Question

Find the area of the region $\left\{\right(x,y);y^2\le 4x,4x^2+4y^2\le 9\}.$

Answer 1

Given curve $y^2=4x$ ...(i)
and circle $4x^2+4y^2=9$ ...(ii)
Which represents a circle whose centre is (0, 0) and radius $\frac{3}{2}.$
For intersecting points of parabola and circle put the value of $y^2$ from
Eq. (i) in Eq. (ii), we get
$4x^2+16x=9$
$\Rightarrow 4x^2+16x-9=0$
$\Rightarrow 4x^2+18x-2x-9=0\Rightarrow 2x(2x+9)-(2x-9)=0\Rightarrow (2x+9)(2x-1)=0$
$\Rightarrow x=\frac{-9}{2},\frac{1}{2}\Rightarrow x=\frac{1}{2}$
$(x\ne \frac{-9}{2}$ because it does not satisfy the given inequality)
On putting $x=\frac{1}{2}$ in Eq. (i) we get $y=\pm \sqrt{2}$
$\therefore$ The points of intersection of both the curves are $A(\frac{1}{2},\sqrt{2})$ and $C(\frac{1}{2},\sqrt{2}).$ Since, the shaded region is symmetrical about X - axis.
$\therefore$ Required area = 2 (Area OABO)
$=2[{\int }_0^{\frac{1}{2}}\sqrt{4x}dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{\frac{9-4x^2}{4}}dx]$
$=2\times 2[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0+2{\int }_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{(\frac{3}{2}{)}^2-x^2}dx=\frac{8}{3}[(\frac{1}{2}{)}^{\frac{3}{2}}-0]+2[\frac{x}{2}\sqrt{\frac{9}{4}-x^2}+\frac{9}{8}si{n}^{-1}\frac{x}{\frac{3}{2}}{]}_{\frac{1}{2}}^{\frac{3}{2}}=\frac{8}{3}.\frac{1}{2\sqrt{2}}+2[0+\frac{9}{8}si{n}^{-1}\left(1\right)-\frac{1}{4}\sqrt{\frac{9}{4}-\frac{1}{4}}-\frac{9}{8}si{n}^{-1}\left(\frac{1}{3}\right)]=\frac{2\sqrt{2}}{3}+\frac{9}{4}.\frac{\pi }{2}-\frac{\sqrt{2}}{2}-\frac{9}{4}si{n}^{-1}(\frac{1}{3})=\frac{9\pi }{8}+\frac{\sqrt{2}}{6}-\frac{9}{4}si{n}^{-1}\frac{1}{3}=\frac{9\pi }{8}+\frac{1}{3\sqrt{2}}-\frac{9}{4}si{n}^{-1}\frac{1}{3}$