Question

Find the area of the region common to the circle $x^2+y^2=9$ and the parabola $y^2=8x$.

Answer 1

Because $x^2+y^2=9;y^2=8x$

So $x^2+8x=9$

$x^2+8x–9=0$

$⟹(x+9)(x-1)=0$

$⟹x=1,x=-9$

Therefore $y=\pm 2\sqrt{2}$

Point of intersections are $p(1,2\sqrt{2},1,-2\sqrt{2})$

$y_1=\sqrt{8x}=2\sqrt{2}{x}^{1/2}$

$y_2=\sqrt{9-x^2}$

Area = Area OPAQO = 2 Area OPAMO

Area = 2 (Area OPMO + Area APMA)

$=2[{\int }_0^1{y}_{1}dx+{\int }_1^3{y}_{2}dx]$

$=2[{\int }_0^{1}2\sqrt{2}{x}^{1/2}dx+{\int }_1^3\sqrt{3^2–x^2}dx]$

$=2\left[\frac{2\sqrt{2}}{3/2}\right(x^{3/2}{)}_0^1+(\frac{x}{2}\sqrt{3^2–x^2}+\frac{9}{2}si{n}^{-1}\frac{x}{3}{)}_1^3]$

$=2[\frac{4\sqrt{2}}{2}.1+\frac{9}{2}si{n}^{-1}1-\frac{1}{2}\times \sqrt{8}-\frac{9}{2}si{n}^{-1}\frac{1}{3}]$

$=2[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\frac{1}{3}]$ sq. Unit