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Theorem

Green's theorem

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Statement

Let \( D \subseteq \mathbb{R}^2 \) be a bounded region whose boundary \( \partial D \) is a piecewise-\( C^1 \), simple closed curve, oriented positively (so that \( D \) lies to the left as the boundary is traversed), and let \( C = \partial D \). Let \( P, Q : \overline{D} \to \mathbb{R} \) be functions of class \( C^1 \) on an open set containing \( \overline{D} \). Then \[ \oint_C \left( P\,dx + Q\,dy \right) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \]

Why it matters

Green's theorem is the two-dimensional prototype of the general Stokes phenomenon: it converts a line integral around a boundary into an area integral of a local differential quantity (the scalar curl \( \partial Q/\partial x - \partial P/\partial y \)). This is the computational engine behind planimeters, the divergence/flux form used in fluid dynamics and electromagnetism, and it is the base case from which the classical Stokes theorem and the divergence theorem are built by the same "boundary = interior accumulation" idea.

It also gives the cleanest route to two facts used throughout vector calculus: a vector field on a simply connected planar domain with \( \partial Q/\partial x = \partial P/\partial y \) everywhere is conservative there, and areas of regions can be computed purely from boundary data, e.g. \( \text{Area}(D) = \tfrac12 \oint_C (x\,dy - y\,dx) \).

Hypotheses
\( D \) is bounded with piecewise-\( C^1 \), simple closed boundary \( \partial D \). If the boundary is not simple (self-intersecting), the "left-hand side" region is ambiguous and orientation can cancel or double-count area; e.g. a figure-eight curve traversed once assigns opposite signs to its two lobes, and the identity as stated fails to match a naive single double integral over the enclosed set. Positive (counterclockwise, interior-on-the-left) orientation of \( C \). Reversing orientation flips the sign of the line integral but not of the area integral, so the stated equality becomes \( \oint_C = -\iint_D \) if \( C \) is traversed clockwise; the theorem as written is an orientation convention, not merely a magnitude statement. \( P, Q \in C^1 \) on an open set containing \( \overline{D} \). If \( \partial P/\partial y \) or \( \partial Q/\partial x \) fails to exist or is discontinuous at even one interior point, the double integral on the right can diverge or take the wrong value. Classic example: \( P = \dfrac{-y}{x^2+y^2} \), \( Q = \dfrac{x}{x^2+y^2} \) is \( C^1 \) on \( \mathbb{R}^2 \setminus \{0\} \) with \( \partial Q/\partial x - \partial P/\partial y \equiv 0 \) there, yet for \( C \) the unit circle \( \oint_C(P\,dx+Q\,dy) = 2\pi \neq 0 = \iint_D 0\,dA \) if one incorrectly includes the origin in \( D \); the theorem requires \( P,Q\) to be \( C^1 \) on all of \( \overline{D} \), and here they are not \( C^1 \) (not even defined) at the origin, so it does not apply to the full disk. \( D \) is a "regular" (Type I ∩ Type II, or finitely decomposable into such) region so that Fubini's theorem applies to the double integral. Without some regularity allowing decomposition into vertically/horizontally simple pieces, the double integral cannot be reduced to iterated single integrals, and the proof strategy below (which builds the result piece-by-piece from Type I and Type II regions) does not directly go through; the theorem is still true for very general Jordan domains via a limiting/measure-theoretic argument, but that is a separate (harder) proof.
Proof

We prove the theorem for a region \( D \) that is simultaneously of Type I and Type II (a "simple region"); the general piecewise-\( C^1 \) simple closed curve case follows by decomposing \( D \) into finitely many such pieces and cancelling the interior boundary contributions, as in the final step.

