Question

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

Answer 1

The given curves are
$x^2+y^2=9$ ..... (i)
and $y^2=8x$ ...... (ii)
To find the points of Intersection of the curves, we solve (i) and (ii), we get
$x^2+8x=9$
$x^2+8x-9=0$
$x^2+9x-x-9=0$
$x(x+9)-1(x+9)=0$
$(x+9)(x-1)=0$,
$x+9=0\Rightarrow x=-9,x-1=0\Rightarrow x=1$
If $x=-9$, then from the equation (ii)
$y^2=8(-9)=-72<0$ which is not possible
When $x=1,y^2=8\left(1\right)=8$
$\Rightarrow y=\pm 2\sqrt{2}$
$\therefore$ the points of Intersection are $P(1,2\sqrt{2})$ and $Q(1,-2\sqrt{2})$ equation (i) represents a circle with center at $(0,0)$ and radius $3$.
Equation (ii) represents a parabola with vertex at $(0,0)$.
The required area is shown shaded
$\therefore$ Required area $=$ Area of the region $OQAPO$
$=2$ (Area of the region $OMAPO)$
$=2\left[\right(\text{Area of the region OMPA)}+\left(\text{Area of the region PMAP}\right]$
Now, Area of the region $OMPO$
$=$ Area under the parabola $y^2=8x\Rightarrow y=2\sqrt{2}\sqrt{x}$
$={\int }_0^{1}2\sqrt{2}\sqrt{x}dx$
$=2\sqrt{2}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$
$=2\sqrt{2}\times \frac{2}{3}\left[\right(1{)}^{\frac{3}{2}}-0]=\frac{4\sqrt{2}}{3}$ .... (iii)
and Area of the region $PMAP$
$=$ Area under the circle $x^2+y^2=9$, i.e., $y=\sqrt{9-x^2}$
$={\int }_1^3\sqrt{9-x^2}dx$
$={[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)]}_1^3$
$=(\frac{3}{2}\sqrt{9-9}+\frac{9}{2}{\sin }^{-2}(1\left)\right)-(\frac{1}{2}\sqrt{9-1}+\frac{9}{2}{\sin }^{-1}\frac{1}{3})$
$=0+\frac{9}{2}\cdot \frac{\pi }{2}-\sqrt{2}-\frac{9}{2}{\sin }^{-1}\frac{1}{3}$ .... (iv)
$\therefore$ required area $=2(\frac{4\sqrt{2}}{3}+\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}{\sin }^{-1}\frac{1}{3})$
$=(\frac{8\sqrt{2}}{3}+\frac{9\pi }{2}-2\sqrt{2}-9{\sin }^{-1}\frac{1}{3})$ square units