# Is there a way to calculate the area of this intersection of four disks without using an integral?

Is there anyway to calculate this area without using integral ?

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What have you tried? Also, it may be more difficult calculating the area using integral. – Quang Hoang Jul 26 '14 at 3:47
It's easy to use integral to solve this. all you have to do is to find the first and second intersection and integrate the circle equation. at the end multiply the result by four. – Shabbeh Jul 26 '14 at 4:09
An old chemist's trick would be to print the image out, cut the area you want to measure out, weigh it, and then weigh a 1cm^2 piece of paper as well. That gives you the density of the paper, which you can divide the weight of the desired area by to get the area in cm^2. – Batman Jul 26 '14 at 15:49
not sure weighing paper would make $1-\sqrt{3}+\frac{\pi}{3}$ clear – John Fernley Jul 29 '14 at 3:05

Assume that the side of the square $\overline{AB}=1$. Consider the diagram

$\hspace{3cm}$

By symmetry, $\overline{EC}=\overline{CD}$; therefore, $\overline{CD}=1/2$. Since $\overline{AC}=1$ and $\overline{AD}\perp\overline{CD}$, we have that $\angle CAD=\pi/6$ ($30$-$60$-$90$ triangle). Similarly, $\angle GAF=\pi/6$, leaving $\angle CAG=\pi/6$.

Since base $\overline{AB}=1$ and altitude $\overline{CD}=1/2$, $\triangle ABC$ has area $1/4$.

Since $\angle CAB=\pi/6$, the circular sector $CAB$ has area $\pi/12$.

Therefore, the area of the purple half-wedge between $B$ and $C$ is $\color{#A050A0}{\dfrac{\pi-3}{12}}$.

Furthermore, $\overline{CG}^2=\overline{BC}^2=\overline{CD}^2+\overline{DB}^2=\left(\frac12\right)^2+\left(1-\frac{\sqrt3}{2}\right)^2=\color{#50B070}{2-\sqrt3}$.

Therefore, the area requested is $\color{#50B070}{2-\sqrt3}+4\left(\color{#A050A0}{\dfrac{\pi-3}{12}}\right)=1+\dfrac\pi3-\sqrt3$

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Thanks for your answer. which software did you use for drawing this ? – Shabbeh Jul 27 '14 at 2:32
I used Intaglio, which is only available for MacOS. – robjohn Jul 27 '14 at 3:22
In my opinion, this answer is clearer than the accepted one. Nice diagram. – qwr Jul 27 '14 at 3:42
@qwr: Thanks. A diagram often makes an explanation clearer. – robjohn Jul 27 '14 at 11:40
i like this one – athos Jul 28 '14 at 1:44

Your "curvilinear square" just cuts the quarter-circles in thirds, so the distance between two adjacent vertices is $2l\sin\frac{\pi}{12}=\frac{\sqrt{3}-1}{\sqrt{2}}l$, given that $l$ is the length of the side of the original square. So the area of the "circular square" is given by $(2-\sqrt{3})l^2$ plus four times the area of a circular segment.

The area of such a circular segment is the difference between the area of a circular sector and the area of an isosceles triangle having base length $l\frac{\sqrt{3}-1}{\sqrt{2}}$ and height $l\cos\frac{\pi}{12}=\frac{\sqrt{3}+1}{2\sqrt{2}}l$, hence: $$S = \left(\frac{\pi}{12}-\frac{1}{4}\right)l^2$$ and the area of the "circular square" is just: $$Q = \left(1-\sqrt{3}+\frac{\pi}{3}\right)l^2.$$

With integrals, by following Shabbeh's suggestion, we have: $$Q = 4l^2\int_{1/2}^{\sqrt{3}/2}\left(\sqrt{1-x^2}-\frac{1}{2}\right)dx = 2l^2\left.\left(x\sqrt{1-x^2}-x+\arcsin x\right)\right|_{1/2}^{\sqrt{3}/2}$$ that obviously leads to the same result. Just a matter of taste, as usual.

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why does the curvilinear square cut the quarter-circles in thirds ? – Shabbeh Jul 26 '14 at 4:12
Because the intersections between quarter-circles lie on the axis of the square and $\arcsin\frac{1}{2}=\frac{\pi}{6}=\frac{1}{3}\cdot\frac{\pi}{2}.$ – Jack D'Aurizio Jul 26 '14 at 4:14
what do you mean by "circular sector"? – Shabbeh Jul 26 '14 at 4:27
The region in a circle between two rays emanating from the center - en.wikipedia.org/wiki/Circular_sector – Jack D'Aurizio Jul 26 '14 at 4:29
And of course, once I use the correct sums, I get the same answer as you. Silly me for trying to do a geometry sum on my mobile... – abiessu Jul 26 '14 at 6:35

Consider the quarter-circle of radius $r$ and a $\pi\over 2$ rotation of it where the two arcs share two common corners and each shares three corners with the square of side length $r$. Call the area not covered by quarter-circle pieces a "counter-arc" (a triangle with missing arc pieces). The area covered by quarter-circles can be broken up into intersecting and non-intersecting sections; note that we can cut in straight lines and get two $\pi\over 12$ arc sections and one equilateral triangle, so we have counter-arc area $$c=r^2-\frac 12 (r)\frac {\sqrt 3}2r-2\cdot \frac 1{12}\pi r^2=r^2\left(1-\frac {\sqrt 3}4-\frac {\pi}6\right)$$ Now we have the counter-arc, we can get the arrow-arc (the area outside one quarter-circle and between two counter-arcs), which is $$a=r^2-{\pi r^2\over 4}-2c=r^2\left(1-\frac {\pi}4-2+\frac{\sqrt 3}2+\frac {\pi}3\right)$$

The arrow-arcs and counter-arcs together make up the outside perimeter of the indicated shape, so we have shaded area

$$r^2-4a-4c=r^2\left(1-4+\sqrt 3+\frac {2\pi}3-4+\pi+8-2\sqrt 3-\frac {4\pi}3\right)=r^2\left(1-\sqrt 3+\frac {\pi}3\right)$$

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But $5\pi-\sqrt{3}-\frac{5}{6}\pi=11.3579\ldots$ looks a bit too big for me. – Jack D'Aurizio Jul 26 '14 at 4:34
@jackdaurizio: I know, it's difficult to add things correctly on my mobile – abiessu Jul 26 '14 at 4:42

Let $R$ be its radius and $D$ its diameter: $R = 5$, $D = 10$.

\begin{align} \text{Area of big square} &= D^2 = 100 \\ \text{Area of circle} &= \frac{\pi D^2}{4} \approx 78.54 \\ \text{Area outside circle} &= 100 - 78.54 = 21.46 \\ \text{Area of 4 petals} &= 78.54 - 21.46 = 57.08 \\ \text{Area of single petal} &= \frac{57.08}{4} = 14.27 \\ \text{Area of small square} &= R^2 = 25 \end{align}

Let $x$ denote the area of the portion selected.

\begin{align} \text{Area of 2 petals} &= 2 \cdot 14.27 = 28.54 \\ 0 &= 25 - 28.54 + x \\ \text{Area of 8 petals} &= 2 \cdot 57.08 = 114.16 \\ \text{OR}\\ 100 - 114.16 + 4 x &= 0 \end{align} $$x = 3.54$$ Here we have a small area which needs to be added, that I found out by modeling to be $4.34$ \begin{align} \text{This gives us the desired area of}\\ \text{4.34+3.54} &\approx 7.88\\ \text{& a percentage of} &\approx 4 \times 7.88\\ \text{that is} &\approx 31.52\% \end{align}

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Please see math notation guide. Also, the answer appears to be incomplete. Where does $57.08$ come from, for the area of four petals? – user147263 Jul 26 '14 at 18:10
Looks like you're making good progress learning the TeX markup! Just a note: Avoid re-posting your answer with every little edit. Use the preview area (under the editor) to see how things look as you type. The StackExchange software is pretty good at remembering the state of your last text entry if you happen to accidentally close your browser window or something, so you don't have to keep "saving" your progress by repeatedly posting. (TeX Tip: Wrap the "words" in your equations with \text{}.) – Blue Jul 27 '14 at 14:31
Is $x=3.54$ supposed to be the area of the shaded region in the question, when the area of the whole square is $100$? In other words, are you saying that the shaded region is less than 4% of the whole square? – Andreas Blass Jul 27 '14 at 15:37
I have now modified the above solution by adding the missing area which I found out by solid modelling and now the percentage area is almost same as that deduced by coordinate geomtry given next. – Berry P J Jul 30 '14 at 16:18

\begin{align} \text{Let r be radius of circle in square, if}\\ \text{ r} &= 10\\ \text{by coordinate geometry marked Area} &= \left( \frac {\pi + 3 - 3 \sqrt 3}{3} \right) \cdot r^2\\ &\approx 31.51 \end{align}

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It seems I have to learn a lot, to come to such level. – Berry P J Jul 29 '14 at 3:32