Area Under the Curve
Area Under Curves
nta_pyq_2025_apr
Grade 12
Question:
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.
Step-by-Step Solution
Key Concept: The region enclosed between $|y|=1-x^2$ (two parabolic arcs) and the unit circle is the part of the disk outside the parabolic region; compute by subtracting $4\int_0^1(1-x^2)dx$ from $\pi$.
The curves $|y|=1-x^2$ enclose a lens inside the circle. By symmetry:
$$\alpha = \pi - 4\int_0^1(1-x^2)dx = \pi - 4\left[x-\frac{x^3}{3}\right]_0^1 = \pi - \frac{8}{3}.$$
$$9\alpha = 9\pi - 24.$$
So $\beta = 9$, $\gamma = -24$, and $|\beta-\gamma| = |9+24| = 33$.
Correct Answer: 2