Question

Sketch the curve \( \{(x, y) : 100x^2 + 25y^2 = 2500\ \) and find the area of the region enclosed by it, using integration.}

Hint:

The area of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( \pi ab \). You can use this to verify your final answer: \( \pi \times 5 \times 10 = 50\pi \).
Always remember to multiply the single quadrant integral by 4 to get the total enclosed area.

Answer 1

Concept:
• Ellipse equation in standard form: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
• Area of a region using integration: \( \text{Area} = 4 \int_{0}^{a} y \, dx \) for an ellipse.
• Integration formula: \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) \).

Step 1: Identify the curve and its key parameters
Given: \( 100x^2 + 25y^2 = 2500 \). Divide by 2500: \[ \frac{x^2}{25} + \frac{y^2}{100} = 1 \Rightarrow \frac{x^2}{5^2} + \frac{y^2}{10^2} = 1 \] This is an ellipse centered at \( (0, 0) \) with vertical major axis length 20 and horizontal minor axis length 10. Vertices are \( (\pm 5, 0) \) and \( (0, \pm 10) \).

Step 2: Set up the integral for the area
Express \( y \) in terms of \( x \): \[ y^2 = 100(1 - x^2/25) = 4(25 - x^2) \Rightarrow y = 2\sqrt{25 - x^2} \] Total Area \( A = 4 \times \text{Area in 1st Quadrant} \): \[ A = 4 \int_{0}^{5} 2\sqrt{25 - x^2} \, dx = 8 \int_{0}^{5} \sqrt{5^2 - x^2} \, dx \]

Step 3: Evaluate the integral
\[ A = 8 \left[ \frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{5} \] Upper limit \( (x=5) \): \( 0 + \frac{25}{2}\sin^{-1}(1) = \frac{25}{2} \cdot \frac{\pi}{2} = \frac{25\pi}{4} \). Lower limit \( (x=0) \): \( 0 + 0 = 0 \). \[ A = 8 \left( \frac{25\pi}{4} \right) = 50\pi \text{ sq. units} \]