Question

Choose the correct answer
The area of the circle $x^2+y^2=16$ exterior to the parabola $y^2=6x$ is
(a) $\frac{4}{3}(4\pi -\sqrt{3})$ (b) $\frac{4}{3}(4\pi +\sqrt{3})$
(c) $\frac{4}{3}(8\pi -\sqrt{3})$ (d) $\frac{4}{3}(8\pi +\sqrt{3})$

Answer 1

The circle $x^2+y^2=16$ and the parabola $y^2=6x$ meet where
$x^2+6x=16$ (Eliminating y)
$\Rightarrow x^2+6x-16=0$
$\Rightarrow (x+8)(x-2)=0\Rightarrow x=-8,2$
$\therefore$ Required area = Area of the circle -2 Area shown in the shaded region..
$=\pi (4{)}^2-2[{\int }_0^2\sqrt{6x}+{\int }_2^4\sqrt{16-x^2}dx]=16\pi -2\sqrt{6}[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0^2-2[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}si{n}^{-1}\frac{x}{4}{]}_2^4=16\pi -\frac{4\sqrt{6}}{3}(2^{\frac{3}{2}}-0)-2[0+8si{n}^{-1}-\sqrt{16-4}-8si{n}^{-1}\frac{1}{2}]=16\pi -\frac{4\sqrt{6}\times 2\sqrt{2}}{3}-2[8\times \frac{\pi }{2}-\sqrt{12}-8\times \frac{\pi }{6}]=16\pi -\frac{16\sqrt{3}}{3}-\frac{16\pi }{3}+4\sqrt{3}=\frac{32\pi }{3}-\frac{4\sqrt{3}}{3}=\frac{4}{3}(8\pi -\sqrt{3})squnit$
So, the correct option is (c)