Question

The area of the region above the x-axis, included between the parabola $y^2=\mathrm{ax}$ and the circle $x^2+y^2=2\mathrm{ax}$is

• A. $a^2\left(\frac{\pi }{4}-\frac{2}{3}\right)squareunits$
• B. $2a^2\left(\frac{\pi }{4}-\frac{2}{3}\right)\mathrm{square}\mathrm{units}$
• C. $a\left(\frac{\pi }{4}-\frac{2}{3}\right)\mathrm{square}\mathrm{units}$
• D. $\frac{a}{2}\left(\frac{\pi }{4}-\frac{2}{3}\right)\mathrm{square}\mathrm{units}$

Answer 1

The correct option is A $a^2\left(\frac{\pi }{4}-\frac{2}{3}\right)squareunits$

Solving the given equations of curves, we have
$x^2+\mathrm{ax}=2\mathrm{ax}\Rightarrow x^2-\mathrm{ax}=0\Rightarrow x\left(x-a\right)=0\Rightarrow x=0\mathrm{or}a\mathrm{which}\mathrm{gives},y=a\mathrm{or}\pm a$
$\mathrm{Area}\mathrm{ODAB}={\int }_0^a\left(\sqrt{2\mathrm{ax}-x^2}-\sqrt{\mathrm{ax}}\right)\mathrm{dx}\mathrm{Let}x=2a{\sin }^{2}t.\mathrm{Then}\mathrm{dx}=4a\sin t\cos t\mathrm{dt}\mathrm{When}x=0\Rightarrow t=0\mathrm{and},\mathrm{When}x=a\Rightarrow t=\frac{\pi }{4}\mathrm{Again},{\int }_0^a\sqrt{2\mathrm{ax}-x^2}\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\left(2a\sin t\cos t\right)\left(4a\sin t\cos t\right)\mathrm{dt}=a^2{\int }_0^{\frac{\pi }{4}}2\times {\left(2\sin t\cos t\right)}^2\mathrm{dt}=a^2{\int }_0^{\frac{\pi }{4}}2{\sin }^{2}2t\mathrm{dt}=a^2{\int }_0^{\frac{\pi }{4}}\left(1-\cos 4t\right)\mathrm{dt}=a^2{\left(t-\frac{\sin 4t}{4}\right]}_0^{\frac{\pi }{4}}=\frac{{\mathrm{\pi a}}^2}{4}\mathrm{Now},{\int }_0^a\sqrt{\mathrm{ax}}\mathrm{dx}=\sqrt{a}\times \frac{2}{3}{\left(x^{\frac{3}{2}}\right]}_0^a=\frac{2}{3}{a}^2\mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{area}=\frac{{\mathrm{\pi a}}^2}{4}-\frac{2}{3}{a}^2=a^2\left(\frac{\pi }{4}-\frac{2}{3}\right)$