Area Under the Curve
Area Under Curves
nta_pyq_2025_apr
Grade 12
Question:
The area of the region bounded by the curves $x(1+y^2) = 1$ and $y^2 = 2x$ is:
$2\!\left(\dfrac{\pi}{2}-\dfrac{1}{3}\right)$
$\dfrac{\pi}{2}-\dfrac{1}{3}$
$\dfrac{\pi}{4}-\dfrac{1}{3}$
$\dfrac{1}{2}\!\left(\dfrac{\pi}{2}-\dfrac{1}{3}\right)$
Step-by-Step Solution
Key Concept: Integrate in $y$: the region is $x \in \left[\tfrac{y^2}{2},\tfrac{1}{1+y^2}\right]$ for $y\in[-1,1]$. Use $\int_{-1}^1 \tfrac{1}{1+y^2}dy = \tfrac{\pi}{2}$ and $\int_{-1}^1 \tfrac{y^2}{2}dy = \tfrac{1}{3}$.
Intersection: $\tfrac{y^2}{2}(1+y^2)=1 \Rightarrow y^4+y^2-2=0 \Rightarrow y^2=1 \Rightarrow y=\pm 1$ (taking $x=\tfrac{1}{2}>0$).
$$\text{Area} = \int_{-1}^{1}\!\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy = \left[\tan^{-1}y - \frac{y^3}{6}\right]_{-1}^{1} = \left(\frac{\pi}{4}-\frac{1}{6}\right)-\left(-\frac{\pi}{4}+\frac{1}{6}\right) = \frac{\pi}{2}-\frac{1}{3}.$$
Correct Answer: 2