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 \(x = 2\).

Show Hint

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

Solution and Explanation

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

Was this answer helpful?
0
1

Questions Asked in CBSE CLASS XII exam

View More Questions