Green's Theorem — Question 2

Section 18-7 · Calculus pages 3463–3464

Show solutionHide solution

Question 2

Let CC be the counterclockwise ellipse x29+y24=1.\frac{x^2}{9}+\frac{y^2}{4}=1. Evaluate 12∮C(−ydx+xdy)\frac 12\oint_C(-y\,dx+x\,dy) and explain its geometric meaning.

Tasks

  1. Apply Green’s Theorem to the given integral.

  2. Compute the enclosed area.

  3. Explain the effect of reversing the ellipse’s orientation.

Question 2 – Solution

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

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

Diagram shown in the original SVG worksheet.

Step 2: Apply Green’s Theorem 12∮C(−ydx+xdy)=∬D1dA=Area⁡(D).\frac 12\oint_C(-y\,dx+x\,dy) =\iint_D1\,dA =\operatorname{Area}(D). The ellipse has semiaxes 33 and 22, so 12∮C(−ydx+xdy)=π(3)(2)=6π.\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 −6π-6\pi. Geometric area itself remains 6π6\pi; the sign records orientation.

Verification The area formula reduces to πab\pi ab, and substituting a=3a=3, b=2b=2 confirms 6π6\pi.