Problem 82 - Calculus of a Single Variable

Question

Area Let \(a>0\) and \(b>0 .\) Show that the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi a b\) (see figure).

Step-by-Step Solution

Answer

The area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is proven to be \(\pi a b\).

Step 1: Setting Up the Integral

The area of ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) can be calculated by integrating with respect to x from -a to a. Rewriting the equation of the ellipse with y as the subject, it gives: \(y = b\sqrt{1-\frac{x^{2}}{a^{2}}}\). The full extent of the ellipse's area is calculated by doubling the result of the integral because y spans from -b to b.

Step 2: Solving the Integral

To solve the integral, substitute \(u = \frac{x}{a}\) to simplify the integral further. The Integral yields: \(2b\int_{-a}^{a}\sqrt{1-\frac{x^{2}}{a^{2}}}dx = 2ab\int_{-1}^{1}\sqrt{1-u^{2}}du = 2ab\left[\frac{u\sqrt{1-u^{2}}+sin^{-1}(u)}{2}\right]_{-1}^{1}\).

Step 3: Applying the Integral Results

Plugging in the integration limits yields: \(2ab\left[\frac{1*\sqrt{1 - 1^{2}} + sin^{-1}(1)}{2} - \frac{-1*\sqrt{1 - (-1)^{2}} - sin^{-1}(-1)}{2}\right] = 2ab\left[\frac{\pi}{2} - (-\frac{\pi}{2})\right] = 2ab\pi = \pi ab\). This shows that the area of the ellipse is indeed \(\pi ab\)