Area Under the Curve
Area between circle and ellipse
Grade 12
Question:
<p>Find the area of the region lying inside \(x^2 + (y-1)^2 = 1\) and outside \(c^2x^2 + y^2 = c^2\), where \(c = \sqrt{2} - 1\).</p>
<p>\(\dfrac{\pi(3\sqrt{2}-1)}{\sqrt{2}} + \dfrac{\sqrt{2}+1}{\sqrt{2}}\)</p>
<p>\(\dfrac{\pi(3\sqrt{2}-1)}{\sqrt{2}} + \dfrac{\sqrt{2}+1}{2\sqrt{2}}\)</p>
<p>\(\dfrac{\pi(3\sqrt{2}-1)}{2\sqrt{2}} + \dfrac{\sqrt{2}+1}{\sqrt{2}}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that x² + (y-1)² = 1 is a circle centered at (0,1) with radius 1, and c²x² + y² = c² is an ellipse. The area is found by integration after determining intersection points and identifying which curve is outer/inner in different regions.
<p><strong>Step 1: Identify the curves</strong></p><p>Circle: x² + (y-1)² = 1 has center (0,1) and radius 1</p><p>Ellipse: c²x² + y² = c², where c = √2 - 1, can be written as x²/(1) + y²/c² = 1</p><p>This is an ellipse with semi-major axis 1 (along x-axis) and semi-minor axis c (along y-axis)</p><p><strong>Step 2: Find intersection points</strong></p><p>From circle: x² = 1 - (y-1)² = 2y - y²</p><p>Substitute into ellipse equation: c²(2y - y²) + y² = c²</p><p>c²(2y - y²) + y² = c² → 2c²y - c²y² + y² = c²</p><p>y²(1 - c²) = c²(1 - 2y) → y²(1 - c²) + 2c²y - c² = 0</p><p><strong>Step 3: Use c = √2 - 1</strong></p><p>Note: c² = (√2 - 1)² = 3 - 2√2, so 1 - c² = 2√2 - 2</p><p>The curves intersect at y = 0 and y = 1 (endpoints of circle's vertical span)</p><p><strong>Step 4: Set up the integral</strong></p><p>Area = ∫₀¹ 2·(x_circle - x_ellipse) dy</p><p>Where x_circle = √(2y - y²) and x_ellipse = √(c²(1 - y²/c²)) = √(c² - y²)</p><p>Area = 2∫₀¹ [√(2y - y²) - √(c² - y²)] dy</p><p><strong>Step 5: Evaluate integrals</strong></p><p>∫₀¹ √(2y - y²) dy = ∫₀¹ √(1 - (y-1)²) dy = π/2 (quarter circle)</p><p>∫₀¹ √(c² - y²) dy = (c²/2)·arcsin(1/c) + (1/2)√(c² - 1) |₀¹</p><p>With c = √2 - 1: Area = π/2 - [π(3-2√2)/4 + (1-√2)/2] = π(√2-1)/2 + (√2-1)/2</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C