Question

Find the area of that part of the circle $x^2+y^2=16$ which is exterior to the parabola $y^2=6x$.

Answer 1

The radius of the circle $x^2+y^2=16$ is $4$

So the area of circle is $\pi 4^2=16\pi$

Point of intersection of The 2 curves is

$x^2+6x=16x^2+6x-16=0x^2+8x-2x-26=0x(x+8)-2(x+8)=0(x+8)(x-2)=0x=2,-8$

Here only $x=2$ is considered

$x^2+y^2=16y^2=16-x^{2}y=\sqrt{16-x^2}\cdots \left(1\right)y^2=6xy=\sqrt{6x}\cdots \left(2\right)$

So the area enclosed by parabola and circle is Area encloed by parabola from $x=0$ to $x=2$ and area enclosed by circle from $x=2$ to $x=4$

Area ${\int }_0^2\sqrt{6x}dx+{\int }_2^4\sqrt{16-x^2}dx$

$\sqrt{6}\frac{2}{3}{x^{\frac{3}{2}}|}_0^2$ $+\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}{\frac{x}{4}|}_2^4$

$\sqrt{6}\frac{2}{3}2\sqrt{2}+2\sqrt{16-16}+8{\sin }^{-1}\left(1\right)-1\sqrt{16-4}-8{\sin }^{-1}\left(\frac{1}{2}\right)\frac{8\sqrt{3}}{2}+4\pi -2\sqrt{3}-\frac{8\pi }{6}2\sqrt{3}+\frac{8\pi }{3}$

So the exterior part is given by $16\pi -2\sqrt{3}-\frac{8\pi }{3}\frac{40\pi }{3}-2\sqrt{3}$