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Recipe

Apply Green's Theorem, tangential or normal form

Use when: you need either \oint_C\mathbf{F}\cdot d\mathbf{r} (circulation) or \oint_C\mathbf{F}\cdot\mathbf{n}\,ds (flux) around a closed, piecewise-smooth curve C bounding a region R

Goal: the circulation or flux value, whichever was asked for, computed as one double integral over R

Taught in Green's theorem, two forms · extends Apply Green's Theorem to a circulation integral

Quick recipe

  1. Confirm orientation
  2. Decide which quantity is being asked for
  3. Compute the divergence integrand P_x+Q_y (flux case only)
  4. Set up and evaluate the double integral over R
  5. Apply the orientation correction, if needed
  1. 1

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

    Why: both forms of the theorem need positive orientation; reversing C negates either side identically.

  2. 2

    Read the question: a circulation or work question needs the tangential form; a flux question needs the normal form.

    • the question asks for circulation or work — use \operatorname{curl}\mathbf{F}=Q_x-P_y as the integrand (Steps 3-5 of the earlier recipe).
    • the question asks for flux — use \operatorname{div}\mathbf{F}=P_x+Q_y as the integrand instead.

    Why: both forms convert a boundary integral to the identical style of double integral over R; only the integrand changes.

  3. 3

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

    Example

    Let \mathbf{F}=(x^3,\ y^3) and let C be the circle of radius 1 centered at the origin. Then P_x = 3x^2 and Q_y=3y^2, so P_x+Q_y = 3x^2+3y^2 = 3(x^2+y^2).

    Why: this is the normal-form analogue of the curl integrand, the local source-or-sink strength Green's Theorem accumulates over R for a flux question.

  4. 4

    Integrate the chosen integrand over R, choosing coordinates that match R's shape.

    Example

    On the unit disk, switch to polar coordinates: x^2+y^2=\rho^2, so \displaystyle\iint_R 3(x^2+y^2)\,dA = \int_0^{2\pi}\!\!\int_0^1 3\rho^2\cdot\rho\,d\rho\,d\theta = 2\pi\cdot\frac{3}{4} = \frac{3\pi}{2}.

    Why: a disk centered at the origin is exactly the case where polar coordinates turn a messy-looking integrand into a one-variable computation.

  5. 5

    If C was originally clockwise, negate the Step 4 result.

    Example

    This C was already counterclockwise, so no correction applies: \displaystyle\oint_C\mathbf{F}\cdot\mathbf{n}\,ds = \frac{3\pi}{2}.

    Why: this is the same orientation fact from Step 1, applied once the integral itself is finished.

Common mistakes

  • Using curl when the question asks for flux, or divergence when it asks for circulation.
  • Forgetting that both forms still need C positively oriented before the theorem applies.
  • Mixing up P_x+Q_y with P_y+Q_x when computing divergence.
  • Recomputing the whole setup from scratch instead of reusing P, Q, and R already identified for whichever form is needed.