Integral Calculus – Area Under Curves – JEE Main 23 Jan 2026 Shift 2

Solution:

Equations of circles:
$$ x^2 + y^2 = 4 \quad \dots(1) $$
$$ x^2 + (y-2)^2 = 4 \quad \dots(2) $$

Solving for intersection:
$$ (1) – (2) \Rightarrow y^2 – (y-2)^2 = 0 \Rightarrow 4y – 4 = 0 \Rightarrow y = 1 $$
$$ \text{At } y = 1, x^2 = 3 \Rightarrow x = \pm\sqrt{3} $$


Due to symmetry, required area is twice the area of the minor segment of circle (1) cut by line $y=1$:
$$ A = 2 \int_{-\sqrt{3}}^{\sqrt{3}} (\sqrt{4-x^2} – 1) dx = 4 \int_{0}^{\sqrt{3}} (\sqrt{4-x^2} – 1) dx $$

$$ = 4 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) – x \right]_{0}^{\sqrt{3}} $$

$$ = 4 \left[ \left( \frac{\sqrt{3}}{2}(1) + 2\left(\frac{\pi}{3}\right) – \sqrt{3} \right) – 0 \right] $$

$$ = 4 \left[ \frac{2\pi}{3} – \frac{\sqrt{3}}{2} \right] = \frac{8\pi}{3} – 2\sqrt{3} $$

$$ = \frac{2}{3}(4\pi – 3\sqrt{3}) $$

Ans. (4)