Curl and Divergence β€” Question 7

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Question 7

For the planar field 𝑭(x,y)=⟨x2βˆ’y,x+y2⟩,\mathbf F(x,y)=\langle x^2-y,\,x+y^2\rangle, compute its planar divergence and scalar curl, then evaluate them at (βˆ’1,1)(-1,1).

Tasks

  1. Compute Px+QyP_x+Q_y.

  2. Compute Qxβˆ’PyQ_x-P_y.

  3. Interpret the two values at the specified point.

Original worksheet page 1: question and worked solution for 6-1-007
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Question 7 – Solution

Strategy. In two dimensions, divergence is scalar and curl is represented by the zz-component of the three-dimensional curl.

Step 1: Divergence With P=x2βˆ’yP=x^2-y and Q=x+y2Q=x+y^2, βˆ‡β‹…π‘­=Px+Qy=2x+2y.\nabla\cdot\mathbf F=P_x+Q_y=2x+2y. At (βˆ’1,1)(-1,1), βˆ‡β‹…π‘­=0.\boxed{\nabla\cdot\mathbf F=0}.

Step 2: Scalar curl curl⁡2D𝑭=Qxβˆ’Py=1βˆ’(βˆ’1)=2.\operatorname{curl}_{2D}\mathbf F=Q_x-P_y=1-(-1)=\boxed{2}. Equivalently, the three-dimensional embedding ⟨P,Q,0⟩\langle P,Q,0\rangle has curl ⟨0,0,2⟩\langle 0,0,2\rangle.

See the diagram in the original worksheet below.

Step 3: Interpret At the point there is no first-order net expansion or contraction, but there is positive counterclockwise rotational tendency. Zero divergence does not imply zero curl.

Verification The scalar curl is constant everywhere, while the divergence vanishes on the line y=βˆ’xy=-x; the point (βˆ’1,1)(-1,1) lies on that line.

Original worksheet page 2: question and worked solution for 6-1-007

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