1
\text{It suffices to prove separately } \oint_C P\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA \ \text{ and } \ \oint_C Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA.
Both the line integral \( \oint_C(P\,dx+Q\,dy) = \oint_C P\,dx + \oint_C Q\,dy \) and the double integral are additive over the integrand, by linearity of the (Riemann) integral. A
2
D = \{ (x,y) : a \le x \le b,\ \phi_1(x) \le y \le \phi_2(x) \}, \qquad \phi_1, \phi_2 \text{ continuous, } \phi_1 \le \phi_2.
This is the definition of \( D \) being a Type I region; by hypothesis \( D \) admits such a description with \( \phi_1, \phi_2 \) piecewise-\( C^1 \) (since \( \partial D \) is piecewise-\( C^1 \)). A
3
\iint_D \frac{\partial P}{\partial y}\,dA = \int_a^b \left[ \int_{\phi_1(x)}^{\phi_2(x)} \frac{\partial P}{\partial y}(x,y)\,dy \right] dx = \int_a^b \Big( P(x,\phi_2(x)) - P(x,\phi_1(x)) \Big)\,dx.
Fubini's theorem reduces the double integral to an iterated integral (legitimate since \( P \in C^1(\overline{D}) \) makes \( \partial P/\partial y \) continuous, hence integrable); the inner integral is then evaluated by the Fundamental Theorem of Calculus applied to \( y \mapsto P(x,y) \). B
4
\partial D = C = C_{\text{bottom}} \cup C_{\text{right}} \cup C_{\text{top}}^{-} \cup C_{\text{left}}, \quad C_{\text{bottom}}: y=\phi_1(x),\ x:a\to b; \ \ C_{\text{top}}: y=\phi_2(x),\ x:a\to b.
Positive orientation of the boundary of a Type I region traverses the graph of \( \phi_1 \) left-to-right along the bottom, then up the right edge \( x=b \), then the graph of \( \phi_2 \) right-to-left (i.e. \( C_{\text{top}} \) reversed) along the top, then down the left edge \( x=a \); this is exactly what "interior on the left" forces for such a region. B
5
\oint_C P\,dx = \int_{C_{\text{bottom}}} P\,dx + \int_{C_{\text{right}}} P\,dx + \int_{C_{\text{top}}^{-}} P\,dx + \int_{C_{\text{left}}} P\,dx = \int_a^b P(x,\phi_1(x))\,dx - \int_a^b P(x,\phi_2(x))\,dx.
On \( C_{\text{right}} \) and \( C_{\text{left}} \), \( x \) is constant so \( dx = 0 \) and those integrals vanish; on \( C_{\text{bottom}} \), parametrizing by \( x \) directly gives \( \int_a^b P(x,\phi_1(x))\,dx \); on \( C_{\text{top}}^{-} \) (top edge traversed backwards, \( x:b\to a \)) the substitution rule for line integrals gives \( -\int_a^b P(x,\phi_2(x))\,dx \). B
6
\oint_C P\,dx = -\int_a^b\Big(P(x,\phi_2(x)) - P(x,\phi_1(x))\Big)\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA.
Combine Step 5 with Step 3: the right-hand side of Step 5 is exactly the negative of the expression computed in Step 3. B
7
\text{Symmetrically, using } D \text{ as a Type II region } D=\{(x,y): c\le y\le d,\ \psi_1(y)\le x\le \psi_2(y)\}: \quad \oint_C Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA.
Identical argument to Steps 2–6 with the roles of \( x,y \) interchanged and the sign flipping once because now the "outer" (right, \( \psi_2 \)) boundary is traversed with \( y \) increasing (matching positive orientation), giving \( +\iint_D \partial Q/\partial x\,dA \) rather than a minus sign; this uses that \( D \) is also Type II (hypothesis). B
8
\oint_C(P\,dx+Q\,dy) = \oint_C P\,dx + \oint_C Q\,dy = -\iint_D \frac{\partial P}{\partial y}\,dA + \iint_D \frac{\partial Q}{\partial x}\,dA = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.
Add the results of Steps 6 and 7, then use linearity of the double integral to combine the two integrands. This proves the theorem for simple (Type I ∩ Type II) regions. A
9
D = \bigcup_{i=1}^n D_i, \quad D_i \text{ simple regions with disjoint interiors}, \qquad \sum_{i=1}^n \oint_{\partial D_i} (P\,dx+Q\,dy) = \oint_{\partial D}(P\,dx+Q\,dy).
General regions bounded by a piecewise-\( C^1 \) simple closed curve can be cut by finitely many auxiliary arcs into simple regions \( D_i \) (a standard but non-trivial decomposition fact for Jordan domains with rectifiable boundary). Each shared auxiliary arc between adjacent \( D_i, D_j \) is traversed once in each direction by the induced positive orientations of \( \partial D_i \) and \( \partial D_j \), so its two contributions to the sum cancel exactly (reversing a path negates the line integral, by the substitution rule); only the contributions along \( \partial D \) itself survive. C
10
\oint_{\partial D}(P\,dx+Q\,dy) = \sum_{i=1}^n \oint_{\partial D_i}(P\,dx+Q\,dy) = \sum_{i=1}^n \iint_{D_i}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dA = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dA.
Apply Step 8 to each simple piece \( D_i \), then use countable (here finite) additivity of the double integral over a partition of \( D \) into pieces with disjoint interiors (measure zero overlap on shared boundaries), which is a standard property of the Riemann/Lebesgue integral. This extends the theorem from simple regions to any region decomposable into finitely many, completing the proof. C
Result
\oint_C \left( P\,dx + Q\,dy \right) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA

