What is the area bounded by the curves x^2 + | vedclass

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(B) Given equations are $x^2 + y^2 = 9$ $(i)$ and $y^2 = 8x$ $(ii)$.Substituting $(ii)$ in $(i)$,we get $x^2 + 8x - 9 = 0$,which factors as $(x + 9)(x

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$\frac{2\sqrt{2}}{3} + \frac{9\pi}{2} - 9\sin^{-1}\left(\frac{1}{3}\right)$

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(B) Given equations are $x^2 + y^2 = 9$ $(i)$ and $y^2 = 8x$ $(ii)$.Substituting $(ii)$ in $(i)$,we get $x^2 + 8x - 9 = 0$,which factors as $(x + 9)(x - 1) = 0$.Thus,$x = 1$ (since $x = -9$ is not possible for $y^2 = 8x$).At $x = 1$,$y^2 = 8$,so $y = \pm 2\sqrt{2}$. The intersection points are $(1, 2\sqrt{2})$ and $(1, -2\sqrt{2})$.The required area is symmetric about the $x$-axis,so Area $= 2 	imes \int_{0}^{1} \sqrt{8x} \, dx + 2 	imes \int_{1}^{3} \sqrt{9 - x^2} \, dx$.$= 2 	imes 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{1} + 2 \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}ight) ight]_{1}^{3}$.$= 4\sqrt{2} \left( \frac{2}{3} ight) + 2 \left[ \left( 0 + \frac{9}{2}\sin^{-1}(1) ight) - \left( \frac{1}{2}\sqrt{8} + \frac{9}{2}\sin^{-1}\left(\frac{1}{3}ight) ight) ight]$.$= \frac{8\sqrt{2}}{3} + 2 \left[ \frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}ight) ight]$.$= \frac{8\sqrt{2}}{3} + \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}\left(\frac{1}{3}ight) = \frac{2\sqrt{2}}{3} + \frac{9\pi}{2} - 9\sin^{-1}\left(\frac{1}{3}ight)$.

What is the area bounded by the curves $x^2 + y^2 = 9$ and $y^2 = 8x$?

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