Question

Using integration, find the area of the ellipse: \[ \frac{x^2}{16} + \frac{y^2}{4} = 1, \] included between the lines \(x = -2\) and \(x = 2\).

Hint:

To find the area of a region using integration, utilize symmetry to simplify calculations and make substitutions to handle square root expressions effectively.

Answer 1

Step 1: Rearranging the equation of the ellipse. The given ellipse equation is: \[ \frac{x^2}{16} + \frac{y^2}{4} = 1. \] Solving for \( y \): \[ \frac{y^2}{4} = 1 - \frac{x^2}{16} \] \[ y^2 = 4 - \frac{x^2}{4} \] \[ y = \pm \sqrt{4 - \frac{x^2}{4}} \] Step 2: Using symmetry to simplify the calculation. Since the ellipse is symmetric about the \(x\)-axis, the required area can be found as twice the area above the \(x\)-axis: \[ \text{Area} = 2 \int_{-2}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx \] Using symmetry about the \(y\)-axis: \[ \text{Area} = 4 \int_{0}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx \] Step 3: Substitution for simplification. Let: \[ u = 4 - \frac{x^2}{4}, \quad \text{so} \quad du = -\frac{x}{2} \, dx \quad \text{and} \quad x \, dx = -2 \, du \] Changing the limits: when \( x = 0 \), \( u = 4 \), and when \( x = 2 \), \( u = 3 \). Thus, the integral becomes: \[ \int_{0}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx = \int_{4}^{3} \sqrt{u} \cdot (-2) \, du \] \[ = 2 \int_{3}^{4} \sqrt{u} \, du \] Step 4: Evaluating the integral. The integral of \( \sqrt{u} \) is: \[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} \] Evaluating from \( u = 3 \) to \( u = 4 \): \[ \int_{3}^{4} \sqrt{u} \, du = \frac{2}{3} \left[ 4^{3/2} - 3^{3/2} \right] \] \[ = \frac{2}{3} \left( 8 - \sqrt{27} \right) \] Step 5: Final area calculation. Substituting back: \[ \text{Area} = 4 \cdot 2 \cdot \frac{2}{3} \left(8 - \sqrt{27}\right) \] \[ = \frac{16}{3} \left(8 - \sqrt{27}\right) \] Final Answer: \[ \text{Area} = \frac{16}{3} \left(8 - \sqrt{27}\right). \]