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

Q : 1         Find the area of the circle  \small 4x^2+4y^2=9  which is interior to the parabola  \small x^2=4y.

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The area bounded by the circle \small 4x^2+4y^2=9 and the parabola  \small x^2=4y.

By solving the equation we get the intersecting point D(-\sqrt{2},\frac{1}{2}) and B(\sqrt{2},\frac{1}{2})
So, the required area (OBCDO)=2 times the area of (OBCO)
Draw a normal on the x-axis (M =\sqrt{2},0)


Thus the area of OBCO = Area of OMBCO - Area of OMBO

\\\int_{0}^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-\int_{0}^{\sqrt{2}}{\frac{x^2}{4}}dx\\ =\frac{1}{2}\int_{0}^{\sqrt{2}}\sqrt{9-4x^2}-\frac{1}{4}\int_{0}^{\sqrt{2}}x^2dx\\ =\frac{1}{4}[x\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}]_0^{\sqrt{2}}-\frac{1}{4}[\frac{x^3}{3}]_0^{\sqrt{2}}\\ =\frac{1}{4}[\sqrt{2}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}]-\frac{1}{12}(\sqrt{2})^3\\ =\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\\ =\frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})
S0, total area = 
\\=2\times \frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})\\ =\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}
 

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manish

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