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Notably absent from the discussion on finding areas between curves was any mention of circular areas. Thinking of a circle as an area between two semi-circular curves, how can we find the area of a circle of radius $r$?
No doubt many readers will quickly say, "It's $\pi r^2$, of course!" -- but how do we know that, beyond simply believing our former geometry teachers who gave us this formula (often without any justification)?
Let us position our circle in the Euclidean plane so that it is centered at the origin.
Given that the equation of this circle is $x^2 + y^2 = r^2$, and solving for $y$, we discover the top semi-circle can be described by the function $f(x) = \sqrt{r^2 - x^2}$. Since the circle is symmetric with respect to both the $x$ and $y$ axes, to find the area of the circe, it suffices to find the area between the top semi-circle and the $x$-axis between $x=0$ and $x=r$ and then multiply the result by $4$.
As such, the area $A$ of a circle with radius $r$ should be given by $$A = \int_0^r \sqrt{r^2 - x^2}\,dx$$ This integral can be a tricky one to evaluate, unless one makes a clever substitution involving a trigonometric function.
First, let us manufacture an appearence of $\sqrt{1-u^2}$ in our integrand by pulling out the (positive) factor of $r$, and then doing a $u$-substitution using $u=\frac{x}{r}$ and the consequent fact that $du=\frac{1}{r}dx$: $$\begin{array}{rcl} A &=& \displaystyle{\int_0^r \sqrt{r^2 \left(1 - \frac{x^2}{r^2}\right)}\,dx}\\\\ &=& \displaystyle{r^2 \int_0^r \sqrt{1 - \left(\frac{x}{r}\right)^2}\,\frac{1}{r}dx}\\\\ &=& \displaystyle{r^2 \int_0^1 \sqrt{1 - u^2}\,du}\\ \end{array}$$ Having manufactured this piece, and noting that $\sqrt{1 - \sin^2(\theta)}$ simplifies down to the far nicer $\cos(\theta)$, when $\theta$ is in the first quadrant, let us make another substitution (called unsurprisingly, a trigonometric substitution) of $u = \sin(\theta)$ along with the consequent $du = \cos(\theta)\,d\theta$, so that we can rewrite $A$ in the following way: $$A = r^2 \int_0^{\pi/2} \cos(\theta) \cdot \cos(\theta)\,d\theta = r^2 \int_0^{\pi/2} \cos^2(\theta)\,d\theta$$From here, the remainder of the integration is more routine. We first use one of the half-angle identities from trigonometry to reduce the exponent on the cosine, and then integrate each of the resulting terms to find $$\begin{array}{rcl} A &=& \displaystyle{r^2 \int_0^{\pi/2} \frac{1+\cos 2\theta}{2}\,d\theta}\\\\ &=& \displaystyle{\frac{r^2}{2} \int_0^{\pi/2} (1+\cos 2\theta)\,d\theta}\\\\ &=& \displaystyle{\frac{r^2}{2} \left[ \int_0^{\pi/2}\,d\theta + \int_0^{\pi/2} \cos 2\theta\,d\theta \right]}\\\\ &=& \displaystyle{\frac{r^2}{2} \left[ \theta\, \Big|_0^{\pi/2} + \frac{1}{2} \int_0^{\pi/2} 2\cos 2\theta\,d\theta \right]}\\\\ &=& \displaystyle{\frac{r^2}{2} \left[ \left(\frac{\pi}{2} - 0\right) + \frac{1}{2} \sin 2\theta\, \Big|_0^{\pi/2} \right]}\\\\ &=& \displaystyle{\frac{r^2}{2} \left[ \frac{\pi}{2} + (0-0) \right]}\\\\ &=& \displaystyle{\frac{\pi r^2}{4}} \end{array}$$
Of course, this represents the area of just a quarter of the circle, so multiplying the result by $4$ yields the result we were anticipating. The area of a circle is given by $\pi r^2$.