Without loss of generality, place the rectangle with its lower-left corner at the origin.

Then use the "formula" $A\={∳}_{C}x \mathrm{dy}$, where, on the two horizontal edges of the rectangle, $\mathrm{dy}\=0$, and where, on the left vertical edge, $x\=0$. The only remaining integration of merit is then along the right vertical edge, with the result ${\int }_0^{b}a \ⅆy$ = $ab$.

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 via the following task template, where the initial point is repeated as the terminal point so that the curve closes on itself.

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.

Applying the factor of $1\/2$ to the calculated value of the flux results in the expected $a b$.

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 in Table 9.10.4(a), where the initial point is repeated as the terminal point so that the curve closes on itself.