Exam

JEE Main

Year

2012

Date

7 May

Mode

Online

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2012 · 7 May · Online · Mathematics · Area Under Curves

The parabola y^2=x divides the circle x^2+y^2=2 into two parts...

The parabola $y^2=x$ divides the circle $x^2+y^2=2$ into two parts whose areas are in the ratio

Step-by-step solution

$$\begin{aligned}& \text{ Area of circle }=\pi(\sqrt{2})^2=2 \pi\\ & \text{ Area of }O C A D O=2\{\text{ Area of }O C A O\} \\ & \quad=2 \text{ area of }O C B+\text{ area of }B C A\} \\ & =2 \int_0^1 y_p d x+2 \int_1^{\sqrt{2}} y_c d x \\ & \text{ where }y_p=\sqrt{x} \text{ and }y_c=\sqrt{2-x^2}\\ & \therefore \text{ Required Area }\\ & =2 \int \frac{1}{x}d x+2 \int_0^{\sqrt{2}} \sqrt{2-x^2}d x \\ & =2\left[\frac{2}{3} \cdot1-0\right]+2\left[\frac{x \sqrt{2-x^2}}{2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}} \end{aligned}$$$$\begin{aligned}& =\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3 \pi+2}{6}\\ & \text{ Bigger area }=2 \pi-\frac{3 \pi+2}{6}=\frac{9 \pi-2}{6}\\ & \therefore \text{ Required Ratio }=\frac{9 \pi-2}{3 \pi+2} \text{ i.e., }9 \pi-2: 3 \pi+2 \end{aligned}$$