# Can a spiral have its centroid at the origin?

A spiral is a curve $\gamma$ with the polar equation $r=f(\theta)$ where $f$ is a continuous positive strictly monotone function on some interval $[a, b]$, $-\infty<a<b<\infty$. Best known examples are the logarithmic spiral and the Archimedean spiral.

Problem: Find a spiral whose centroid is the origin of the coordinate system.

Progress so far: We want $$\int_\gamma x\,ds = \int_\gamma y \,ds = 0 \tag1$$ Note that $x = f(\theta)\cos\theta$, $y = f(\theta)\sin\theta$, and $ds = \sqrt{(f'(\theta))^2 + f(\theta)^2}\,d\theta$. Thus, we need the function $$g(\theta) = f(\theta) \sqrt{(f'(\theta))^2 + f(\theta)^2}$$ to be orthogonal to both $\cos \theta$ and $\sin\theta$ on the interval $[a, b]$, meaning $$\int_a^b g(\theta)\cos\theta\,d\theta = \int_a^b g(\theta)\sin\theta\,d\theta = 0\tag2$$ A natural way to satisfy (2) is to take $[a, b] = [0, 2\pi]$ and $g$ to be constant (say $g\equiv 1$ as scaling does not matter). However this fails, because solving the equation $g\equiv 1$ for $f$ (as an autonomous ODE) yields $f(\theta) = \sqrt{\sin 2\theta}$ (up to a shift), which is not even defined, let alone monotone, on any interval of length $2\pi$.

Note: It is not required for $[a, b]$ to have length $2\pi$ or a multiple of $2\pi$; it can be any nontrivial finite interval.

• I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case. – Mark Bennet Jul 25 '18 at 19:39
• @MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.) – Brian Tung Jul 25 '18 at 19:41
• For $[a,b] = [0,2\pi]$ (and likewise for multiples of $2\pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $\theta$ between $0$ and $\pi$, the spiral point corresponding to $\theta+\pi$ is farther below the $x$-axis than the point corresponding to $\theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $\theta$ in $(\pi,2\pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider. – Blue Jul 25 '18 at 19:42
• (continued) Note that your spirals are inherently "left heavy" over the interval $(0,\pi)$. Therefore, if the vertical-balancing $\theta$ (from my previous comment) is between $\pi$ and $3\pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant. – Blue Jul 25 '18 at 19:53
• @Blue That is the intuition I was trying to capture. – Mark Bennet Jul 25 '18 at 20:11

I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 \cos \theta, 2\sin \theta)$ and a line segment tangent to the circle of length $L$. I find that for $\theta = 4.95, L\approx 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $\theta$ but they were easy to calculate and we can approach them arbitrarily closely. 