The area of the smaller part between the circ | vedclass

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(A) The equation of the circle is $x^2 + y^2 = 4$,which represents a circle with center $(0,0)$ and radius $r = 2$.The line is $x = 1$.The area of the

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$\frac{4\pi}{3} - \sqrt{3}$

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(A) The equation of the circle is $x^2 + y^2 = 4$,which represents a circle with center $(0,0)$ and radius $r = 2$.The line is $x = 1$.The area of the smaller part is symmetric about the $x$-axis.Area $= 2 \int_{1}^{2} y \, dx = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$:Area $= 2 \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2}) \right]_{1}^{2}$$= 2 \left[ (\frac{2}{2} \sqrt{4 - 4} + 2 \sin^{-1}(1)) - (\frac{1}{2} \sqrt{4 - 1} + 2 \sin^{-1}(\frac{1}{2})) \right]$$= 2 \left[ (0 + 2 \cdot \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} + 2 \cdot \frac{\pi}{6}) \right]$$= 2 \left[ \pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3} \right]$$= 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right] = \frac{4\pi}{3} - \sqrt{3}$.

The area of the smaller part between the circle $x^2 + y^2 = 4$ and the line $x = 1$ is

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