Does anyone know how to solve this? Geometric problem, conflicting sine wave and parabola shape.

Question

Alright, I'll explain. I have created a shape by graphing an inequality of the space in between a parabola and sine wave with matching vertexes and neutral points. I cannot find anything online about this shape so for the time being I thought I would call it a spatio-inter curve, for "space in between." Anyways, back to the point. Here is a link to a desmos graph with the equation: https://www.desmos.com/calculator/gc2gl1k4k7. I do not want to write it out as it is hard to format. I am posting this here to put some record of this and also I want to find out a bit about it. I am not very far in math, as I am a high school student. I was wondering if anyone smarter than me could figure anything out about this, such as area, width of different part, honestly anything, this shape is pretty cool looking so yeah. Just wanted to post this to put it out on the internet and see if anyone else can figure anything out about it.

Answer 1

Essentially what we have here are two curves:$$f(x)=\sin x$$ and $$g(x)=1-(\frac {2x}{\pi}-1)^2$$ First thing to notice is that, both functions are symmetric about $x=\frac {\pi}{2}$. This means, for both functions, the property $h(x+\frac {\pi}{2})=h(\frac {\pi}{2}-x)$ holds. This means that any analysis that we do for the first "part" of the curve, would hold for the second part too, via symmetry. So i shall confine myself to only the first "part" of the curve. I shall assume you know calculus for the rest of the solution, because almost any analysis will hinge on it. The area of the curve can be found out via integration: $$A=\int_0^{\frac {\pi}{2}} |f(x)-g(x)|dx$$ The integration is rather elementary, I that no manipulation is required other than knowledge of the standard integrals, and I shall leave you to it as an exercise. One could also attempt to calculate the perimeter of the figure, the general process for which is as follows: For a function $y=h(x)$, the perimeter of a curve from $x=a$ to $x=b$ is given by: $$P=\int_a^b \sqrt {h'(x)^2+1} dx$$

Note: This comes from the Pythagoras Theorem, try to prove it.

Hence, to calculate perimeter of first part of the figure, you should add up the lengths of the curves due to $f(x)$ and $g(x)$ in the interval, using the algorithm I laid out. Is there anything else you wish to know about it? You mentioned "width", but that doesn't make much sense because not only is the width changing, finding out the maximum width is also meaningless as $f(x)$ and $g(x)$ aren't symmetrically placed about $y=\frac {\pi x}{2}$.