Let the area enclosed between the curves |y|= | vedclass

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(D) The given curves are $C_1: |y|=1-x^2$ and $C_2: x^2+y^2=1$.Due to symmetry,the area $\alpha$ is four times the area in the first quadrant between

Vedclass · Accepted Answer

$33$

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(D) The given curves are $C_1: |y|=1-x^2$ and $C_2: x^2+y^2=1$.Due to symmetry,the area $\alpha$ is four times the area in the first quadrant between the circle $y=\sqrt{1-x^2}$ and the parabola $y=1-x^2$.$\alpha = 4 \int_0^1 (\sqrt{1-x^2} - (1-x^2)) dx$$\alpha = 4 \left[ \int_0^1 \sqrt{1-x^2} dx - \int_0^1 (1-x^2) dx \right]$Using the standard integral $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,we get:$\alpha = 4 \left[ \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right)_0^1 - \left( x - \frac{x^3}{3} \right)_0^1 \right]$$\alpha = 4 \left[ (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (1 - \frac{1}{3}) \right]$$\alpha = 4 \left[ \frac{\pi}{4} - \frac{2}{3} \right] = \pi - \frac{8}{3}$Given $9\alpha = \beta\pi + \gamma$,we have $9(\pi - \frac{8}{3}) = 9\pi - 24$.Comparing,$\beta = 9$ and $\gamma = -24$.Thus,$|\beta - \gamma| = |9 - (-24)| = |9 + 24| = 33$.

Let the area enclosed between the curves $|y|=1-x^2$ and $x^2+y^2=1$ be $\alpha$. If $9\alpha=\beta\pi+\gamma$,where $\beta$ and $\gamma$ are integers,then the value of $|\beta-\gamma|$ equals

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