Question

Using integration, find the area bounded by the ellipse \( 9x^2 + 25y^2 = 225 \), the lines \( x = -2, x = 2 \), and the X-axis.

Hint:

When integrating to find areas involving ellipses or circles, use symmetry and standard integral formulas for \( \sqrt{a^2 - x^2} \).

Answer 1

Step 1: Rewrite the equation of the ellipse
The equation of the ellipse is: \[ 9x^2 + 25y^2 = 225 \implies y = \pm \frac{3}{5} \sqrt{25 - x^2} \] Step 2: Set up the integral for the area
The area of the region bounded by the ellipse, the X-axis, and the lines \( x = -2 \) and \( x = 2 \) is given by: \[ \text{Area} = 2 \int_{0}^{2} \frac{3}{5} \sqrt{25 - x^2} \, dx \] Step 3: Simplify the integral
Let \( I = \int \sqrt{a^2 - x^2} \, dx \), where \( a = 5 \). Using the standard formula: \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \] Step 4: Evaluate the integral
Substitute \( a = 5 \) and evaluate \( \int_{0}^{2} \sqrt{25 - x^2} \, dx \): \[ \int_{0}^{2} \sqrt{25 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{2} \] At \( x = 2 \): \[ \frac{2}{2} \sqrt{25 - 2^2} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right) = \sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right) \] At \( x = 0 \): \[ \frac{0}{2} \sqrt{25 - 0^2} + \frac{25}{2} \sin^{-1}(0) = 0 \] Thus: \[ \int_{0}^{2} \sqrt{25 - x^2} \, dx = \sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right) \] Step 5: Final area calculation
Multiply by \( \frac{6}{5} \) to account for \( \frac{3}{5} \) and the factor of 2: \[ \text{Area} = \frac{6}{5} \left(\sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right)\right) \] Step 6: Final result
The area bounded by the ellipse, the X-axis, and the lines \( x = -2 \) and \( x = 2 \) is: \[ \frac{6\sqrt{21}}{5} + 15 \sin^{-1}\left(\frac{2}{5}\right) \]