If the area of the circle x^2+y^2=2 is divide | vedclass

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(A) Given equations of curves are $x^2+y^2=2$ and $y=x^2$.For the point of intersection,substitute $x^2=y$ into the circle equation:$y+y^2=2 \Rightarr

Vedclass · Accepted Answer

$\frac{3 \pi}{2}-\frac{1}{3}$

Vedclass · Answer

(A) Given equations of curves are $x^2+y^2=2$ and $y=x^2$.For the point of intersection,substitute $x^2=y$ into the circle equation:$y+y^2=2 \Rightarrow y^2+y-2=0$$(y+2)(y-1)=0$Since $y=x^2 \ge 0$,we have $y=1$. Thus,$x^2=1 \Rightarrow x=\pm 1$.The area of the circle is $A_{circle} = \pi r^2 = 2\pi$.The area of the smaller part $A_{small}$ is the area under the circle minus the area under the parabola between $x=-1$ and $x=1$:$A_{small} = \int_{-1}^{1} (\sqrt{2-x^2} - x^2) dx = 2 \int_{0}^{1} \sqrt{2-x^2} dx - 2 \int_{0}^{1} x^2 dx$Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:$A_{small} = 2 [\frac{x}{2}\sqrt{2-x^2} + \frac{2}{2}\sin^{-1}(\frac{x}{\sqrt{2}})]_0^1 - 2[\frac{x^3}{3}]_0^1$$A_{small} = 2 [(\frac{1}{2}\sqrt{1} + \sin^{-1}(\frac{1}{\sqrt{2}})) - 0] - \frac{2}{3} = 2(\frac{1}{2} + \frac{\pi}{4}) - \frac{2}{3} = 1 + \frac{\pi}{2} - \frac{2}{3} = \frac{\pi}{2} + \frac{1}{3}$.The area of the larger part is $A_{large} = A_{circle} - A_{small} = 2\pi - (\frac{\pi}{2} + \frac{1}{3}) = \frac{3\pi}{2} - \frac{1}{3}$ sq units.Thus,option $A$ is correct.

If the area of the circle $x^2+y^2=2$ is divided into two parts by the parabola $y=x^2$,then the area (in sq units) of the larger part is

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