Find This Area

Since arc $AC$ is one fourth of circumference of center $D$ , then $\angle CDA=90^\circ$ . Since arc $AB$ is $\dfrac{1}{8}$ of circumference of center $C$ , $\angle ACB=45^\circ$ .

By Pythagorean theorem, $AC=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$ .

It follows that $BC=AC=2\sqrt{2}$ .

Consider sector $CDA:$ The area of sector $CDA$ is $A=\dfrac{1}{4}\pi (2^2)=\pi$

Consider sector $ACB:$ The area of sector $ACB$ is $A=\dfrac{45}{360}\pi(2\sqrt{2})^2=\dfrac{1}{8}\pi(4)(2)=\pi$

Area of the yellow segment is $\pi-\dfrac{1}{2}(2)(2)=\pi-2$

Finally, the area of the green region is $\pi-(\pi-2)=\pi-\pi+2=$ $\boxed{2}$

Unshaded area:

$=\frac { 1 }{ 8 } { \left( 2\sqrt { 2 } \right) }^{ 2 }\pi -\left( \frac { 1 }{ 4 } \cdot { 2 }^{ 2 }\pi -\frac { 1 }{ 2 } \cdot { 2 }^{ 2 } \right) \\ =\frac { 1 }{ 8 } \cdot 8\pi -\left( \frac { 1 }{ 4 } \cdot 4\pi -2 \right) \\ =\pi -(\pi -2)\\ =\pi -\pi +2\\ =2$