The area of the region,enclosed by the circle | vedclass

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(B) The circle is given by $x^{2}+y^{2}=2$,so its radius $r = \sqrt{2}$. The area of the circle is $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$.The region c

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$\frac{1}{6}(12 \pi-1)$

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(B) The circle is given by $x^{2}+y^{2}=2$,so its radius $r = \sqrt{2}$. The area of the circle is $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$.The region common to the parabola $y^{2}=x$ and the line $y=x$ is bounded by the points of intersection. Setting $y^{2}=x$ and $y=x$ gives $x^{2}=x$,so $x(x-1)=0$,meaning $x=0$ and $x=1$. The points of intersection are $(0,0)$ and $(1,1)$.The area $A$ of the region bounded by the parabola and the line is given by the integral:$A = \int_{0}^{1} (\sqrt{x} - x) dx$$A = \left[ \frac{2}{3}x^{3/2} - \frac{x^{2}}{2} ight]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.The required area is the area of the circle minus the common area $A$:$	ext{Required Area} = 2\pi - \frac{1}{6} = \frac{12\pi - 1}{6} = \frac{1}{6}(12\pi - 1)$.

The area of the region,enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x$,is

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