Compute the area bounded by an ellipse \(\frac{x^{2}}{4} +\frac{y^{2}}{9}=1\) in two different ways outlined. (a) By finding an appropriate change of co-ordinates and then using the Jacobian to compute the integral. (b) Using Green’s theorem.

Solution:-

(a) \(\frac{x^{2}}{4} +\frac{y^{2}}{9}=1\)

\(a=2\) and \(b=3\) from the above relation

by changing the variable

taking, \(u= r\cos\Theta\) and \(v=rsin\Theta\)

Jacobian (\(J)\)\(=\)

\(=\)\(\frac{du}{dr}\) \(\frac{du}{d\theta}\)

\(\frac{dv}{dr}\) \(\frac{dv}{d\theta}\)

\(=\)\(\cos\) \(-r\sin\theta\)

\(\sin\theta\) \(r\cos\theta\)

\(=\) \(r\cos^{2}\theta+r\sin^{2}\theta\)

\(=\)\(r\)\((sin^{2}\theta+\cos^{2}\theta)\)

\(J\)\(=\)\(r\).\(1\)

\(J\)\(=\)\(r\)

Limits are, \(r=0\) to\(1\) and 0 to \(2\pi\) since radius is \(1\)

Area \(=\) \(ab\) \(\int_{0}^{2\pi}\int_{0}^{1}\)\(rdrd\theta\)

\(=\)\(2.3\int_{0}^{2\pi}[r^{2}]_{0}^{1}\)

\(=\)\(6\int_{0}^{2\pi}(1-0)d\theta\)

\(=\)\(6.[\theta]_{0}^{2\pi}\)

\(=\)\(\frac{12\Pi}{2}\)

\(=6\pi\)

(b) Solution:- The ellipse has parametric equations \(x=acost\) and \(y=bsint\), Where \(0\)\(\leq\)\(t\)\(\leq2\Pi\)

By using Green’s theorem

A\(=\)\(\frac{1}{2}\int_{c}(xdy-ydx)\)

\(=\)\(\frac{1}{2}\int_{0}^{2\pi}(acost)(bcost)dt-(bsint)(-asint)dt\)

\(=\)\(\frac{ab}{2}\int_{0}^{2\pi}dt\)

\(=\)\(\pi\)\(ab\)

\(=\)\(2.3\)

\(=\)\(6\pi\) Sq units