Mathematics · Area under the Curves

JEE Main 2024 — 8 April, Shift 2 — Question 11

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

Answer: B

Required area $=\operatorname{Ar}($ circle from 0 to 2 $)-\operatorname{ar}($ para from 0 to 2 $)$

$=\int_{0}^{2} \sqrt{8-x^{2}} d x-\int_{0}^{2} \sqrt{2 x} d x$

$=\left[\frac{x}{2} \sqrt{8-x^{2}}+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}}\right]_{0}^{2}-\sqrt{2}\left[\frac{x \sqrt{x}}{3 / 2}\right]_{0}^{2}$

$=\frac{2}{2} \sqrt{8-4}+\frac{8}{2} \sin ^{-1} \frac{2}{2 \sqrt{2}}-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}-0)$

$\Rightarrow 2+4 \cdot \frac{\pi}{4}-\frac{8}{3}=\pi-\frac{2}{3}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.