Area Under the Curve
Regions with Constraints
Grade 12
Question:
<p>The area (in sq units) of the region \(A = \{(x, y):|x| + |y| \leq 1, 2y^2 \geq |x|\}\) is</p>
<p>(a) \(\frac{1}{3}\)</p>
<p>(b) \(\frac{7}{6}\)</p>
<p>(c) \(\frac{1}{6}\)</p>
<p>(d) \(\frac{5}{6}\)</p>
Step-by-Step Solution
Key Concept: Use symmetry to simplify the problem and set up an integral with proper bounds determined by the intersection of the boundary curves.
<p><strong>Solution:</strong></p><p>The region is bounded by $|x| + |y| \leq 1$ and $2y^2 \geq |x|$.</p><p>By symmetry, we can compute the area in one quadrant and multiply by 4.</p><p>From $2y^2 = |x|$, we get $y = \pm\frac{1}{2}$ when $x = \frac{1}{2}$.</p><p>$A = 4\int_0^{1/2} \left(1 - x - \sqrt{\frac{x}{2}}\right) dx$</p><p>$= 4\left[x - \frac{x^2}{2} - \frac{2x^{3/2}}{3\sqrt{2}}\right]_0^{1/2}$</p><p>$= 4\left[\frac{1}{2} - \frac{1}{8} - \frac{2}{3\sqrt{2}} \cdot \frac{1}{2\sqrt{2}}\right]$</p><p>$= 4\left[\frac{3}{8} - \frac{1}{12}\right] = \frac{5}{6}$</p><p>∴ Answer is (d).</p>
Correct Answer: D