Question

What is the area enclosed by 4 circle in graph?
Given that the verticles of square is center of circle.
And the radius of circle = 1 units
the sides of squares = 1 units

Answer 1

With a little luck, we can exploit some symmetries. I suggest placing this whole figure on a set of coordinate axes with the Origin at the very center. This makes the equations of the four circles

A: \((x - 1/2)^{2} + (y - 1/2)^2 = 1\)

B: \((x - 1/2)^{2} + (y + 1/2)^2 = 1\)

C: \((x + 1/2)^{2} + (y - 1/2)^2 = 1\)

D: \((x + 1/2)^{2} + (y + 1/2)^2 = 1\)

There is an intersection on the positive x-axis that is of interest. It requires solving between C and D. You should find that \(x = \dfrac{\sqrt{3} - 1}{2}\)

With symmetry, there is a similar intersection point between B and D on the positive y-axis. We don't really need this, but it's interesting to notice.

We're ready, then: \(4\cdot\int\limits_{0}^{\dfrac{\sqrt{3} - 1}{2}} -\dfrac{1}{2}+\sqrt{1-\left(x + 1/2\right)^2}\;dx\)

Simple as that! :-)

There may be some algebraic simplifications in there.

Answer 2

Anyway, I get \(\dfrac{\pi}{3} + 1 - \sqrt{3}\), a little less than 1/3, which appears to be reasonable.

Answer 3

@tkhunny
Got the same result integrating over 1/4 the required area.
\[4\int\limits_{\frac{1}{2} \left(2-\sqrt{3}\right)}^{\frac{1}{2}} \left(\sqrt{2 x-x^2}-\frac{1}{2}\right) \, dx=\frac{1}{3} \left(3-3 \sqrt{3}+\pi \right)=1-\sqrt{3}+\frac{\pi }{3} \]