What is the total area belonging to only one of four unit circles?

Question

Please forgive the crudeness of this diagram.

(I took an image from some psychobabble website and tried to delete the larger circle that's not relevant to my question).

Let's say these are four unit circles joined together such that each circle shares some area with two other circles.

Obviously the total area not shared with other circles is four times the area of this (again please forgive the crude diagram)

Or I could calculate the total are of a single "petal" and multiply that by $4$. But I have truly forgotten all the calculus and trigonometry I was taught more than half a century ago.

Am I on the right track? Is there a better way than either of the two ideas I've had so far?

P.S. Not sure if osculating circle tag applies.

Answer 1

For one "petal" only, consider this:

$$\text{area of petal} = \text{area of square} = \left(\;\sqrt{2}\;r\;\right)^2 = 2 r^2$$

For the entire figure:

$$\text{total area} \;=\; 4 r^2 + 4\cdot \frac{1}{2} \pi r^2 \;=\; 4 r^2 + 2 \pi r^2$$

Answer 2

If the circle has radius $r$, the area of the "black diagram" is the area of the circle minus the area of two "petals". The area of two "petals" can be obtained as the area of a circle minus the area of the inscribed square. Therefore the area of the black diagram is just the area of the inscribed square $4(r^2/2))=2r^2.$

P.S. My "petal" is the intersection of two circles.

P.P.S. The diagram is composed by two full circles and two black diagrams so its total area is $2\pi r^2+2\cdot 2r^2=(2\pi+4)r^2.$

P.P.P.S. The picture says it all:

Answer 3

Let's say the circles each have radius $1$. Rotating a bit for convenience, their centres could be at $[1,0]$, $[0,1]$, $[-1,0]$, $[0,-1]$. The area of your black region is the area of a circle minus twice the area common to two adjacent circles. See e.g. here

Answer 4

Draw two lines from the center of a circle, one to the center of the diagram, one to the center of the diagram, the other to a tip of the petal.

This sector has an angle of 90 degrees. The area of the sector = $\frac 14$ the area of the cirlce or $\frac \pi4 r^2$ Now join the endpoints, creating an isosceles right triangle. The area of the triangle is $\frac 12 r^2.$

$\frac 12$ petal = $(\frac \pi4 - \frac 12) r^2$

There are 8 half petals.

The total area then is $4 \pi r^2 - 8(\frac \pi4 - \frac 12) r^2 = (2\pi + 4) r^2$

Answer 5

Another approach:

Consider the square enclosed in red. It has side length $r$, and therefore area $r^2$. Notice that a quarter portion of two different circles overlap in this region to create a single petal, which has area $p$.

So we have $r^2 = 2\left(\frac{1}{4}\pi r^2\right) - p$, which rearranges to $p = (\frac{\pi}{2}-1) r^2$

Then the unshared area belonging to a single circle has area $\pi r^2 - 2p = 2r^2$

The area of the entire diagram is $4\pi r^2 - 4p = (2\pi+4) r^2$