Exam

JEE Main

Year

2024

Date

8 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Area Under Curves

The area of the region in the first quadrant inside the circle...

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2 x$ is equal to :

Step-by-step solution

Required area $=\operatorname{Ar}($ circle from 0 to 2$)- \operatorname{ar}($ para from 0 to 2$)$$\begin{aligned} & =\int_0^2 \sqrt{8-x^2} d x-\int_0^2 \sqrt{2 x} d x \\ & =\left[\frac{x}{2} \sqrt{8-x^2}+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}}\right]_0^2-\sqrt{2}\left[\frac{x \sqrt{x}}{3 / 2}\right]_0^2 \\ & =\frac{2}{2} \sqrt{8-4}+\frac{8}{2} \sin ^{-1} \frac{2}{2 \sqrt{2}}-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}-0) \\ & \Rightarrow 2+4 \cdot \frac{\pi}{4}-\frac{8}{3}=\pi-\frac{2}{3}\end{aligned}$$$