The circle of radius 3 have an area \(9\pi\). We sign r as the radius of  the pictured quadrant of the done cỉcle, then

\(r=3\sqrt{2}+3\). Put x is the area to calculate, we have  

               \(\dfrac{1}{4}\pi r^2=2x+\pi.3^2+\left(3^2-\dfrac{1}{4}.\pi.3^2\right)=2x+9\left(1+\dfrac{3\pi}{4}\right)\)

      ​ ​\(\Leftrightarrow\dfrac{\pi}{4}\left(3\sqrt{2}+3\right)^2=2x+9\left(1+\dfrac{3\pi}{4}\right)\Leftrightarrow2x=\dfrac{\left(27+18\sqrt{2}\right)\pi}{4}-\dfrac{36+27\pi}{4}\)

       \(\Leftrightarrow x=\dfrac{9\sqrt{2}\pi-36}{4}\)

We have    \(y=3^2-\left(\dfrac{1}{4}\pi3^2\right)\) and  \(r-3=3\sqrt{2}\Rightarrow r=3+3\sqrt{2}\).

The circle of radius 3 have an area 9π

. We sign r as the radius of  the pictured quadrant of the done cỉcle, then

r=3√2+3

. Put x is the area to calculate, we have  

               14πr2=2x+π.32+(32−14.π.32)=2x+9(1+3π4)

      ​ ​⇔π4(3√2+3)2=2x+9(1+3π4)⇔2x=(27+18√2)π4−36+27π4

       ⇔x=9√2π−364