areacircles
In the diagram,the curved paths are arcs of circles centered at vertices $A$ and $B$ of a square of side $6$. Find the area of the shaded section $BCD$.
I've been stuck on this problem for days. I have tried to get the shaded section by puzzling with the figures but failed. I've tried to add more lines to get a set of equations but that failed. But I still want to be able to solve it by myself so if you guys can give me only hints, that would be best.
In my opinion, solving this problem by line integrals and Green's theorem is overkill. The area of the region of intersection can be found by extremely simple geometry. Note that the triangle formed by the vertices $\{\big(0,0\big), \big(1,0\big), \big(\frac{1}{2}, \frac{\sqrt{3}}{2}\big)\}$ is an equilateral triangle with unit side length, so its area is $\frac{\sqrt{3}}{4}$. Next note that the area of this triangle plus one of the circular segments is a unit circle sector with angle $60^{\circ}$, or in other words, $\frac16$ the area of a unit circle, i.e., $\frac{\pi}{6}$. Thus, the area of one circular segment must be $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$. Altogether, the area of the triangle plus the area of three circular segments is,
$$A=\frac{\sqrt{3}}{4}+3\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=\frac{\pi-\sqrt{3}}{2}.$$
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:
$$\text{total area} \;=\; 4 r^2 + 4\cdot \frac{1}{2} \pi r^2 \;=\; 4 r^2 + 2 \pi r^2$$
Best Answer
Here is a hint with double integrals. Split the area in two:
Then
\begin{align} S_{CBED}&= S_{CED}+S_{CBE} = \int_{x_1}^{x_2}\int_{f_2(x)}^{f_1(x)} dy dx + \int_{x_2}^{x_3}\int_{f_3(x)}^{f_1(x)} dy dx \end{align} where
\begin{align} f_1(x)&=\sqrt{6^2-(x-6)^2}, \\ f_2(x)&=6-\sqrt{6^2-(x-6)^2}, \\ f_3(x)&=x, \\ x_1&=6-3\sqrt{3}, \\ x_2&=6-3\sqrt{2}, \\ x_3&=6. \end{align}
$S_{CBED}=\tfrac{15}{2}\pi-9\sqrt{3}\approx 7.97348763$.
Another way to split the area:
suggests a geometric solution as a sum of the sector $BCD$ and a difference between the sector $ABD$ and $\triangle ABD$.
$\triangle ABD$ is equilateral (why?), so
\begin{align} S_1&=\tfrac12 \cdot 36(\tfrac\pi3-\tfrac\pi4)=\tfrac32 \pi \\ S_2&=\tfrac12 \cdot 36\tfrac\pi3=6\pi \\ S_3&=9\sqrt3 \end{align}
And the answer is $\tfrac{15}2\pi-9\sqrt3$, as above.