Let the area enclosed between the curves y =1- x 2 and x 2 +y 2 =1 be α . If 9 α=β π+γ ; β, γ are integers, then the value of β-γ equals

Mathematics · Area under the Curves

JEE Main 2025 — 29 January, Evening Shift — Question 44

Let the area enclosed between the curves $|y|=1-$ $x^{2}$ and $x^{2}+y^{2}=1$ be $\alpha$ .

If $9 \alpha=\beta \pi+\gamma ; \beta, \gamma$ are integers, then the value of $|\beta-\gamma|$ equals

$C_{1}:|y|=1-x^{2}$

$\mathrm{C}_{2}: \mathrm{x}^{2}+\mathrm{y}^{2}=1$

$\therefore$ Required Area $=\alpha=4\left[\right.$

Area of circle in $1^{\text {st }}$ quad.

$\left.-\int_{0}^{1}\left(1-x^{2}\right) \mathrm{dx}\right]$

$=4\left[\frac{\pi}{4}-\left[x-\frac{x^{3}}{3}\right]_{0}^{1}\right]$

$\alpha=\pi-\frac{8}{3}$

$\therefore 3 \alpha=3 \pi-8$

$\therefore 9 \alpha=9 \pi-24$

$\therefore \beta=9, \gamma=-24$

$\therefore|\beta-\gamma|=33$

Answer key and solution verified before publishing.

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