Putnam 2021, Problem B-3

Let $h(x,y)$ be a real-valued function that is twice continuously differentiable throughout $\mathbb{R}^2$, and define
\[
\rho(x,y) = y h_x - x h_y.
\]
Prove or disprove: For any positive constants $d$ and $r$ with $d > r$, there is a circle $\mathcal{S}$ of radius $r$ whose center is a distance $d$ away from the origin such that the integral of $\rho$ over the interior of $\mathcal{S}$ is zero.

Solution:
We prove the given statement.

For any circle $\mathcal{S}$ of radius $r$ whose center is at distance $d$ from the origin, express the integral in polar coordinates $s, \theta$:
$$\iint_{\mathcal{S}} \rho = \int_{s_1}^{s_2} \int_{\theta_1(s)}^{\theta_2(s)} (yh_x - xh_y)(s\sin\theta, s\cos\theta) s d\theta ds.$$

For fixed $s$, the integral over $\theta$ is a line integral of $\text{grad}h$, which evaluates to $h(P_2) - h(P_1)$ where $P_1, P_2$ are the endpoints of the endpoints of the arc of the circle of radius $s$ centered at the origin lying within $\mathcal{S}$. If we now fix $r$ and $d$ and integrate $\iint_{\mathcal{S}} \rho$ over all choices of $\mathcal{S}$ (this amounts to a single integral over an angle in the range $[0, 2\pi]$), we may interchange the order of integration to first integrate over $\theta$, then over the choice of $\mathcal{S}$, and at this point we get 0 for every $s$. We conclude that the integral of $\iint_{\mathcal{S}} \rho$ over all choices of $\mathcal{S}$ vanishes; since the given integral varies continuously in $\mathcal{S}$, by the intermediate value theorem there must be some $\mathcal{S}$ where the given integral is 0.