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

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:

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

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 $4\pi r^2 - 4p = (2\pi+4) r^2$

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: