This is something that started as an exercise, shifted to a small art project, and then shifted back into an exercise. The first step is to find the Fourier series associated with the curve $y=x^2-x$ in the interval $[0,1]$. After a bit of work, we see that this is equal to $y=-\frac{1}{6}+\Sigma_{n=0}^\infty\frac{1}{n^2\pi^2}\cos{(2n\pi x)}$.
What I then started to do, seeing that the full curve looked like a child's drawing of the sea, was to play around with it. Specifically, I started using Desmos to look at the function $y\leq-\frac{1}{6}+\Sigma_{n=0}^\infty\frac{1}{n^2\pi^2}\cos{(2n\pi(x-b))}+a$ when $a$ and $b$ are continuously changed to give the appearance of waves rising and falling (click here for an interactive demonstration). Then, keeping in with a nautical theme, I added a "porthole" to restrict the viewing area of the waves to a circle, by adding the restriction $x^2+y^2\leq1$. An image of the resulting figure is shown below.
Here is where it circled back to math - how would I figure out what the shaded area here is? For $a=b=0$, the solution is simple. Since the radius of the circle is $1$, the area is simply $\pi-\frac{\pi}{2}-2|\int_0^1(x^2-x)dx|=\frac{\pi}{2}-\frac{1}{3}$. However, what if we change those variables? What if, for example, $a=\frac{1}{4}$ and $b=\frac{1}{3}$? How do we generalize the area of the region bounded by $y\leq-\frac{1}{6}+\Sigma_{n=0}^\infty\frac{1}{n^2\pi^2}\cos{(2n\pi(x-b))}+a$ and $x^2+y^2\leq1$ for $a\in[-1,\frac{5}{4}]$ and $b\in[0,1]$? (The reason for these intervals being that the absolute minimum when $a=0$ is $y=-\frac{1}{4}$, meaning if $a>1+\frac{1}{4}$ then the region is simply a circle, and the function is periodic with period $1$.)