Green's Theorem — Question 9

Question 9 – Solution

Strategy. Make the Green’s Theorem integrand $Q_x-P_y$ equal to $1$ in three different ways.

Step 1: Formula using $x\,dy$ Choose $P=0$, $Q=x$. Then $Q_x-P_y=1$, so $$\operatorname{Area}(D)=\iint_D1\,dA=\boxed{\oint_Cx\,dy}.$$

Step 2: Formula using $-y\,dx$ Choose $P=-y$, $Q=0$. Again $Q_x-P_y=1$, giving $$\operatorname{Area}(D)=\boxed{-\oint_Cy\,dx}.$$

Step 3: Symmetric formula Choose $P=-y/2$ and $Q=x/2$. Then $$Q_x-P_y=\frac 12-\left(-\frac 12\right)=1,$$ and therefore $$\boxed{\operatorname{Area}(D)=\frac 12\oint_C(x\,dy-y\,dx)}.$$ This is also the average of the first two formulas.

Verification Green’s Theorem uses positive, counterclockwise orientation. Clockwise traversal negates all three line integrals, so a minus sign would be needed to recover positive geometric area.