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

The Correct Option is C

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.