Reading. The total circulation of the vector field \( \mathbf{F}=(P,Q) \) around the positively oriented boundary \( C \) of \( D \) equals the sum, over the interior, of the local rotation (scalar curl) of \( \mathbf{F} \) at every point. Circulation is a boundary measurement; curl is a pointwise, infinitesimal one, and the theorem says the boundary measurement is exactly the accumulation of the infinitesimal ones.

Scope. Applies to any bounded planar region with piecewise-\( C^1 \) simple closed boundary, positively oriented, and any \( P,Q \) that are \( C^1 \) on an open neighbourhood of \( \overline{D} \) (not merely on \( D \) or on \( C \)). It does not directly apply if \( P,Q \) fail to be \( C^1 \) at even one point of \( \overline{D} \), nor to boundaries that are not simple, without further care (multiply connected domains require the "boundary with holes, each oriented consistently" extension).

Corollaries & converses
  • Area formula. Taking \( P=-y/2,\ Q=x/2 \) gives \( \partial Q/\partial x - \partial P/\partial y = 1 \), so \( \text{Area}(D) = \tfrac12\oint_C(x\,dy-y\,dx) \); also \( \text{Area}(D)=\oint_C x\,dy = -\oint_C y\,dx \) (each a special case).
  • Conservative field test. If \( D \) is simply connected and \( \partial Q/\partial x = \partial P/\partial y \) everywhere on \( D \), then \( \oint_C(P\,dx+Q\,dy)=0 \) for every simple closed \( C \) in \( D \), which (given path-connectedness) implies \( \mathbf{F}=(P,Q) \) is conservative on \( D \).
  • Multiply connected extension. On a region with holes, Green's theorem still holds if the boundary is taken to be the outer curve (positively oriented) together with each inner boundary curve oriented negatively (clockwise), obtained by cutting the region into simply connected pieces as in Step 9.
  • Converse fails in general. \( \partial Q/\partial x - \partial P/\partial y = 0 \) at every point of \( D \) does not imply \( \oint_C(P\,dx+Q\,dy)=0 \) if \( D \) is not simply connected — the field \( P=-y/(x^2+y^2), Q=x/(x^2+y^2) \) on the punctured plane has zero curl everywhere it is defined, yet circulation \( 2\pi \) around any loop encircling the puncture once; the theorem itself is an equality, not a two-way implication about curl-free fields on arbitrary domains.
  • Green's theorem is the \( n=2 \) case of the general (Kelvin–)Stokes theorem, and its normal-flux form \( \oint_C \mathbf{F}\cdot\mathbf{n}\,ds = \iint_D \nabla\cdot\mathbf{F}\,dA \) is the planar case of the divergence theorem.
Fails without
  • Dropping \( C^1 \) on all of \( \overline{D} \): with \( P=-y/(x^2+y^2),\ Q=x/(x^2+y^2) \) and \( C \) the unit circle, \( \partial Q/\partial x-\partial P/\partial y=0 \) wherever defined, so a naive application gives \( \iint_D 0\,dA = 0\), but the true circulation is \( \oint_C(P\,dx+Q\,dy)=2\pi \); the discrepancy is entirely due to the singularity of \( P,Q \) at the origin, which lies in \( \overline{D} \).
  • Dropping simplicity of the boundary: a figure-eight curve \( C \) (self-intersecting at the origin) bounds two lobes traversed with opposite orientations; the line integral \( \oint_C(P\,dx+Q\,dy) \) picks up the curl-integral of one lobe minus the other, not the sum \( \iint_{D}(\partial Q/\partial x-\partial P/\partial y)\,dA \) over the union of both lobes as plain point-sets — the "region enclosed" is not even well-defined without specifying which lobe counts positively.
  • Reversing orientation: traversing \( C \) clockwise while still calling the enclosed region \( D \) with its usual (positive, planar) area measure flips the sign, turning the claimed identity into \( \oint_C(P\,dx+Q\,dy) = -\iint_D(\partial Q/\partial x-\partial P/\partial y)\,dA \); e.g. for \( \mathbf F=(-y,x) \) on the unit disk, clockwise circulation is \( -2\pi \) while \( \iint_D 2\,dA=2\pi \).
Common errors
  • Forgetting to check orientation and applying the formula with a clockwise-traversed boundary, silently picking up a sign error.
  • Applying the theorem when \( P \) or \( Q \) has a singularity strictly inside \( D \) (as with the standard "angle field" example), without first excising a small disk around the singularity and accounting for the extra boundary circle.
  • Computing \( \partial Q/\partial x - \partial P/\partial y \) as \( \partial P/\partial x - \partial Q/\partial y \) or otherwise swapping which partial derivative goes with which variable — the asymmetry (curl, not divergence) is easy to garble under time pressure.
  • Using the area formula \( \text{Area}=\tfrac12\oint(x\,dy-y\,dx) \) but parametrizing \( C \) clockwise, producing a negative "area".
  • Treating Green's theorem as applicable to a multiply connected region (an annulus, say) without orienting the inner boundary clockwise (negatively) relative to the outer one.
  • Confusing the two equivalent forms — circulation-curl \( \oint\mathbf F\cdot d\mathbf r = \iint(\nabla\times\mathbf F)\cdot\hat{\mathbf k}\,dA \) versus flux-divergence \( \oint\mathbf F\cdot\mathbf n\,ds=\iint\nabla\cdot\mathbf F\,dA \) — and mismatching \( (P,Q) \) with \( \partial x, \partial y \) accordingly.
Discussion

