The area enclosed by the curves y = sqrt{x} a | vedclass

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(D) The curves are $y = \sqrt{x} \Rightarrow x = y^2$ and $x = -\sqrt{y} \Rightarrow y = x^2$ (for $x < 0$).The intersection points with the circle $x

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$\frac{\pi}{2}$

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(D) The curves are $y = \sqrt{x} \Rightarrow x = y^2$ and $x = -\sqrt{y} \Rightarrow y = x^2$ (for $x The intersection points with the circle $x^2 + y^2 = 2$ are $B(1, 1)$ and $A(-1, 1)$.The area is the region bounded by the circle $y = \sqrt{2 - x^2}$ and the curves $y = x^2$ (for $x  0$).The area $A$ is given by:$A = \int_{-1}^{0} (\sqrt{2 - x^2} - x^2) dx + \int_{0}^{1} (\sqrt{2 - x^2} - \sqrt{x}) dx$Evaluating the integrals:$\int_{-1}^{0} \sqrt{2 - x^2} dx = [\frac{x}{2}\sqrt{2 - x^2} + \frac{2}{2}\arcsin(\frac{x}{\sqrt{2}})]_{-1}^{0} = 0 - (-\frac{1}{2} - \frac{\pi}{4}) = \frac{1}{2} + \frac{\pi}{4}$$\int_{-1}^{0} -x^2 dx = [-\frac{x^3}{3}]_{-1}^{0} = 0 - \frac{1}{3} = -\frac{1}{3}$$\int_{0}^{1} \sqrt{2 - x^2} dx = [\frac{x}{2}\sqrt{2 - x^2} + \arcsin(\frac{x}{\sqrt{2}})]_{0}^{1} = (\frac{1}{2} + \frac{\pi}{4}) - 0 = \frac{1}{2} + \frac{\pi}{4}$$\int_{0}^{1} -\sqrt{x} dx = [-\frac{2}{3}x^{3/2}]_{0}^{1} = -\frac{2}{3}$Summing these: $A = (\frac{1}{2} + \frac{\pi}{4} - \frac{1}{3}) + (\frac{1}{2} + \frac{\pi}{4} - \frac{2}{3}) = \frac{\pi}{2} + 1 - 1 = \frac{\pi}{2}$.

The area enclosed by the curves $y = \sqrt{x}$ and $x = -\sqrt{y}$,and the circle $x^2 + y^2 = 2$ above the $x$-axis,is

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