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Chapter 7: Problem 82

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).

Short Answer

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

Step by step solution

01

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.
02

Solving the Integral

To solve the integral, substitute \(u = \frac{x}{a}\) to simplify the integral further. The Integral yields: \(2b\int_{-a}^{a}\sqrt{1-\frac{x^{2}}{a^{2}}}dx = 2ab\int_{-1}^{1}\sqrt{1-u^{2}}du = 2ab\left[\frac{u\sqrt{1-u^{2}}+sin^{-1}(u)}{2}\right]_{-1}^{1}\).
03

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 \(\pi ab\)

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Most popular questions from this chapter

Volume of an Ellipsoid Consider the plane region bounded by the graph of $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$ where \(a>0\) and \(b>0 .\) Show that the volume of the ellipsoid formed when this region is revolved about the \(y\) -axis is $$\frac{4}{3} \pi a^{2} b$$ What is the volume when the region is revolved about the \(x\) -axis?

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