The parabola $y^2=4 x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:

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Aicharya · Accepted Answer

$\begin{aligned} & y^2=4 x \ & x^2+y^2=5\end{aligned}$ $	herefore$ Area of shaded region as shown in the figure will be $\begin{aligned} & \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \ & =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}ight]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}ight]_1^{\sqrt{5}} \ & =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}ight) \ & 	herefore 	ext { Required Area }=2 \mathrm{~A}_1 \ & =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}ight) \ & =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}ight) \ & =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} \ & =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}ight)\end{aligned}$

The parabola $y^2=4 x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part…

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