Where does πr² come from?

Part 1 defined area, Part 2 defined π, Part 3 defined length. The claim "the disc of radius $r$ has area $\pi r^2$" now names a specific number on each side, and the library proves the two are equal (A disc of radius r has Riemann area pi r squared; in particular the unit disc has area pi) in five numbered steps. This part walks them.

The area is an integral

By Part 1's definition, the closed disc of radius $r$ is the region between the graphs $y=-\sqrt{r^2-x^2}$ and $y=+\sqrt{r^2-x^2}$, so its area is the integral of the height (Riemann area between two continuous graphs and the disc as a vertically simple region):

$\mathrm{area}=2{\int }_{-r}^r\sqrt{r^2-x^2}dx\mathrm{.}$

The r² is pure scaling

Substitute $x=ru$ (Substitution: if $\phi$ is differentiable on $[c,d]$ with ${\phi }^{\mathrm{\prime }}$ integrable and $f$ is continuous on an interval containing $\phi ([c,d])$, then ${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }}$). The radius factors out of the square root and the $dx$, leaving $r^2$ times the unit-disc integral. The theorem splits in two: $r^2$ measures nothing but scale, and the entire question is why the unit disc has area $\pi$.

Where π enters

Substitute $x=\sin t$ on $[-\pi \mathrm{/}2,\pi \mathrm{/}2]$. There cosine is nonnegative, so $\sqrt{1-{\sin }^{2}t}=\cos t$ by the Pythagorean identity (Parity and the Pythagorean identity for sine and cosine), and the area becomes $2{\int }_{-\pi \mathrm{/}2}^{\pi \mathrm{/}2}{\cos }^{2}tdt$. The addition formulas (The addition formulas for sine and cosine) split ${\cos }^{2}t$ into $\frac{1}{2}+\frac{1}{2}\cos 2t$. The $\cos 2t$ half integrates to zero by the fundamental theorem (The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$), since $\frac{1}{2}\sin 2t$ vanishes at both endpoints. The constant half integrates to half the length of the interval (If $m\le f\le M$ on $[a,b]$ then $m(b-a)\le L(f,P)\le \underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P)\le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with ${\int }_a^{b}c=c(b-a)$), and the leading factor of $2$ turns that into the full length. The unit disc has area π.

That is where π enters: as the length of $[-\pi \mathrm{/}2,\pi \mathrm{/}2]$. The substitution's limits are $\pm \pi \mathrm{/}2$, and $\pi \mathrm{/}2$ is exactly the first positive zero of cosine, which Part 2 took as the definition of π.

The same answer by hand

Archimedes ran the inner-and-outer squeeze on the circle two thousand years before integrals. The library proves the polygon squeeze: regular $n$-gons inscribed in and circumscribed about the unit circle have perimeters $I_n=2n\sin (\pi \mathrm{/}n)$ and $O_n=2n\tan (\pi \mathrm{/}n)$ with $I_n<2\pi <O_n$, increasing and decreasing respectively, both converging to the circumference $2\pi$ (Inscribed regular-polygon perimeters increase to 2 pi, while circumscribed perimeters decrease to 2 pi). The number itself can be computed from proved series and products: Gregory-Leibniz with an explicit remainder bound (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...), Wallis (Wallis's product: pi over two is the limit of the finite Wallis products), Viète (Viete's nested-radical product: two over pi is the limit of the finite cosine products).

One door left

The four requirements of Part 1 assign an area to every region the squeeze can trap. Which sets is that, and is there a set the squeeze can never trap? That is the last part of this rabbit hole: how deep does it go?