Question

Use the parametric equations of an ellipse, x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π, to find the area that it encloses.

Answer 1

The area of the ellipse $E$ is

$\displaystyle\iint_E\mathrm dx\,\mathrm dy$

Parameterize $E$ by

$\begin{cases}x=ar\cos\theta\\y=br\sin\theta\end{cases}$

with $0\le r\le1$ and $0\le\theta\le2\pi$. The Jacobian for this transformation is

$J=\begin{bmatrix}a\cos\theta&-ar\sin\theta\\b\sin\theta&br\cos\theta\end{bmatrix}$

with determinant $abr\cos^2\theta+abr\sin^2\theta=abr$. Then the integral is equivalent to

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=ab\iint_Er\,\mathrm dr\,\mathrm d\theta=ab\int_0^{2\pi}\int_0^1r\,\mathrm dr\,\mathrm d\theta=\boxed{\pi ab}$

Answer 2

Area of encloses ellipse is πab.

Given that

x = a cos θ

y = b sin θ

0 ≤ θ ≤ 2π

Find:

The area encloses

Computation:

$x = a cos \theta \\\\cos \theta = \frac{x}{a} \\\\y = b sin \theta\\\\Sin \theta = \frac{y}{b}$

By squaring and adding both term

$\frac{x^2}{a^2} + \frac{y^2}{B^2} = Sin^2\theta + Cos^2\theta\\\\\frac{x^2}{a^2} + \frac{y^2}{B^2} = 1$

So, Area of encloses ellipse = πab

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