Question ID: #546
Let the area enclosed between the curves $|y| = 1 – x^{2}$ and $x^{2} + y^{2} = 1$ be $\alpha$. If $9\alpha = \beta\pi + \gamma$, where $\beta, \gamma$ are integers, then the value of $|\beta – \gamma|$ equals
- (1) 27
- (2) 18
- (3) 15
- (4) 33
Solution:
The curve $C_1: |y| = 1 – x^2$ represents two parabolas: $y = 1 – x^2$ (opening down) and $y = x^2 – 1$ (opening up).
The curve $C_2: x^2 + y^2 = 1$ is a circle with radius 1.
The required area $\alpha$ lies between the circle and the parabolas. Due to symmetry, we calculate the area in the first quadrant and multiply by 4.
Area in 1st quadrant = Area of quarter circle – Area under parabola $y = 1 – x^2$.
$\alpha = 4 \left[ \frac{\pi(1)^2}{4} – \int_{0}^{1} (1 – x^2) dx \right]$
$\alpha = 4 \left[ \frac{\pi}{4} – \left[ x – \frac{x^3}{3} \right]_{0}^{1} \right]$
$\alpha = 4 \left[ \frac{\pi}{4} – \left( 1 – \frac{1}{3} \right) \right] = 4 \left[ \frac{\pi}{4} – \frac{2}{3} \right] = \pi – \frac{8}{3}$
Given $9\alpha = \beta\pi + \gamma$:
$9(\pi – \frac{8}{3}) = 9\pi – 24$
Comparing terms, $\beta = 9$ and $\gamma = -24$.
$|\beta – \gamma| = |9 – (-24)| = |9 + 24| = 33$
Ans. (4)
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