Without loss of generality, the circle can be centered at the origin, so it will have the Cartesian representation $x^2\+y^2\=a^2$. To continue working in Cartesian coordinates, obtain $y_{\pm }\=\pm \sqrt{a^2-x^2}$. Apply the "formula" $A\=-{∳}_{C}y \mathrm{dx}$ to obtain

Now an antiderivative for $y_{\+}$ is $\left(x y_{\+}\+a^2 \arctan \left(x\/y_{\+}\right)\right)\/2$. Since this will then be evaluated at the endpoints where $y_{\+}$vanishes, it is better to use the two-argument form of the arctangent function. Consequently, the computed value for $A$ will be

$a^2\left(\arctan \left(a\,0\right)-\arctan \left(-a\,0\right)\right)\=a^2\left(\mathrm{\π}\/2-\left(-\mathrm{\π}\/2\right)\right)\=\mathrm{\π} a^2$

as expected.

An alternative evaluation for the integral of $y_{\+}$starts with the trigonometric substitution $x\=a \sin \left(u\right)$.

Alternatively, 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.

Applying the factor of $1\/2$ to the calculated value of the flux results in the expected $\mathrm{\π} a^2$.

The alternate solution based on a direct calculation of flux is contained in Table 9.10.3(a).