Exam

JEE Main

Year

2026

Date

22 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Area Under Curves

The area of the region A=\ (x, y): 4 x^ 2 +y^ 2 ≤ 8. and .y^ 2 ≤ 4...

The area of the region $\mathrm{A}=\left\{(x, y): 4 x^{2}+y^{2} \leqslant 8\right. and \left.y^{2} \leqslant 4 x\right\}$ is:

Step-by-step solution

$A = \int_{0}^{2} 2\sqrt{x} \, dx + 2\int_{1}^{\sqrt{2}} \sqrt{8 - 4x^2} \, dx$
$= \dfrac{8}{3} \left( x^{\frac{3}{2}} \right) \bigg|_{0}^{1} + 4\int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx$
$= \dfrac{8}{3} + 4 \times \dfrac{1}{2} \left[ x\sqrt{2 - x^2} + 2\sin^{-1}\left( \dfrac{x}{\sqrt{2}} \right) \right] \bigg|_{1}^{\sqrt{2}}$
$= \dfrac{8}{3} + 2 \left[ 2 \times \dfrac{\pi}{2} - 1 - 2 \times \dfrac{\pi}{4} \right]$
$= \dfrac{8}{3} + 2\pi - 2 - \pi = \pi + \dfrac{2}{3}$ sq. units