4050.12 – Area External to Two Circles

Let $\mathcal{A}$ be the shaded area. Then

$$\mathcal{A} = 2(\Delta ADB - \text{I} - \text{II})$$

For the area of $\Delta ADB$, we need the length $AD$:

$$AD = \sqrt{(3r)^2 - (2r)^2} = r\sqrt{5}$$

$$\Delta ADB = \dfrac{r\sqrt{5} \cdot (2r)}{2} = r^2\sqrt{5}$$

For the area of sectors $\text{I}$ and $\text{II}$ we need the angle $\Theta$:

$$\Theta = \arcsin\left(\dfrac{2r}{3r}\right) = \arcsin\left(\dfrac{2}{3}\right)$$

$$I = \dfrac{\Theta}{360} \pi r^2, \ II = \dfrac{90-\Theta}{360} \pi (2r)^2$$

Putting this all together:

$$\mathcal{A} = r^2 \left( 2 \sqrt{5} + \pi \left(\dfrac{\arcsin{\left(\dfrac{2}{3}\right)}}{360} + \dfrac{90 - \arcsin{\left(\dfrac{2}{3}\right)}}{90}\right) \right)$$