Opening your learning space
Starting the app …
Opening your learning space
Starting the app …
Green's theorem relates a positively oriented line integral around the boundary of a planar region to an area integral of curl over its interior.
Free to browse · flashcards or a multiple-choice quiz · import from Essential
Formula
Both sides have units of field quantity times length.
| Symbol | Meaning | Unit |
|---|---|---|
| Region with piecewise smooth boundary | area unit | |
| Positively oriented boundary | length unit | |
| Components of the vector field | field unit | |
| Scalar planar curl | field/length |
\oint_{\partial D} P\,dx+Q\,dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dAThe positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed. The field must be sufficiently smooth on the region; holes require boundary components with the corresponding opposite orientation. The theorem is the two-dimensional relative of Stokes' theorem.
The positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed. The field must be sufficiently smooth on the region; holes require boundary components with the corresponding opposite orientation. The theorem is the two-dimensional relative of Stokes' theorem.
Green's theorem relates a positively oriented line integral around the boundary of a planar region to an area integral of curl over its interior.
Area from a boundary integral
Applies to a positively oriented simple boundary.
Rework the example for “Green's Theorem” and explain why the check is valid.
For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π.
The boundary integral is also π.
Check: both integrals agree.
Ignoring an assumption of the formula: The positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed
Check units and model conditions before substitution; then verify the result in the original relation or a limiting case.
Formula cluster
Related relations in the formula sheet: rotation-curl-vektorfeld, vektorielles-linienintegral, stokes-satz.
For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π. The boundary integral is also π. Check: both integrals agree.
Compute areas from boundary curves and relate circulation to local curl.
Here is how to work through a typical Green's Theorem (∮∂D P dx+Q dy = ∬D(Qₓ−Pᵧ)dA) task step by step:
Task
Rework the example for “Green's Theorem” and explain why the check is valid.
Solution path
For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π.
The boundary integral is also π.
Check: both integrals agree.
Green's theorem relates a positively oriented line integral around the boundary of a planar region to an area integral of curl over its interior. The notation ∮∂D P dx+Q dy = ∬D(Qx−Pᵧ)dA shows which quantities are connected. Here, D: Region with piecewise smooth boundary (area unit); ∂D: Positively oriented boundary (length unit); P,Q: Components of the vector field (field unit). Both sides have units of field quantity times length. An application depends on whether the supplied quantities match the definitions and reference directions. Common uses include Compute areas from boundary curves and relate circulation to local curl.. Read the variables and their dimensions before substituting numbers; this often reveals whether the result should be a length, rate, probability, or dimensionless statistic.
Start with a diagram or a list of known quantities, assign each one a symbol, and convert units to a consistent form. Then choose the appropriate equation and substitute values only afterward. Example: For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π. The boundary integral is also π. Check: both integrals agree. Verification is part of the calculation: The positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed Also compare the magnitude and unit with the context. Round only at the end, since early rounding can affect multi-step calculations, especially when small differences or sums are involved.
The positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed. The field must be sufficiently smooth on the region; holes require boundary components with the corresponding opposite orientation. The theorem is the two-dimensional relative of Stokes' theorem. These are not minor footnotes: when assumptions fail, the formula can produce a precise but inappropriate result. Before calculating, identify which values are observed, estimated, or modeled, and whether special cases change the equation. Compute areas from boundary curves and relate circulation to local curl. For numerical methods, also check step size and stability; for statistics, sampling design and distribution; for calculus, differentiability, boundary, and orientation. Use another method when these conditions do not hold.
Both relations belong to the same topic, but answer different questions. “Green's Theorem”: Green's theorem relates a positively oriented line integral around the boundary of a planar region to an area integral of curl over its interior. The companion relation “Curl of a Vector Field” The curl maps a differentiable vector field to a vector field describing local rotation direction and strength. Choose based on the quantity sought and the model: Compute areas from boundary curves and relate circulation to local curl. The worked example For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π. The boundary integral is also π. Check: both integrals agree. shows which quantities actually enter this relation. Compare both formulas using the same data and identify the companion’s extra assumption. This clarifies which relation fits.
Local assumptions and boundary or orientation conventions are part of the formula. Check them before differentiating or integrating; a valid local expression must not be extended to an entire domain without justification. The symbols must also match their definitions exactly: D: Region with piecewise smooth boundary (area unit); ∂D: Positively oriented boundary (length unit); P,Q: Components of the vector field (field unit). Do not calculate mechanically; write the assumption beside the setup, then check dimensions and sign. Also note: The positive orientation is counterclockwise, keeping the region on the left as the boundary is traversed The example For F=(-y/2,x/2) on the unit disk, Qx−Pᵧ=1, so the area integral is π. The boundary integral is also π. Check: both integrals agree. lets you follow these checks directly. If the assumptions do not fit, choose another relation from the chapter instead of forcing values into an unsuitable formula.
Quanta exam set
Curated exam set for "Green's Theorem":
Question (front)
Which formula describes Green's Theorem?
Answer in your set
Question (front)
How do you rearrange Green's Theorem for Area from a boundary integral?
Answer in your set
Question (front)
Which common mistake happens with Green's Theorem?
Answer in your set
+ 7 more cards: units, variables, derivation, example, exam task
The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.
This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.
Create a curated FSRS exam set for ∮∂D P dx+Q dy = ∬D(Qₓ−Pᵧ)dA: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
Curated formula set · LaTeX · FSRS spaced repetition · import from Essential