The general equation of an ellipse centered at the origin with its semiaxes coinciding with the coordinate axes is given by
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]
Substituting [tex]x=a\cos\theta[/tex] and [tex]y=b\sin\theta[/tex] into the above equation gives the Pythagorean identity, so we can use polar coordinates quite nicely to our advantage.
If [tex]\mathcal E[/tex] is the ellipse with the equation above, the area is given by the double integral
[tex]\displaystyle\iint_{\mathcal E}\mathrm dA=\iint_{(x/a)^2+(y/b)^2\le1}\mathrm dx\,\mathrm dy[/tex]
Let [tex]x(r,\theta)=ar\cos\theta[/tex] and [tex]y(r,\theta)=br\sin\theta[/tex], so that the Jacobian matrix is
[tex]\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}[/tex]
and the magnitude of its determinant is [tex]|\det\mathbf J|=|abr|=abr[/tex]
since in polar coordinates we use the convention that [tex]r\ge0[/tex], and [tex]a,b>0[/tex] because they are lengths.
Now, the area is given by
[tex]\displaystyle\iint_{\mathcal E}\mathrm dA=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}abr\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\pi a b\int_{r=0}^{r=1}r\,\mathrm dr[/tex]
[tex]=\pi a b[/tex]
