Question

Find the area common to the circle x² + y² = 16 a² and the parabola y² = 6 ax.

OR

Find the area of the region {(x, y) : y² ≤ 6ax} and {(x, y) : x² + y² ≤ 16a²}.

Answer 1

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

$$\begin{gathered} x^2+y^2=16a^2\mathrm{and}{y}^2=6ax \\ \Rightarrow x^2+6ax=16a^2 \\ \Rightarrow x^2+6ax-16a^2=0 \\ \Rightarrow \left(x+8a\right)\left(x-2a\right)=0 \\ \Rightarrow x=2a\mathrm{or}x=-8a,x=-8a\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{possible}\mathrm{solution}. \\ \therefore \mathrm{When}x=2a,\mathrm{y}=\pm \sqrt{6a\times 2a}=\pm \sqrt{12}a=\pm 2\sqrt{3}a \\ \therefore B\left(2a,2\sqrt{3a}\right)\mathrm{and}B\text{'}\left(2a,-2\sqrt{3}a\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{and}\mathrm{circle}. \\ \mathrm{Now},\mathrm{Required}\mathrm{area}=\mathrm{area}\left(OBAB\text{'}O\right) \\ =2\times \mathrm{area}\left(OBAO\right) \\ =2\left\{\mathrm{area}\left(OBDO\right)+\mathrm{area}\left(DBAD\right)\right\} \\ =2\times \left[{\int }_0^{2a}\sqrt{6ax}dx+{\int }_{2a}^{4a}\sqrt{16a^2-x^2}dx\right] \\ =2\times \left\{{\left[\sqrt{6a}\frac{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}_0^{2a}+{\left[\frac{1}{2}x\sqrt{16a^2-x^2}+\frac{1}{2}\times 16a^2{\sin }^{-1}\left(\frac{x}{4a}\right)\right]}_{2a}^{4a}\right\} \\ =2\times \left\{\left(\sqrt{6a}\times \frac{2}{3}\times {\left(2a\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0\right)+\left(\frac{1}{2}\times 4a\sqrt{16a^2-{\left(4a\right)}^2}+\frac{1}{2}\times 16a^2{\sin }^{-1}\frac{4a}{4a}-\frac{1}{2}\times 2a\sqrt{16a^2-{\left(2a\right)}^2}-\frac{1}{2}\times 16a^2{\sin }^{-1}\frac{2a}{4a}\right)\right\} \\ =2\times \left\{\left(\sqrt{6a}\times \frac{2}{3}\times 2a\sqrt{2a}\right)+0+8a^2{\sin }^{-1}\left(1\right)-2\sqrt{3}{a}^2-8a{\sin }^{-1}\left(\frac{1}{2}\right)\right\} \\ =2\times \left[\frac{8a^2\sqrt{3}}{3}+8a^2\times \frac{\mathrm{\pi }}{2}-2\sqrt{3}{a}^2-8a^2\frac{\mathrm{\pi }}{6}\right] \\ =2\left\{\left(\frac{8\sqrt{3}-6\sqrt{3}}{3}\right)a^2+8\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}\right)a^2\right\} \\ =2\left\{\frac{2\sqrt{3}}{3}{a}^2+8a^2\left(\frac{2\mathrm{\pi }}{6}\right)\right\} \\ =\frac{4\sqrt{3}}{3}{a}^2+\frac{16\mathrm{\pi }}{3}{a}^2 \\ =\frac{4a^2}{3}\left(4\mathrm{\pi }+\sqrt{3}\right) \end{gathered}$$