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 \( \sqrta^2 - x^2 \).

Answer 1

Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is: \[ 9x^2 + 25y^2 = 225. \] Solving for \( y \): \[ y = \pm \frac{5}{3} \sqrt{25 - x^2}. \]
Step 2: Set up the integral for the area
The area of the region bounded by the ellipse, the X-axis, and the vertical lines \( x = -2 \) and \( x = 2 \) is given by: \[ \text{Area} = 2 \int_0^2 \frac{5}{3} \sqrt{25 - x^2} \, dx. \]
Step 3: Use the standard integral formula
The standard integral formula for: \[ \int \sqrt{a^2 - x^2} \, dx \] is given by: \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left(\frac{x}{a}\right) + C. \] Here, \( a = 5 \), so we evaluate: \[ \int_0^2 \sqrt{25 - x^2} \, dx. \]
Step 4: Evaluate the integral
Using the formula: \[ \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1} \left(\frac{x}{5} \right) \right]_0^2. \] 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 \( x = 0 \): \[ \frac{0}{2} \sqrt{25 - 0^2} + \frac{25}{2} \sin^{-1} (0) = 0. \] So the integral evaluates to: \[ \sqrt{21} + \frac{25}{2} \sin^{-1} \left(\frac{2}{5} \right). \]
Step 5: Compute the final area
Multiply by \( \frac{10}{3} \) (accounting for \( \frac{5}{3} \) and the factor of 2 from symmetry): \[ \text{Area} = \frac{10}{3} \left(\sqrt{21} + \frac{25}{2} \sin^{-1} \left(\frac{2}{5} \right) \right). \]
Step 6: Final result
The required area is: \[ \boxed{\frac{10}{3} \sqrt{21} + \frac{125}{6} \sin^{-1} \left(\frac{2}{5} \right)}. \]