Green's Theorem — Question 2

Question 2 – Solution

Strategy. The chosen coefficients make $Q_x-P_y=1$, so the line integral measures area.

Step 1: Set $P$ and $Q$ Write $$P=-\frac y2, \qquad Q=\frac x2.$$ Then $$Q_x-P_y=\frac 12-\left(-\frac 12\right)=1.$$

Step 2: Apply Green’s Theorem $$\frac 12\oint_C(-y\,dx+x\,dy) =\iint_D1\,dA =\operatorname{Area}(D).$$ The ellipse has semiaxes $3$ and $2$, so $$\boxed{\frac 12\oint_C(-y\,dx+x\,dy)=\pi(3)(2)=6\pi}.$$

Step 3: Reverse orientation Clockwise traversal negates the line integral, producing $-6\pi$. Geometric area itself remains $6\pi$; the sign records orientation.

Verification The area formula reduces to $\pi ab$, and substituting $a=3$, $b=2$ confirms $6\pi$.