Question

The area of the circle $x^2+y^2=16$ exterior to the parabola $y^2=6x$ is

• A. $\frac{4}{3}(4π-\sqrt{3})$
• B. $\frac{4}{3}(4π+\sqrt{3})$
• C. $\frac{4}{3}(8π-\sqrt{3})$
• D. $\frac{4}{3}(4π+\sqrt{3})$

Answer 1

Answer: (C)

$\frac{4}{3}(8π-\sqrt{3})$

Answer 2

The correct answer is C:$=\frac{4}{3}(8π-\sqrt{3})units$
The given equations are
$x^2+y^2=16...(1)$
$y^2=6x...(2)$

Area bounded by the circle and parabola
$=2[Area(OADO)+Area(ADBA)]$
$=2[∫^2_0\sqrt{16x}dx+∫^4_2\sqrt{16-x^2}dx]$
$=2[\sqrt{6}\bigg[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\bigg]^2_0]+2\bigg[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}sin^{-1}\frac{x}{4}\bigg]^4_2$
$=2\sqrt{6}\times\frac{2}{3}\bigg[x^{\frac{3}{2}}\bigg]^2_0+2[8.\frac{π}{2}-\sqrt{16-4}-8sin^{-1}(\frac{1}{2})]$
$=\frac{4\sqrt{6}}{3}(2\sqrt{2})+2[4π-\sqrt{12}-8\frac{π}{6}]$
$=\frac{16\sqrt{3}}{3}+8π-4\sqrt{3}-\frac{8}{3}π$
$=\frac{4}{3}[4\sqrt{3}+6π-3\sqrt{3}-2π]$
$=\frac{4}{3}[\sqrt{3}+4π]$
$=\frac{4}{3}[4π+\sqrt{3}]units$
Area of circle$=π(r)^2$
$=π(4)^2$
$=16πunits$
∴Required area$=16π-\frac{4}{3}[4π+\sqrt{3}]$
$=\frac{4}{3}[4\times3π-4π-\sqrt{3}]$
$=\frac{4}{3}(8π-\sqrt{3})units$
Thus,the correct answer is C.