# On an integral calculus approach to finding the area of a quadrarc

I recently gave a ‘stretch and challenge’ problem to some of my GCSE maths students which involved finding the area of a quadrarc. A quadrarc, shown shaded in the diagram, is the shape enclosed by four arcs having identical radius but with centres at the four corners of a square. Finding the formula for the area of a quadrarc is in principle just within the reach of a high-attaining GCSE maths student since it can be done using only plane geometry techniques covered in the later stages of a GCSE course. However, it is a challenging problem and although some of my students got quite close none of them managed to solve it fully. The same plane geometry techniques are covered more thoroughly in the foundation year of A-level maths courses so this approach should be more feasible for first-year A-level maths students.

When exploring alternative ways to solve the problem myself, I employed an integral calculus approach that struck me as also being potentially very pedagogically useful for A-level students. It involves a number of techniques from coordinate geometry and integration that are in principle within the reach of such students in the later stages of their studies, so the quadrarc problem is also eminently useful as a ‘stretch and challenge’ revision problem for high-attaining A-level maths students if it is specified that integral calculus is to be used (rather than plane geometry). I have not been able to find any integral calculus solutions to the quadrarc problem in the literature which bring out the potentially useful pedagogical features at A-level, so I want to record my solution in the present note emphasising these features. The reader should try to solve the problem of finding the area of a quadrarc for himself/herself before reading any further.

In what follows we will always take the main square within which the quadrarc lies to be of side length $r$. I will begin by quickly setting out a plane geometry solution which involves finding the areas of an equilateral triangle, a sector, and a segment (diagrams A, B and C below).

Given that the main square has side length $r$, the area of the equilateral triangle in diagram A is (using the sine rule for area)

$\frac{1}{2} r^2 \sin 60^{\circ} = \frac{\sqrt{3}}{4} r^2$

The area of the sector in diagram B is

$\frac{60^{\circ}}{360^{\circ}} \pi r^2 = \frac{\pi}{6} r^2$

so the area of the segment shaded in diagram C is

$\frac{\pi}{6} r^2 - \frac{\sqrt{3}}{4} r^2$

Finally we obtain the shaded area in diagram D by deducting a triangle and two segments from the area of a quarter circle:

$\frac{\pi}{4} r^2 - \frac{\sqrt{3}}{4} r^2 - 2\big( \frac{\pi}{6} r^2 - \frac{\sqrt{3}}{4} r^2 \big) = \frac{\sqrt{3}}{4} r^2 - \frac{\pi}{12} r^2$

The area of the quadrarc is then the area of the square minus four of the areas in diagram D:

$r^2 - 4 \big( \frac{\sqrt{3}}{4} r^2 - \frac{\pi}{12} r^2 \big)$

$= \big( \frac{\pi}{3} - (\sqrt{3} - 1) \big) r^2$

This same formula for the area of a quadrarc, $\big( \frac{\pi}{3} - (\sqrt{3} - 1) \big) r^2$, can be derived using integral calculus as follows.

Let the origin be at the bottom left corner of the square with increasing x-direction to the right and increasing y-direction upwards. Based on the equation of a circle with centre at $(r, 0)$ and radius $r$, the equation of the arc through $R$ and $P$ is

$y_{RP} = \sqrt{r^2 - (x - r)^2}$

Similarly, based on the equation of a circle with centre at $(r, r)$, the equation of the arc through $R$ and $S$ is

$y_{RS} = r - \sqrt{r^2 - (x - r)^2}$

The x-coordinate of the point $R$ is given by the intersection of $y_{RP}$ and $y_{RS}$, i.e.,

$r - \sqrt{r^2 - (x - r)^2} = \sqrt{r^2 - (x - r)^2}$

$\iff$

$4x^2 - 8xr + r^2 = 0$

$\implies$

$x = r \pm \frac{\sqrt{3}}{2}r$

Therefore the x-coordinate of the point $R$ is

$r \big(1 - \frac{\sqrt{3}}{2} \big)$

Note that we need to discard the other solution, namely $r \big(1 + \frac{\sqrt{3}}{2} \big)$, because it exceeds $r$ and therefore lies outside the main square. By symmetry, the x-coordinate of both the points $P$ and $S$ is $\frac{r}{2}$. Therefore half of the area of the quadrarc should be given by the integral

$\int_{r(1 - \sqrt{3}/2)}^{r/2} (y_{RP} - y_{RS}) dx = \int_{r(1 - \sqrt{3}/2)}^{r/2} (2\sqrt{r^2 - (x - r)^2} - r) dx$

$= 2 \int_{r(1 - \sqrt{3}/2)}^{r/2} \sqrt{r^2 - (x - r)^2} dx - \frac{(\sqrt{3} - 1)}{2}r^2$

From tables of standard integrals we have that

$\int\sqrt{a^2 - z^2} dz = \frac{1}{2} z \sqrt{a^2 - z^2} + \frac{1}{2} a^2 \arctan\big( \frac{z}{\sqrt{a^2 - z^2}}\big)$

Therefore let’s employ the change of variable $z = x - r$. Then $dz = dx$ and we have $z = -\frac{\sqrt{3}}{2}r$ when $x = r\big(1 - \frac{3}{2}\big)$ whereas $z = -\frac{r}{2}$ when $x = \frac{r}{2}$. Our integral becomes

$2 \int_{- (\sqrt{3}/2)r}^{-r/2} \sqrt{r^2 - z^2} dz = \bigg[ z\sqrt{r^2 - z^2} + r^2 \arctan\big( \frac{z}{\sqrt{r^2 - z^2}}\big)\bigg]_{- (\sqrt{3}/2)r}^{-r/2}$

$= -\frac{r}{2}\sqrt{r^2 - \frac{r^2}{4}} + r^2 \arctan\bigg( \frac{-\frac{r}{2}}{\sqrt{r^2 - \frac{r^2}{4}}}\bigg) + \frac{\sqrt{3}}{2}r \sqrt{r^2 - \frac{3r^2}{4}} + r^2 \arctan\bigg( \frac{\frac{\sqrt{3}r}{2}}{\sqrt{r^2 - \frac{3r^2}{4}}}\bigg)$

$= r^2 \bigg\{ \arctan(\sqrt{3}) - \arctan\big( \frac{1}{\sqrt{3}}\big) \bigg\}$

$= \big\{ \frac{\pi}{3} - \frac{\pi}{6}\big\} r^2 = \frac{\pi}{6}r^2$

Therefore half the area of the quadrarc is

$\int_{r(1 - \sqrt{3}/2)}^{r/2} (y_{RP} - y_{RS}) dx$

$= 2 \int_{r(1 - \sqrt{3}/2)}^{r/2} \sqrt{r^2 - (x - r)^2} dx - \frac{(\sqrt{3} - 1)}{2}r^2$

$= \bigg(\frac{\pi}{6} - \frac{(\sqrt{3} - 1)}{2}\bigg)r^2$

The full area of the quadrarc is then

$\big( \frac{\pi}{3} - (\sqrt{3} - 1) \big) r^2$

as before.