Q. The parabola $y^2=x$ divides the circle $x^2 +y^2 = 2$ into two parts whose areas are in the ratio
AIEEEAIEEE 2012Application of Integrals
Solution:
$Area of circle =\pi\left(\sqrt{2}\right)^{2}=2\pi$
Area of $OCADO = 2 \left\{Area\, of\, OCAO\right\}$
$=2\left\{area of OCB + area of BCA\right\}$
$=2\int\limits^{1}_{{0}}y_pdx+2$$\int\limits^{{{\sqrt{2}}}}_{{1}}y_cdx$
where $y_{p}=\sqrt{x}$ and $y_{c}=\sqrt{2-x^{2}}$
$\therefore $ Required Area
$=2\int\limits^{1}_{{0}}$$\sqrt{x} dx+2 $$\int\limits^{{{\sqrt{2}}}}_{{1}}$$\sqrt{2-x^{2}}dx$
$=\left[\frac{2}{3}.1-0\right]+2\left[\frac{x\sqrt{2-x^{2}}}{2}+sin^{-1} \frac{x}{\sqrt{2}}\right]^{^{\sqrt{2}}}_{_{_1}}$
$=\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi }{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3\pi+2}{6}$
Bigger area $=2\pi-\frac{3\pi+2}{6}=\frac{9\pi-2}{6}$
$\therefore $ Required Ratio $= \frac{9\pi-2}{3\pi+2} i.e., 9\pi-2 : 3\pi+2$
