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Calculus IIIChapter 5: Line IntegralsSection 5-7

Green's Theorem — Question 1

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Question 1

Let CC be the positively oriented boundary of the rectangle D={(x,y):0≤x≤2,0≤y≤3}.D=\{(x,y):0\le x\le 2,\ 0\le y\le 3\}. Evaluate ∮C(x2−y)dx+(x+y2)dy\oint_C (x^2-y)\,dx+(x+y^2)\,dy using Green’s Theorem.

Tasks

  1. Identify PP, QQ, and the required derivative difference.

  2. Convert the line integral to a double integral.

  3. Evaluate and check the positive orientation.

Original worksheet page 1: question and worked solution for 5-7-001
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Question 1 – Solution

Strategy. Apply the circulation form ∮CPdx+Qdy=∬D(Qx−Py)dA\oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.

Step 1: Differentiate Here P=x2−y,Q=x+y2,P=x^2-y, \qquad Q=x+y^2, so Qx−Py=1−(−1)=2.Q_x-P_y=1-(-1)=2.

See the diagram in the original worksheet below.

Step 2: Integrate over the region ∮CPdx+Qdy=∬D2dA=2Area⁡(D)=2(2)(3)=12.\oint_C P\,dx+Q\,dy =\iint_D2\,dA =2\operatorname{Area}(D)=2(2)(3)=\boxed{12}.

Step 3: Orientation Positive orientation means counterclockwise traversal, with the region on the left. That is the orientation for which Green’s Theorem gives the displayed positive sign.

Verification The derivative difference is the constant 22 and the rectangle has area 66, so the value must be 1212; reversing CC would give −12-12.

Original worksheet page 2: question and worked solution for 5-7-001

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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