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 first isolate \( y \) as a function 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 enclosed by the curve from \( x = -3 \) to \( x = 3 \) is computed by the integral: \[A = 2 \int_{-3}^{3} 2 \sqrt{1 - \frac{x^2}{9}} \, dx\] Applying the substitution \( x = 3 \sin(\theta), \, dx = 3 \cos(\theta) \, d\theta \), the integral transforms to:\[A = 36 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta\] Utilizing the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \), the integral simplifies to:\[A = 36 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta\] The evaluation of this integral yields:\[A = 18\pi\]