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The region is bounded by the circle $x^2 + y^2 = 4$ (upper semi-circle $y = \sqrt{4-x^2}$) and the line $y = 2 - x$. The condition $y \geq 0$ restricts the region to the first and second quadrants.
Substitute $y = 2 - x$ into $x^2 + y^2 = 4$: $$x^2 + (2-x)^2 = 4$$ $$x^2 + 4 - 4x + x^2 = 4$$ $$2x^2 - 4x = 0 \implies 2x(x-2) = 0$$ The intersection points are $x = 0$ and $x = 2$. Corresponding $y$ values are $y = 2$ and $y = 0$.
The area is the integral of the upper curve minus the lower curve from $x=0$ to $x=2$: $$Area = \int_{0}^{2} (\sqrt{4-x^2} - (2-x)) dx$$
Split the integral: $$\int_{0}^{2} \sqrt{4-x^2} dx - \int_{0}^{2} (2-x) dx$$ Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$: $$[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}(\frac{x}{2})]_0^2 - [2x - \frac{x^2}{2}]_0^2$$ $$= (0 + 2\sin^{-1}(1)) - (0 + 2\sin^{-1}(0)) - (4 - 2) = 2(\frac{\pi}{2}) - 2 = \pi - 2$$
Final Answer: (\pi - 2) \text{ square units}
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