Area Under the Curve
Area Between Circle and Parabola
nta_pyq_2023_apr
Grade 12
Question:
Area of the region $\{(x,y):\ x^2+(y-2)^2\leq 4,\ x^2\geq 2y\}$ is
$\pi+\dfrac{8}{3}$
$2\pi+\dfrac{16}{3}$
$\pi-\dfrac{8}{3}$
$2\pi-\dfrac{16}{3}$
Step-by-Step Solution
Key Concept: The circle has centre $(0,2)$ and radius $2$. It intersects $x^2=2y$ at $(\pm 2,2)$. The required region is inside the circle and outside (above) the parabola, so subtract the parabolic segment from the semicircle area.
Area $=2\!\left[\pi-\int_0^2\sqrt{2y}\,dy\right]=2\!\left[\pi-\frac{\sqrt{2}\cdot 2^{3/2}\cdot 2}{3}\right]=2\pi-\frac{16}{3}$.
Correct Answer: 4