoints, therefore proves the fundamental theorem for line integrals:On a closed curve $\mathbf{r}(a)=\mathbf{r}(b)$, so $\boxed{{\oint }_C\nabla \varphi \cdot d\mathbf{r}=0}$.

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.