Green's theorem is usually attributed to George Green, who stated a version of it (without a rigorous proof, in modern terms) in his 1828 essay on electricity and magnetism; the result was independently arrived at and made rigorous through the 19th century alongside the broader development of vector calculus by Thomson (Kelvin), Stokes, and others, and it now sits as the \( \mathbb{R}^2 \) instance of the generalized Stokes theorem for differential forms, \( \int_{\partial M}\omega = \int_M d\omega \), with \( \omega = P\,dx+Q\,dy \) a 1-form and \( d\omega = (\partial Q/\partial x-\partial P/\partial y)\,dx\wedge dy \) its exterior derivative.

The proof strategy above — verify on "nice" (rectangle-like) regions using Fubini plus the Fundamental Theorem of Calculus, then patch together arbitrary regions by cutting them into nice pieces and cancelling shared internal boundaries — is the archetypal proof technique for every theorem in the Stokes family. The cancellation in Step 9 (adjacent regions induce opposite orientations on their shared cut) is the two-dimensional seed of the "boundary of a boundary is empty" principle (\( \partial\partial = 0 \)) that underlies homology theory.

Conceptually, the theorem converts between a global, extensive quantity (circulation, measured only along the edge) and a local, intensive density (curl, measured pointwise) integrated over the interior — the same conversion pattern as the Fundamental Theorem of Calculus itself, of which Green's theorem is sometimes described as a higher-dimensional avatar: FTC exchanges a "boundary evaluation" \( f(b)-f(a) \) for an interior integral of \( f' \), exactly as Green's theorem exchanges a boundary line integral for an interior area integral of a derivative-like quantity.

A subtlety worth flagging: Green's theorem as proved here assumes \( C^1 \) regularity of \( P,Q\) on a neighbourhood of the closed region, which is stronger than needed for the conclusion to hold — there exist refined versions (e.g. requiring only that \( P,Q\) be continuous on \( \overline D\), differentiable on \( D\), with \( \partial Q/\partial x - \partial P/\partial y\) merely Riemann integrable) proved via different techniques (e.g. approximating \( D\) by finite unions of small squares), but these lie beyond the elementary Type-I/Type-II decomposition argument given above.

Common misconceptions. A frequent misconception is that "zero curl everywhere in \( D \)" always implies zero circulation around \( \partial D \); this needs \( D \) simply connected (see Corollaries). Another is treating Green's theorem as only a computational shortcut for otherwise-hard line integrals, rather than recognizing it as the statement that curl is precisely the density whose integral reconstructs circulation — the conceptual content, not just the algebraic convenience, is what generalizes to Stokes' theorem in \( \mathbb{R}^3 \) and beyond.

