Question

The area of the region enclosed between by the $x^2+y^2=16$ and the parabola $y^2=6x$.

• A. $\frac{2}{3}(\sqrt{3}+4\pi )$ sq. units
• B. $\frac{4}{3}(\sqrt{3}+4\pi )$ sq. units
• C. $\frac{2}{3}(\sqrt{3}+8\pi )$ sq. units
• D. $\frac{4}{3}(\sqrt{3}+8\pi )$ sq. units

Answer 1

The correct option is A $\frac{2}{3}(\sqrt{3}+4\pi )$ sq. units

Point of intersection of the parabola and the circle is obtained by solving the equations:

$x^2+y^2=16$ and $y^2=6x$

$\Rightarrow x^2+6x-16=0$

$\Rightarrow x^2+8x-2x-16=0$

$\Rightarrow x(x+8)-2(x+8)=0$

$\Rightarrow (x-2)(x+8)=0$

$\Rightarrow x=2,x=-8$

$\therefore x=2$ is the only possible solution(from the fig.)

$\therefore$ when $x=2,y=\pm \sqrt{6\times 2}=\pm 2\sqrt{3}$

$\therefore B(2,2\sqrt{3})$ and $B^{\prime }(2,-2\sqrt{3})$ are the points of intersection of parabola and the circle.

Required area$=areaofOBA{B}^{\prime }O=2areaofOBAO$

$=2[areaofOBDO+areaofDBAD]$

$=2\left[{\int }_0^2\sqrt{6x}dx+{\int }_2^4\sqrt{16-x^2}dx\right]$

$=2[\sqrt{6}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^2+{[\frac{1}{2}x\sqrt{16-x^2}+\frac{1}{2}\times 16{\sin }^{-1}\frac{x}{4}]}_2^4]$

$=2[[\sqrt{6}\times \frac{2}{3}\times 2^{\frac{3}{2}}-0]+[\frac{1}{2}\times 4\sqrt{16-16}+\frac{1}{2}\times 16{\sin }^{-1}\frac{4}{4}-\frac{1}{2}\times 2\sqrt{16-4}-\frac{1}{2}\times 16{\sin }^{-1}\frac{1}{2}]]$

$=2[\frac{8\sqrt{3}}{3}+\frac{8\pi }{2}-2\sqrt{3}-\frac{8\pi }{6}]$

$=2[\frac{8\sqrt{3}-6\sqrt{3}}{3}+8(\frac{\pi }{2}-\frac{\pi }{6})]$

$=2[\frac{2\sqrt{3}}{3}+8\times \frac{2\pi }{6}]$

$=\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}$