The area of the region described by {(x, y) m | vedclass

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(C) The region is bounded by the circle $x^2+y^2=1$ and the parabola $y^2=1-x$. To find the intersection points,substitute $y^2=1-x$ into $x^2+y^2=1$:

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$\frac{\pi}{2}+\frac{4}{3}$

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(C) The region is bounded by the circle $x^2+y^2=1$ and the parabola $y^2=1-x$. To find the intersection points,substitute $y^2=1-x$ into $x^2+y^2=1$: $x^2 + (1-x) = 1 \implies x^2 - x = 0 \implies x(x-1) = 0$. So,$x=0$ or $x=1$. For $x=0$,$y^2=1 \implies y=\pm 1$. For $x=1$,$y^2=0 \implies y=0$. The area is symmetric about the $x$-axis. Area $= 2 \left[ \int_{-1}^0 \sqrt{1-x^2} dx + \int_0^1 \sqrt{1-x} dx \right]$. Evaluating the integrals: $2 \left[ \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x \right]_{-1}^0 = 2 \left[ (0 + 0) - (0 - \frac{\pi}{4}) \right] = \frac{\pi}{2}$. $2 \int_0^1 (1-x)^{1/2} dx = 2 \left[ \frac{-2}{3} (1-x)^{3/2} \right]_0^1 = 2 \left[ 0 - (-\frac{2}{3}) \right] = \frac{4}{3}$. Total Area $= \frac{\pi}{2} + \frac{4}{3}$.

The area of the region described by $\{(x, y) \mid x^2+y^2 \leq 1\}$ and $\{y^2 \leq 1-x\}$ is

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