Question

Use double integrals to calculate the area inside the ellipse whose semiaxes have length a and b

Answer 1

The general equation of an ellipse centered at the origin with its semiaxes coinciding with the coordinate axes is given by

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Substituting $x=a\cos\theta$ and $y=b\sin\theta$ into the above equation gives the Pythagorean identity, so we can use polar coordinates quite nicely to our advantage.

If $\mathcal E$ is the ellipse with the equation above, the area is given by the double integral

$\displaystyle\iint_{\mathcal E}\mathrm dA=\iint_{(x/a)^2+(y/b)^2\le1}\mathrm dx\,\mathrm dy$

Let $x(r,\theta)=ar\cos\theta$ and $y(r,\theta)=br\sin\theta$, so that the Jacobian matrix is

$\mathbf J=\begin{bmatrix}x_r&x_\theta\\y_r&y_\theta\end{bmatrix}=\begin{bmatrix}a\cos t&-ar\sin t\\b\sin t&br\cos t\end{bmatrix}$

and the magnitude of its determinant is $|\det\mathbf J|=|abr|=abr$

since in polar coordinates we use the convention that $r\ge0$, and $a,b>0$ because they are lengths.

Now, the area is given by

$\displaystyle\iint_{\mathcal E}\mathrm dA=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}abr\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi a b\int_{r=0}^{r=1}r\,\mathrm dr$
$=\pi a b$