Area Under the Curve
Area Between Parabola and Absolute Value Parabola with Lower Bound
nta_pyq_2023_apr
Grade 12
Question:
The area of the region $\left\{x,y:\ x^2\leq y\leq|x^2-4|,\ y\geq 1\right\}$ is
$\dfrac{4(4\sqrt{2}-1)}{3}$
$\dfrac{4(4\sqrt{2}+1)}{3}$
$\dfrac{3(4\sqrt{2}+1)}{4}$
$\dfrac{3(4\sqrt{2}-1)}{4}$
Step-by-Step Solution
Key Concept: By symmetry, integrate over $y$ from $1$ to $2$ (where $y=4-x^2$ bounds above) and from $2$ to $4$ (where $y=4-x^2$ still applies on part). Find the horizontal extent $x=\sqrt{y}$ on left.
Area $=2\!\left[\frac{2}{3}y^{3/2}\Big|_1^2-\frac{2}{3}(4-y)^{3/2}\Big|_2^4\right]=\frac{4}{3}(4\sqrt{2}-1)$.
Correct Answer: 1