# 3Blue1Brown Calculus Video: Why are the unraveled rings trapezoids?

In this video https://www.youtube.com/watch?v=WUvTyaaNkzM&list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr, at 2:45, why is the unraveled ring a trapezoid when it should clearly be a rectangle? I was thinking that in real life, the unraveled ring actually is a rectangle, but Grant is saying it's a trapezoid so that he can prove a point. The point he's proving is that if you increase the number of rings by a finite amount, the unraveled rings will look less like trapezoids and more like rectangles. And when you set the number of rings to approach infinity, the unraveled rings are rectangles, and so to find the area of the circle, you just add up an infinite amount of rectangles, which is also the area under the function $$2\pi r$$, which is $$\pi r^2$$. So, in real life, are the unraveled rings actually rectangles or not?

Also at 3:00, what does Grant mean when he says the circumference formula $$2\pi r$$ is the definition of pi?

I was thinking that in real life, the unraveled ring actually is a rectangle

Think of it like a racetrack. The outer track is always slightly longer than the inner track so when you flatten it out, you approximately have quadrilateral with one edge longer than the opposite one; i.e. like a trapezoid.

So, in real life, are the unraveled rings actually rectangles or not?

Not perfectly but they become more rectangular when we take thinner rings. So an "infinitely thin ring" would be perfectly rectangular but that's a bit murky to define. When we unravel the rings of the circle, we get trapezoids.* But the thinner your rings are, the more the trapezoids look like rectangles.

This is a ring of width $$1$$ from a circle

Compared with a ring of width $$\frac1{10}$$

When we make the ring width as thin as $$\frac1{100}$$, it's pretty much just a very thin rectangle.

The essence of integration is that the thinner you make your rings or strips, the less the curve of the edge matters and the more they look like thin rectangles.

* We can show that the shape of the rings is trapezoidal since each they are the set of all infinitesimally thin circumferences, $$2\pi r$$; which are linear with respect to the radii $$r$$. Alternatively, the rings are the set of curves described by double the arc length of $$\sqrt{r^2-x^2}$$. We can use the formula for arc-length to calculate this as $$2\int_{-r}^r\sqrt{1+\left(\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{r^2-x^2}\right)^2}\ \mathrm{d}x=2\int_{-r}^r\frac{r}{\sqrt{r^2-x^2}}\ \mathrm{d}x=2r\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\ \mathrm{d}x=2\pi r$$.

• Do you know what Grant meant at 3:00? – user532874 Jul 2 '19 at 20:22
• @user532874 Pi is defined as the ratio of a circle's circumference, $c$, to its diameter, which double its radius ($2r$). So $\pi=\frac{c}{2r}$. It's a simple rearrangement to solve for $c$. – Jam Jul 2 '19 at 20:28
• Sorry, the rings are perfect trapezoids; I made an algebra error. – Jam Jul 2 '19 at 20:37
• Here's a desmos graph with sliders that shows how the unraveled rings form trapezoids. desmos.com/calculator/zohoruatoz – Jam Jul 3 '19 at 11:28

The ring has inner radius $$r$$ and outer radius $$R$$ with $$R>r$$. Thus, the edge of the quadrilateral corresponding with the inner edge of the ring will be shorter than the edge of the quadrilateral corresponding with the outer edge of the ring, which is why the quadrilateral is a trapezoid.

For your second question, $$\pi$$ is defined to be the ratio of a circle's circumference $$C$$ to its diameter $$d$$, so, by defintion of $$\pi$$, $$C = \pi d = 2\pi r$$