Find the area

Taking only the portion between vertical central line and line from the center at 45 degrees to the first line. One eighth of the whole.
$\text{The area covered by an arc= Quarter circle - half sqare.} \\ =\dfrac 1 4 *\pi*r^2-\dfrac1 2 *r^2.\\ Outer ~~ ~ arc:- Area=\dfrac 1 4 *\pi*3^2-\dfrac1 2 *3^2 .......................(1)\\ Middle~ arc:- Area=\dfrac 1 4 *\pi*2^2-\dfrac1 2 *2^2 .......................(2)\\ Inner ~arc:- ~~Area=\dfrac 1 4 *\pi*1^2-\dfrac1 2 *1^2 .......................(3)\\ Shaded~Area~=~(1)~-~(2)~+~(3)=\dfrac 1 4 *\pi*6-3 \\ \text{So for the whole shaded area }=8* \{\dfrac 1 4 *\pi*6-3\}\\ \Large \color{#D61F06}{ 13.68 }$

Looking at the case of one quadrant and one arc :

The shaded area is $r^2 - \frac{\pi r^2}{4}$

Therefore when both arcs are present the total area of one petal : p(r) {the area in one quadrant bounded by the two arcs of radius r) :

$p(r) = r^2 - 2(r^2 - \frac{\pi r^2}{4}) = r^2 - (2r^2 - \frac{\pi r^2}{2}) = r^2 - r^2 (2 - \frac{\pi}{2}) = r^2(1-(2 - \frac{\pi}{2})) = r^2 (\frac{\pi }{2} -1)$

Looking now at the full problem - the shaded area there is :

$4(p(3)-p(2)+p1) = 4(9(\frac{\pi}{2} -1) - 2(\frac{\pi}{2} -1) + (\frac{\pi}{2} -1) )=$ $4((\frac{\pi}{2} -1)(9-4+1) = 4*6*((\frac{\pi}{2} -1) = 12\pi - 24 = 13.68$

The precise area is 13.69911184307

consider half of the figure. Area of semicircle is πr2/2 subtract the right Δ from that the sides of that are 2r, r√2 & r√2. Area is r2. So one petal area is r2 ( π/2 – 1). Total area of 4 petals 4*(r12 – r22 + r32 ) (π/2 – 1) = 4(9 – 4 + 1)(π/2 -1) = 13.7 cm2