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  • Find the area of the region {(x, y) : y^2 ≤ 4x, 4x^2 + 4y^2 ≤ 9}

Q : 15 Find the area of the region  \small \left \{ (x,y);y^2\leq 4x,4x^2+4y^2\leq 9 \right \}.

Answers (1)

best_answer

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

For the first quadrant

\\4x^{2}+4y^{2}=9\\ y=\sqrt{\frac{9}{4}-x^{2}}

\\y^{2}=4x\\ y=2\sqrt{x}

In the first quadrant, the curves intersect at a point \left ( \frac{1}{2},\sqrt{2} \right )

Area of  the unshaded region in the first quadrant is

\\\int_{0}^{\frac{1}{2}}\left ( \sqrt{\frac{9}{4}-x^{2}} -2\sqrt{x}\right )dx\\ =[\frac{x}{2}\sqrt{\frac{9}{4}-x^{2}}+\frac{9}{8}sin^{-1}\frac{2x}{3}]{_{0}}^{\frac{1}{2}}-4[\frac{x^{3/2}}{3}]_0^{1/2}\\ =\frac{\sqrt{2}}{4}+\frac{9}{8}sin^{-1}\frac{1}{3}-\frac{\sqrt{2}}{3}

The total area of the shaded region is-
= Area of half circle - area of the shaded region in the first quadrant

\\\frac{\pi }{2}\times (\frac{3}{2})^{2}-2\left ( \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{3}+\frac{9}{8}sin^{-1}\frac{1}{3}\right )\\ =\frac{9 }{8}\left ( \pi-2sin^{-1}\frac{1}{3} \right )+\frac{\sqrt{2}}{6}\ units

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Sayak

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