Area Under the Curve
Area Between Parabola and Circle Arc
nta_pyq_2023_apr
Grade 12
Question:
Let $y=p(x)$ be the parabola passing through the points $(-1,0)$, $(0,1)$ and $(1,0)$. If the area of the region $\{(x,y):(x+1)^2+(y-1)^2\leq 1,\ y\leq p(x)\}$ is $A$, then $12(\pi-4A)$ is equal to ________.
Step-by-Step Solution
Key Concept: The parabola through the three points is $y=1-x^2$. The region is inside the circle $(x+1)^2+(y-1)^2\leq 1$ and below the parabola, which intersects the circle at $(-1,0)$ and $(0,1)$.
Parabola: $y=1-x^2$. $A=\int_{-1}^0\!\left(\sqrt{1-(x+1)^2}-x^2\right)dx=\frac{\pi}{4}-\frac{1}{3}$. $12(\pi-4A)=12\!\left(\pi-\pi+\frac{4}{3}\right)=16$.
Correct Answer: 16