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Recipe

Apply Green's Theorem to a circulation integral

Use when: you need \oint_C\mathbf{F}\cdot d\mathbf{r} around a closed, piecewise-smooth curve C that bounds a region R

Goal: the circulation value, computed as a double integral of curl over R instead of a line integral around C

Taught in Green's theorem · a later section extends this recipe with more cases

Quick recipe

  1. Confirm orientation
  2. Identify P, Q, and the enclosed region R
  3. Compute the curl integrand Q_x-P_y
  4. Set up and evaluate the double integral over R
  5. Apply the orientation correction from Step 1, if needed
  1. 1

    Check whether C is traversed counterclockwise (positively oriented, with R on the left of the direction of travel).

    • C is already counterclockwise — proceed normally through the rest of the recipe.
    • C is given clockwise — run the rest of the recipe as if it were counterclockwise, then negate the final answer.

    Why: Green's Theorem as stated needs positive orientation, and reversing C negates both sides of the identity identically.

  2. 2

    Write \mathbf{F}=(P,Q) and identify the region R that C encloses.

    Example

    Let \mathbf{F}=(xy,\ x^2) and let C be the boundary of the rectangle R=[0,2]\times[0,1], traversed counterclockwise.

    Why: the theorem trades the boundary integral for one specific double integral, over exactly this region.

  3. 3

    Differentiate Q with respect to x and P with respect to y, then subtract.

    Example

    Q_x = \dfrac{\partial}{\partial x}(x^2) = 2x and P_y = \dfrac{\partial}{\partial y}(xy) = x, so Q_x-P_y = 2x-x = x.

    Why: this is exactly the quantity the paddle-wheel instrument reads at every interior point, per this section's theorem.

  4. 4

    Integrate Q_x-P_y over R, choosing coordinates that match R's shape.

    Example

    \displaystyle\iint_R x\,dA = \int_0^2\!\!\int_0^1 x\,dy\,dx = \int_0^2 x\,dx = 2.

    • R is a disk or a sector centered at the origin — polar coordinates usually simplify the integral.
    • R is a rectangle or another axis-aligned region — rectangular coordinates are usually simplest.

    Why: Green's Theorem only supplies the integrand; choosing coordinates that match R's shape is what makes the double integral tractable.

  5. 5

    If C was originally clockwise, negate the Step 4 result. Otherwise the Step 4 result already stands as the answer.

    Example

    Since this C was already counterclockwise, no correction is needed: \oint_C\mathbf{F}\cdot d\mathbf{r} = 2.

    • Not yet: converting a FLUX integral, instead of a circulation integral, to a double integral of divergence — covered in Green's theorem, two forms

    Why: this is the same orientation fact from Step 1, applied to the finished number instead of the setup.

Common mistakes

  • Forgetting to check C's orientation before integrating, then reporting the wrong sign.
  • Computing P_x-Q_y or P_y-Q_x instead of Q_x-P_y.
  • Setting up the double integral over the wrong region, such as the curve itself instead of the area it encloses.
  • Choosing rectangular coordinates for a circular region and getting stuck in unnecessary algebra.