Find the area of the circle 4x^{2}+4y^{2}=9 w | vedclass

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(A) The required area is represented by the shaded region $OBCDO$.Solving the equations $4x^{2}+4y^{2}=9$ and $x^{2}=4y$,we substitute $x^{2}=4y$ into

Vedclass · Accepted Answer

$\left[\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right]$

Vedclass · Answer

(A) The required area is represented by the shaded region $OBCDO$.Solving the equations $4x^{2}+4y^{2}=9$ and $x^{2}=4y$,we substitute $x^{2}=4y$ into the circle equation:$4(4y)+4y^{2}=9 \implies 4y^{2}+16y-9=0$.Using the quadratic formula,$y = \frac{-16 \pm \sqrt{256 - 4(4)(-9)}}{8} = \frac{-16 \pm \sqrt{400}}{8} = \frac{-16 \pm 20}{8}$.Since $y \ge 0$ for the parabola,$y = \frac{4}{8} = \frac{1}{2}$.Then $x^{2} = 4(\frac{1}{2}) = 2$,so $x = \pm \sqrt{2}$. The intersection points are $B(\sqrt{2}, \frac{1}{2})$ and $D(-\sqrt{2}, \frac{1}{2})$.The area is symmetric about the $y$-axis,so Area $= 2 	imes \int_{0}^{\sqrt{2}} (y_{circle} - y_{parabola}) dx$.$y_{circle} = \sqrt{\frac{9-4x^{2}}{4}} = \frac{1}{2}\sqrt{9-4x^{2}}$ and $y_{parabola} = \frac{x^{2}}{4}$.Area $= 2 \int_{0}^{\sqrt{2}} (\frac{1}{2}\sqrt{9-4x^{2}} - \frac{x^{2}}{4}) dx = \int_{0}^{\sqrt{2}} \sqrt{9-4x^{2}} dx - \frac{1}{2} \int_{0}^{\sqrt{2}} x^{2} dx$.Using $\int \sqrt{a^{2}-u^{2}} du = \frac{u}{2}\sqrt{a^{2}-u^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{u}{a})$,let $u=2x, du=2dx$:$= \frac{1}{2} [\frac{2x}{2}\sqrt{9-4x^{2}} + \frac{9}{2}\sin^{-1}(\frac{2x}{3})]_{0}^{\sqrt{2}} - \frac{1}{2} [\frac{x^{3}}{3}]_{0}^{\sqrt{2}}$$= [\frac{x}{2}\sqrt{9-4x^{2}} + \frac{9}{4}\sin^{-1}(\frac{2x}{3})]_{0}^{\sqrt{2}} - \frac{1}{6}(\sqrt{2})^{3}$$= (\frac{\sqrt{2}}{2}\sqrt{9-8} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3})) - \frac{2\sqrt{2}}{6}$$= \frac{\sqrt{2}}{2} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3}) - \frac{\sqrt{2}}{3} = \frac{\sqrt{2}}{6} + \frac{9}{4}\sin^{-1}(\frac{2\sqrt{2}}{3})$ sq. units.

Find the area of the circle $4x^{2}+4y^{2}=9$ which is interior to the parabola $x^{2}=4y$.

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