Worked examples
1
\text{Evaluate } \oint_C (xy^2\,dx + (x^2y+2x)\,dy), \quad C: \text{ boundary of the square } [0,1]\times[0,1], \text{ positively oriented.}
Set \( P=xy^2,\ Q=x^2y+2x \); both are polynomials, hence \( C^1 \) on all of \( \mathbb{R}^2 \supseteq \overline D \), and \( D=[0,1]^2\) is a simple (Type I ∩ Type II) region with piecewise-\( C^1\) simple closed boundary, so Green's theorem applies. A
2
\frac{\partial Q}{\partial x} = 2xy+2, \qquad \frac{\partial P}{\partial y}=2xy, \qquad \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 2.
Direct differentiation. A
3
\oint_C(P\,dx+Q\,dy) = \iint_{[0,1]^2} 2\,dA = 2\cdot\text{Area}([0,1]^2) = 2.
Apply Green's theorem, then evaluate the double integral of the constant \( 2 \) as \( 2 \) times the area of the unit square. A
\oint_C (xy^2\,dx+(x^2y+2x)\,dy) = 2

Reading. Computing this line integral directly would require four separate parametrizations (one per side); Green's theorem replaces that with a single elementary area integral.

1
\text{Find the area enclosed by the ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a,b\gt 0, \text{ using } \mathbf r(t) = (a\cos t, b\sin t),\ t\in[0,2\pi].
The ellipse is a simple closed \( C^1 \) curve; parametrize it counterclockwise (positive orientation, since \( \mathbf r(t)\) traces it with the interior on the left as \( t\) increases from \( 0\) to \( 2\pi\)). A
2
\text{Area}(D) = \frac12\oint_C(x\,dy-y\,dx), \qquad \text{taking } P=-\tfrac{y}{2},\ Q=\tfrac{x}{2}\ \Rightarrow\ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\tfrac12+\tfrac12=1.
This is the area corollary of Green's theorem: with \( P=-y/2,\ Q=x/2\) (both \( C^1\) on \( \mathbb R^2\)), the curl density is identically \( 1\), so \( \iint_D 1\,dA=\text{Area}(D)\) equals the boundary integral. B
3
x\,dy - y\,dx = (a\cos t)(b\cos t)\,dt - (b\sin t)(-a\sin t)\,dt = ab\cos^2 t\,dt + ab\sin^2 t\,dt = ab\,dt.
Substitute \( x=a\cos t,\ dx=-a\sin t\,dt,\ y=b\sin t,\ dy=b\cos t\,dt\) and use \( \cos^2t+\sin^2t=1\). A
4
\text{Area}(D) = \frac12\int_0^{2\pi} ab\,dt = \frac12\cdot ab\cdot 2\pi = \pi ab.
Evaluate the resulting elementary integral over \( t\in[0,2\pi]\). A
\text{Area of the ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ is } \pi a b

Reading. The familiar ellipse area formula falls out of Green's theorem as a boundary computation, with no need to set up the corresponding double integral in \( x,y\) directly.

