Question

Calculate the area of the region bounded by the curve \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] and the x-axis using integration.

Hint:

For calculating areas under curves, especially for bounded regions involving symmetric shapes like ellipses, using trigonometric substitution often simplifies the integral significantly.

Answer 1

We start by solving for \( y \) in terms of \( x \):
\[ \frac{y^2}{4} = 1 - \frac{x^2}{9} \quad \Rightarrow \quad y^2 = 4 \left( 1 - \frac{x^2}{9} \right) \quad \Rightarrow \quad y = \pm 2 \sqrt{1 - \frac{x^2}{9}} \] The area under the curve from \( x = -3 \) to \( x = 3 \) is given by the integral: \[ A = 2 \int_{-3}^{3} 2 \sqrt{1 - \frac{x^2}{9}} \, dx \] Using the substitution \( x = 3 \sin(\theta), \, dx = 3 \cos(\theta) \, d\theta \), we get: \[ A = 36 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta \] Using \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \), the integral becomes: \[ A = 36 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \] After solving, we get: \[ A = 18\pi \]