Question

The area of the circle x² + y² = 16 enterior to the parabola y² = 6x is
(a) $\frac{4}{3}\left(4\mathrm{\pi }-\sqrt{3}\right)$

(b) $\frac{4}{3}\left(4\mathrm{\pi }+\sqrt{3}\right)$

(c) $\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)$

(d) $\frac{4}{3}\left(8\mathrm{\pi }+\sqrt{3}\right)$

Answer 1

(c) $\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)$



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

$$\begin{gathered} x^2+y^2=16\mathrm{and}{y}^2=6x \\ \Rightarrow x^2+6x=16 \\ \Rightarrow x^2+6x-16=0 \\ \Rightarrow \left(x+8\right)\left(x-2\right)=0 \\ \Rightarrow x=2\mathrm{or}x=-8 \\ x\mathrm{can}\mathrm{not}\mathrm{be}-8\mathrm{as}\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{outside}\mathrm{circle}. \\ \therefore x=2 \\ \therefore \mathrm{When}x=2,y=\pm \sqrt{6\times 2}=\pm \sqrt{12}=\pm 2\sqrt{3} \\ \therefore B\left(2,2\sqrt{3}\right)\mathrm{and}B\text{'}\left(2,-2\sqrt{3}\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{and}\mathrm{circle}. \\ \mathrm{Required}\mathrm{area}=\mathrm{Area}\left(OB\text{'}C\text{'}A\text{'}CBO\right)=\mathrm{area}\mathrm{of}\mathrm{circle}-\mathrm{area}\left(OBAB\text{'}O\right) \\ \mathrm{Area}\mathrm{of}\mathrm{circle}\mathrm{with}\mathrm{radius}4=\mathrm{\pi }\times 4^2=16\mathrm{\pi } \\ \mathrm{Now}, \\ \mathrm{Area}\left(OBAB\text{'}O\right)=2\mathrm{area}\left(OBAO\right) \\ =2\left[\mathrm{area}\left(OBDO\right)+\mathrm{area}\left(DBAD\right)\right] \\ =2\times \left[{\int }_0^2\sqrt{6x}dx+{\int }_2^4\sqrt{16-x^2}dx\right] \\ =2\times \left\{{\left[\sqrt{6}\frac{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}_0^2+{\left[\frac{x}{2}\sqrt{16-x^2}+\frac{1}{2}\times 16{\sin }^{-1}\left(\frac{x}{4}\right)\right]}_2^4\right\} \\ =2\times \left\{\left(\sqrt{6}\times \frac{2}{3}\times 2^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0\right)+\left(\frac{1}{2}4\sqrt{16-{\left(4\right)}^2}+\frac{1}{2}\times 16{\sin }^{-1}\frac{4}{4}-\frac{2}{2}\sqrt{16-2^2}-\frac{1}{2}\times 16{\sin }^{-1}\frac{2}{4}\right)\right\} \\ =2\times \left[\left(\sqrt{6}\times \frac{2}{3}\times 2\sqrt{2}\right)+0+8{\sin }^{-1}\left(1\right)-\sqrt{12}-8{\sin }^{-1}\left(\frac{1}{2}\right)\right] \\ =2\times \left[\frac{8\sqrt{3}}{3}+8\times \frac{\mathrm{\pi }}{2}-2\sqrt{3}-8\frac{\mathrm{\pi }}{6}\right] \\ =2\left\{\frac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}\right)\right\} \\ =2\left\{\frac{2\sqrt{3}}{3}+8\left(\frac{2\mathrm{\pi }}{6}\right)\right\} \\ =\frac{4\sqrt{3}}{3}+\frac{16\mathrm{\pi }}{3} \\ \mathrm{Shaded}\mathrm{area}=16\mathrm{\pi }-\left(\frac{4\sqrt{3}}{3}+\frac{16\mathrm{\pi }}{3}\right) \\ =\frac{48\mathrm{\pi }-16\mathrm{\pi }}{3}-\frac{4\sqrt{3}}{3} \\ =\frac{32\mathrm{\pi }}{3}-\frac{4\sqrt{3}}{3} \\ =\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)\mathrm{sq}\mathrm{units} \end{gathered}$$