Problems
  1. Use Green's theorem to evaluate \( \oint_C (2xy\,dx + x^2\,dy) \) where \( C \) is the boundary of the region bounded by \( y=x^2 \) and \( y=x \) for \( 0\le x\le1 \), positively oriented.
    SolutionWith \( P=2xy,\ Q=x^2\): \( \partial Q/\partial x-\partial P/\partial y = 2x-2x=0\). Both are polynomials so \( C^1\) everywhere, and \( D=\{(x,y):0\le x\le1,\ x^2\le y\le x\}\) is a simple region, so Green's theorem applies directly: \( \oint_C(P\,dx+Q\,dy)=\iint_D 0\,dA = 0\). (This reflects that \( \mathbf F=(2xy,x^2)=\nabla(x^2y)\) is conservative on all of \( \mathbb R^2\), consistent with the corollary.)
  2. Use the area corollary of Green's theorem to find the area enclosed by the astroid \( x=\cos^3t,\ y=\sin^3t,\ t\in[0,2\pi]\).
    SolutionArea \( =\tfrac12\oint(x\,dy-y\,dx)\). Compute \( dx=-3\cos^2t\sin t\,dt,\ dy=3\sin^2t\cos t\,dt\). Then \( x\,dy-y\,dx = \cos^3t\cdot3\sin^2t\cos t\,dt - \sin^3t\cdot(-3\cos^2t\sin t)\,dt = 3\cos^4t\sin^2t\,dt+3\sin^4t\cos^2t\,dt = 3\sin^2t\cos^2t(\cos^2t+\sin^2t)\,dt=3\sin^2t\cos^2t\,dt=\tfrac34\sin^2(2t)\,dt\). So Area \( =\tfrac12\int_0^{2\pi}\tfrac34\sin^2(2t)\,dt = \tfrac38\int_0^{2\pi}\sin^2(2t)\,dt=\tfrac38\cdot\pi = \tfrac{3\pi}{8}\), using \( \int_0^{2\pi}\sin^2(2t)\,dt=\pi\). Area \( =\dfrac{3\pi}{8}\).
  3. Let \( P=-y/(x^2+y^2),\ Q=x/(x^2+y^2) \), and let \( C_R \) be the circle of radius \( R \) centred at the origin, positively oriented. Compute \( \oint_{C_R}(P\,dx+Q\,dy) \) directly, and explain why this does not contradict Green's theorem giving \( 0 \).
    SolutionParametrize \( x=R\cos t,\ y=R\sin t\). Then \( P\,dx+Q\,dy = \dfrac{-R\sin t\cdot(-R\sin t)+R\cos t\cdot R\cos t}{R^2}\,dt = \dfrac{R^2\sin^2t+R^2\cos^2t}{R^2}\,dt = dt\), independent of \( R\). So \( \oint_{C_R}(P\,dx+Q\,dy) = \int_0^{2\pi}dt = 2\pi\), for every \( R\gt0\). Green's theorem does NOT apply to the disk \( D_R=\{x^2+y^2\le R^2\}\) because \( P,Q\) are not defined (hence not \( C^1\)) at the origin, which lies in \( \overline{D_R}\); the hypothesis "\(P,Q\in C^1\) on an open set containing \( \overline D\)" fails, so there is no contradiction — the theorem simply does not license computing \( \iint_{D_R}0\,dA\) here.
  4. Region \( D \) is the annulus \( 1\le x^2+y^2\le 4 \). Using the multiply-connected extension of Green's theorem, express \( \iint_D \left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA \) in terms of line integrals over the two boundary circles, for general \( C^1 \) \( P,Q \) on a neighbourhood of \( D \), and verify consistency with Problem 3's field restricted to the annulus (where it is \( C^1\)).
    SolutionLet \( C_1\) be the inner circle \( x^2+y^2=1\) and \( C_2\) the outer circle \( x^2+y^2=4\), both traversed counterclockwise as curves in their own right. The correctly oriented boundary of the annulus is \( C_2\) (counterclockwise, positive) together with \( C_1\) traversed clockwise (negative), since the interior of the annulus must stay on the left throughout. Then \( \iint_D\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA = \oint_{C_2}(P\,dx+Q\,dy) - \oint_{C_1}(P\,dx+Q\,dy)\). For the field of Problem 3, which is \( C^1\) throughout the annulus (the origin is excluded), the curl is \( 0\) everywhere on \( D\), so the left side is \( 0\); indeed by Problem 3's computation \( \oint_{C_2}=2\pi\) and \( \oint_{C_1}=2\pi\) (the value \( 2\pi\) held for every radius \( R\)), so the right side is \( 2\pi-2\pi=0\), confirming consistency.
  5. Let \( \mathbf F=(P,Q) \) with \( P=e^{x}\sin y - ky,\ Q=e^{x}\cos y\), for a constant \( k \). For which \( k \) is \( \mathbf F \) conservative on \( \mathbb R^2\)? For general \( k\), compute \( \oint_C(P\,dx+Q\,dy)\) where \( C\) is the unit circle, positively oriented, using Green's theorem.
    Solution\( \partial Q/\partial x = e^x\cos y\). \( \partial P/\partial y = e^x\cos y - k\). So \( \partial Q/\partial x-\partial P/\partial y = k\). Since \( \mathbb R^2\) is simply connected and \( P,Q\) are \( C^1\) (indeed \( C^\infty\)) everywhere, \( \mathbf F\) is conservative iff this curl is identically \( 0\), i.e. \( k=0\). For general \( k\), Green's theorem applies (\( D=\)unit disk is a simple region, \( P,Q\in C^1(\mathbb R^2)\)) and gives \( \oint_C(P\,dx+Q\,dy)=\iint_D k\,dA = k\cdot\pi(1)^2=k\pi\).