Question

The area of the region bounded by $1 - y^2 = |x|$ and $|x| + |y| = 1$ is

• A. $\frac{1}{3}$ sq unit
• B. $\frac{2}{3}$ sq unit
• C. $\frac{4}{3}$ sq units
• D. $1$ sq unit

Hint:

For $|x|, |y|$, always reduce to first quadrant and multiply by symmetry.

Answer 1

The Correct Option is B

Concept: Use symmetry and integrate in first quadrant.

Step 1: Consider first quadrant.
\[ x = 1 - y^2,\quad x = 1 - y \]

Step 2: Find intersection.
\[ 1 - y^2 = 1 - y \Rightarrow y(y-1)=0 \Rightarrow y=0,1 \]

Step 3: Set up area.
\[ \text{Area} = 4 \int_0^1 [(1-y) - (1-y^2)] dy \] \[ = 4 \int_0^1 (y^2 - y) dy \]

Step 4: Integrate.
\[ = 4\left[\frac{y^3}{3} - \frac{y^2}{2}\right]_0^1 = 4\left(\frac{1}{3} - \frac{1}{2}\right) \] \[ = 4 \cdot \left(-\frac{1}{6}\right) = \frac{2}{3} \] Conclusion:
Area = $\frac{2}{3}$