Question

Use the parametric equations of an ellipse, \(x = acos\theta ,y = bsin\theta ,0 \le \theta \le 2\pi \) to find the area that it encloses.

Short answer

The area enclosed of the given ellipse is given as: \(A = ab\pi \)

Step-by-step solution

Draw the graph of the given expression

The parametric equation of an ellipse is:

\(\begin{array}{l}x = a\cos \theta \\y = b\sin \theta ,0 \le \theta \le 2\pi \end{array}\)

The graph of the ellipse when is shown below:

Find the area

The area enclosed by the curve\(x = f(\theta ),y = g(\theta )\)as \(\theta \)increases from\(\alpha \)to\(\beta \)is:

\(A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \)

Here\(f(\theta ) = a\cos \theta ,g(\theta ) = b\sin \theta ,\alpha = 0,\beta = 2\pi \)

Find\(f'(\theta )\)

\(\begin{array}{l}f(\theta ) = a\cos \theta \\f'(\theta ) = - a\sin \theta \end{array}\)

The area of the ellipse is:

\(\begin{array}{l}A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \\A = \int\limits_\alpha ^\beta {b\sin \theta ( - a\sin \theta )} d\theta \\A = - ab\int\limits_0^{2\pi } {{{\sin }^2}\theta } d\theta \\A = - ab\int\limits_0^{2\pi } {\frac{{1 - \cos 2\theta }}{2}} d\theta \\A = \frac{{ - ab}}{2}\int\limits_0^{2\pi } {1 - \cos 2\theta d\theta } \\A = \frac{{ - ab}}{2}\left( {\theta - \frac{{\sin 2\theta }}{2}} \right)_0^{2\pi }\\A = \frac{{ - ab}}{2}\left( { - 2\pi + \frac{{\sin 4\pi }}{2} - \frac{{\sin 0}}{2}} \right)\\A = \frac{{ - ab}}{2}\left( { - 2\pi } \right)\\A = ab\pi \end{array}\)

Hence, the area enclosed of the given ellipse is given as: \(A = ab\pi \).