Write the ellipse in polar coordinates to obtain $r^2\left({\cos }^2\left(t\right)\/a^2\+{\sin }^2\left(t\right)\/b^2\right)\=1$, so

$r\left(t\right)\=\frac{a b}{\sqrt{b^2{\cos }^2\left(t\right)\+a^2{\sin }^2\left(t\right)}}$ 

Use the "formula" $A\={∳}_{C}x \mathrm{dy}$, with $x\=r\left(t\right) \cos \left(t\right)$ and

 $\mathrm{dy}\=\frac{d}{\mathrm{dt}}\left(r\left(t\right) \sin \left(t\right)\right) \mathrm{dt}\=\left(r\prime \left(t\right) \sin \left(t\right)\+r\left(t\right) \cos \left(t\right)\right) \mathrm{dt}$ 

where, after simplification, $r\prime \left(t\right)\=\frac{a b \left(b^2-a^2\right) \cos \left(t\right)\sin \left(t\right)}{{\left(a^2{\sin }^2\left(t\right)\+b^2{\cos }^2\left(t\right)\right)}^{3\/2}}$.

Putting this all together gives

$A\={\int }_0^{2 \mathrm{\π}}r\left(t\right)\cos \left(t\right)\left(r\prime \left(t\right) \sin \left(t\right)\+r\left(t\right) \cos \left(t\right)\right) \mathrm{dt}$ = $\mathrm{\π} a b$ 

Of course, the actual integration is complex enough to warrant the use of Maple, and even then, Maple must know that both $a$ and $b$ are positive before it can complete the calculation.

Recall that the "formula" $A\=\frac{1}{2}{∳}_{C}x \mathrm{dy}-y \mathrm{dx}$ derives from the divergence-form of Green's theorem, so that the line integral is the flux of the field $\mathbf{F}\mathit{\=}x \mathbf{i}\+y \mathbf{j}$ through the curve $C$. This flux is obtained with the following task template.

Should the "Clear All and Reset" button in the Task Template be pressed, all the data that has been input to the template will be lost. In that event, the reader should simply re-launch the example to recover the appropriate inputs to the template.

The Simplify button has been pressed to simplify the flux integral. Since both $a$ and $b$ are positive, $\mathrm{csgn}\left(a\/b\right)\=1$, and applying the factor of $1\/2$ to the calculated value of the flux results in the expected $\mathrm{\π} a b$.

Recall that the "formula" $A\=\frac{1}{2}{∳}_{C}x \mathrm{dy}-y \mathrm{dx}$ derives from the divergence-form of Green's theorem, so that the line integral is the flux of the field $\mathbf{F}\mathit{\=}x \mathbf{i}\+y \mathbf{j}$ through the curve $C$. This flux is obtained by the direct calculation contained in Table 9.10.5(a).