Find the area (in square units) of the region | vedclass

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(A) Step $1$: Find the intersection points of $x^2 + y^2 = 9$ and $y^2 = 8x$. Substituting $y^2 = 8x$ into the circle equation: $x^2 + 8x - 9 = 0 \imp

Vedclass · Accepted Answer

$8\frac{\sqrt{2}}{3} + \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}\frac{1}{3}$

Vedclass · Answer

(A) Step $1$: Find the intersection points of $x^2 + y^2 = 9$ and $y^2 = 8x$. Substituting $y^2 = 8x$ into the circle equation: $x^2 + 8x - 9 = 0 \implies (x+9)(x-1) = 0$. Since $x \geq 0$, $x = 1$. Then $y^2 = 8$, so $y = \pm 2\sqrt{2}$.Step $2$: The area is symmetric about the $x$-axis. Area $= 2 \left[ \int_{0}^{1} \sqrt{8x} \, dx + \int_{1}^{3} \sqrt{9-x^2} \, dx \right]$.Step $3$: Calculate the first integral: $2 \int_{0}^{1} 2\sqrt{2} x^{1/2} \, dx = 4\sqrt{2} [\frac{2}{3} x^{3/2}]_0^1 = \frac{8\sqrt{2}}{3}$.Step $4$: Calculate the second integral: $2 [\frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3})]_1^3 = 2 [(\frac{3}{2}(0) + \frac{9}{2}\sin^{-1}(1)) - (\frac{1}{2}\sqrt{8} + \frac{9}{2}\sin^{-1}(\frac{1}{3}))] = 2 [\frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}(\frac{1}{3})] = \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}(\frac{1}{3})$.Step $5$: Total Area $= \frac{8\sqrt{2}}{3} + \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}(\frac{1}{3})$.

Find the area (in square units) of the region bounded by the circle $x^2 + y^2 = 9$ and the parabola $y^2 \leq 8x$.

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