The area (in sq. units) of the smaller region lying above the X-axis and bounded between the circle $x^2+y^2=2 a x$ and the parabola $y^2=a x$ is

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Aicharya · Accepted Answer

Given the curve $x^2+y^2=2 a x \Rightarrow(x-a)^2+y^2=a^2$ and, $y^2=a x$

Since, area of shaded part $=\frac{\pi a^2}{4}-\int_0^a \sqrt{a x} d x$ $\begin{aligned} & =\frac{\pi a^2}{4}-\sqrt{a} \int_0^a \sqrt{x} d x=\frac{\pi a^2}{4}-\sqrt{a} \times\left(\frac{x^{3 / 2}}{\frac{3}{2}}\right)_0^a \\ & =\frac{a^2 \pi}{4}-\frac{2 a^2}{3}=a^2\left(\frac{\pi}{4}-\frac{2}{3}\right) \text { sq. units. }\end{aligned}$

The area (in sq. units) of the smaller region lying above the X-axis and bounded between the circle…

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