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[ASSESS]

[ADVANCE]

[VERIFY]

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Let \(A\) and \(E\) be the centers of circles \(\omega_{1}\) and \(\omega_{2}\) with radii \(r\) and \(R\), respectively, where \(r < R\). According to the problem, \(A \in \omega_{2}\), and \(B\) and \(C\) are the intersection points of these circles. Note that their common chord \(BC\) is perpendicular to the segment \(AE\). We have:
\[ AB = AC = r, \quad EB = EC = EA = R. \]
Denote \(\angle BAE = \angle CAE = \alpha\) and \(\angle BEA = \angle CEA = \gamma\).

Write the formula for calculating the area of the union of the disks \(\Omega_{1}\) and \(\Omega_{2}\) enclosed by circles \(\omega_{1}\) and \(\omega_{2}\):
\[ S_{\Omega_{1} \cup \Omega_{2}} = S_{\Omega_{1}} + S_{\Omega_{2}} - S_{\Omega_{1} \cap \Omega_{2}}, \]
where \(S_{\Omega_{1}} = \pi r^{2}\) and \(S_{\Omega_{2}} = \pi R^{2}\), the areas of the disks \(\Omega_{1}\) and \(\Omega_{2}\), and \(S_{\Omega_{1} \cap \Omega_{2}}\) is the area of their intersection.

Find \(S_{\Omega_{1} \cap \Omega_{2}}\). The intersection \(\Omega_{1} \cap \Omega_{2}\) consists of two segments \(BDC \in \Omega_{1}\) and \(BAC \in \Omega_{2}\). Denote the areas of these segments by \(S_{1}\) and \(S_{2}\), respectively.

The area \(S_{1}\) of segment \(BDC\) is the difference between the area of sector \(BDCA\) and the area of \(\triangle ABC\):
\[ S_{1} = \frac{(2 \alpha) r^{2}}{2} - \frac{1}{2} r^{2} \sin 2 \alpha \]

Given that \(\triangle ABE\) is isosceles (\(AE = BE\)), we have \(\cos \alpha = \frac{r}{2R}\). Thus:
\[ \alpha = \arccos \frac{r}{2R}, \quad \sin \alpha = \sqrt{1 - \left(\frac{r}{2R}\right)^{2}} \]
\[ \sin 2 \alpha = 2 \sin \alpha \cos \alpha = \frac{r}{2R} \cdot \sqrt{1 - \left(\frac{r}{2R}\right)^{2}} \]

Given \(r = 1\), \(R = 2\), substitute these values into the above formulas:
\[ \cos \alpha = \frac{1}{4}, \quad \alpha = \arccos \frac{1}{4}, \quad \sin \alpha = \frac{\sqrt{15}}{4}, \quad \sin 2 \alpha = \frac{\sqrt{15}}{8} \]
Thus:
\[ S_{1} = \arccos \frac{1}{4} - \frac{\sqrt{15}}{16} \]

The area \(S_{2}\) of segment \(BAC\) is the difference between the area of sector \(EABC\) and the area of \(\triangle CBE\):
\[ S_{2} = \frac{(2 \gamma) R^{2}}{2} - \frac{1}{2} R^{2} \sin 2 \gamma \]

Since \(\triangle ABE\) is isosceles (\(AE = BE\)), \(\gamma = \pi - 2\alpha\). Thus:
\[ \cos \gamma = \cos (\pi - 2\alpha) = -\cos 2\alpha = \sin^{2} \alpha - \cos^{2} \alpha = \frac{15}{16} - \frac{1}{16} = \frac{7}{8} \]
\[ \sin \gamma = \sqrt{1 - \cos^{2} \gamma} = \frac{\sqrt{15}}{8}, \quad \sin 2 \gamma = \frac{7 \sqrt{15}}{32} \]
Thus:
\[ S_{2} = 4 \left( \pi - 2 \arccos \frac{1}{4} \right) - \frac{7 \sqrt{15}}{16} \]

Calculate the area of the intersection of disks \(\Omega_{1}\) and \(\Omega_{2}\):
\[ S_{\Omega_{1} \cap \Omega_{2}} = S_{1} + S_{2} = 4 \pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2} \]

Substitute this result into the formula:
\[ S_{\Omega_{1} \cup \Omega_{2}} = \pi r^{2} + \pi R^{2} - (S_{1} + S_{2}) \]
\[ S_{\Omega_{1} \cup \Omega_{2}} = \pi + 4 \pi - \left( 4 \pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2} \right) \]
Therefore:
\[ S_{\Omega_{1} \cup \Omega_{2}} = \pi + 7 \arccos \frac{1}{4} + \frac{\sqrt{15}}{2} \]

Present the answer in LaTex format: \boxed{Your answer}
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