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

The equation of the ellipse is: \[ \frac{x^2}{16} + \frac{y^2}{4} = 1. \] 
Rearrange to solve for \(y^2\): \[ \frac{y^2}{4} = 1 - \frac{x^2}{16}. \] \[ y^2 = 4\left(1 - \frac{x^2}{16}\right) = 4 - \frac{x^2}{4}. \] \[ y = \pm \sqrt{4 - \frac{x^2}{4}}. \]

 Step 1: Use symmetry to simplify the calculation. The ellipse is symmetric about the \(x\)-axis. The area between \(x = -2\) and \(x = 2\) can be calculated as twice the area above the \(x\)-axis: \[ \text{Area} = 2 \int_{-2}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx. \] 

Step 2: Change the limits and integrate. Since the integrand is even (symmetric about the \(y\)-axis), we can further simplify: \[ \text{Area} = 4 \int_{0}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx. \] 

Step 3: Substitution for simplification. 
Let: \[ u = 4 - \frac{x^2}{4}, \quad \text{so} \quad du = -\frac{x}{2} \, dx \quad \text{and} \quad x \, dx = -2 \, du. \] When \(x = 0\), \(u = 4\), and when \(x = 2\), \(u = 4 - \frac{2^2}{4} = 3\). 

The integral becomes: \[ \int_{0}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx = \int_{4}^{3} \sqrt{u} \cdot (-2) \, du. \] Simplify: \[ \int_{0}^{2} \sqrt{4 - \frac{x^2}{4}} \, dx = 2 \int_{3}^{4} \sqrt{u} \, du. \] 

Step 4: Evaluate the integral. The integral of \(\sqrt{u}\) is: \[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2}. \] Evaluate from \(u = 3\) to \(u = 4\): \[ \int_{3}^{4} \sqrt{u} \, du = \frac{2}{3} \left[4^{3/2} - 3^{3/2}\right]. \]
Simplify: \[ 4^{3/2} = (2^2)^{3/2} = 2^3 = 8, \quad 3^{3/2} = \sqrt{3^3} = \sqrt{27}. \] 
Thus: \[ \int_{3}^{4} \sqrt{u} \, du = \frac{2}{3} \left[8 - \sqrt{27}\right]. \] 

Step 5: Final area. Substitute back into the expression for the area: \[ \text{Area} = 4 \cdot 2 \cdot \frac{2}{3} \left[8 - \sqrt{27}\right] = \frac{16}{3} \left[8 - \sqrt{27}\right]. \] 

Final Answer: \[ \text{Area} = \frac{16}{3} \left[8 - \sqrt{27}\right]. \]