Question

Using integration, find the area of the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{bounded between the lines} \quad x = -\frac{a}{2} \quad \text{to} \quad x = \frac{a}{2}. \]

Hint:

The area of an ellipse can be found by using integration, and the formula for the area of the ellipse over a specific interval can be derived using standard integral results for ellipses.

Answer 1

We are given the equation of an ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] First, solve for \(y\) in terms of \(x\): \[ \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \quad \Rightarrow \quad y^2 = b^2 \left( 1 - \frac{x^2}{a^2} \right) \] Taking the square root of both sides: \[ y = \pm b \sqrt{1 - \frac{x^2}{a^2}} \] Now, to find the area between the lines \(x = -\frac{a}{2}\) and \(x = \frac{a}{2}\), we need to integrate the positive half of the ellipse over this interval. The area is given by: \[ A = 2 \int_{-\frac{a}{2}}^{\frac{a}{2}} b \sqrt{1 - \frac{x^2}{a^2}} \, dx \] Simplify: \[ A = 2b \int_{-\frac{a}{2}}^{\frac{a}{2}} \sqrt{1 - \frac{x^2}{a^2}} \, dx \] This is a standard integral, and its solution is known. The result of the integral is: \[ \int \sqrt{1 - \frac{x^2}{a^2}} \, dx = \frac{a}{2} \left( x \sqrt{1 - \frac{x^2}{a^2}} + \arcsin\left( \frac{x}{a} \right) \right) \] Evaluating this from \(x = -\frac{a}{2}\) to \(x = \frac{a}{2}\): \[ A = 2b \left[ \frac{a}{2} \left( x \sqrt{1 - \frac{x^2}{a^2}} + \arcsin\left( \frac{x}{a} \right) \right) \right]_{-\frac{a}{2}}^{\frac{a}{2}} \] After performing the integration and simplifying, the total area is: \[ A = \frac{\pi a b}{2} \] Thus, the area of the ellipse bounded between the lines \(x = -\frac{a}{2}\) and \(x = \frac{a}{2}\) is: \[ \boxed{\frac{\pi a b}{2}} \]