Green's Theorem

Suppose an ant walks once counterclockwise around the boundary of a muddy field. At every point of the boundary, you record how strongly the wind pushes the ant forward. Is there a way to predict the total push — the circulation — just by looking at how the wind swirls inside the field? This could be done without ever walking the boundary yourself. Green's Theorem answers yes: it converts a one-dimensional line integral around a closed curve into a two-dimensional double integral. That integral involves a single scalar quantity, the scalar curl, taken over the region enclosed. This is the two-dimensional prototype of one of the deepest ideas in mathematics. The behavior of a field on the boundary of a region is controlled by an accumulated local quantity in its interior. The same idea, dressed in different notation, becomes the Fundamental Theorem of Calculus in one variable. It also appears as Stokes' Theorem on curved surfaces in three-space, and The Divergence Theorem for flux. Green's Theorem is the first place a student sees this "boundary = interior sum" philosophy in genuinely two-dimensional form. It is also the computational workhorse for turning nasty line integrals into easy double integrals (and vice versa). The theorem also gives, almost as a bonus, one of the most elegant formulas in calculus. The area of a planar region can be computed entirely from an integral over its boundary,

A=12∮∂D(x dy−y dx).A = \frac{1}{2}\oint_{\partial D} (x\,dy - y\,dx).

This is not a coincidence. It is the special case of Green's Theorem applied to the vector field (−y/2,x/2)(-y/2, x/2). This explains why planimeters (mechanical area-measuring devices that trace only the boundary of a shape) actually work. This article gives the precise statement of Green's Theorem for simple, positively oriented regions. It includes a complete proof by direct computation on regions that are simultaneously xx-simple and yy-simple. It also covers the area corollary, two fully worked examples of contrasting flavor, and a counterexample. That counterexample shows exactly what fails when the region is not simply connected.

Key takeaways
  • Green's Theorem converts a line integral of P dx+Q dyP\,dx + Q\,dy around a positively oriented simple closed curve into a double integral of Qx−PyQ_x - P_y over the enclosed region.
  • The quantity Qx−PyQ_x - P_y is the scalar curl: it measures the local rotational tendency (circulation density) of the vector field F=(P,Q)\mathbf{F} = (P, Q).
  • The proof for "simple" regions (both xx-simple and yy-simple) reduces to two applications of the Fundamental Theorem of Calculus, one per variable.
  • A special choice of (P,Q)(P,Q) turns the theorem into a boundary formula for area, A=12∮(x dy−y dx)A = \tfrac12\oint(x\,dy - y\,dx).
  • The theorem requires the curve to be simple, closed, positively oriented, and piecewise-C1C^1, and P,QP, Q to be C1C^1 on an open set containing D‾\overline{D} — dropping simple-connectedness (a hole in the region) breaks the naive statement unless the boundary is taken with all components correctly oriented.
  • Key idea: circulation around a boundary equals the integral of local rotation over the interior.

Where this lives today

Green's Theorem is the two-dimensional special case of a single unifying principle — the generalized Stokes' Theorem — that also produces Stokes' Theorem (circulation of a vector field around the boundary of a curved surface equals the flux of its curl through the surface) and The Divergence Theorem (flux through a closed surface equals the integral of divergence over the enclosed volume). All three are instances of ∫∂Mω=∫Mdω\int_{\partial M}\omega = \int_M d\omega for differential forms, the modern language that subsumes the Fundamental Theorem of Calculus itself. In physics, Green's Theorem underlies two-dimensional fluid dynamics (relating circulation around an airfoil cross-section to vorticity inside it, central to lift computations), electrostatics in the plane, and heat flow problems. In engineering, the planimeter, a mechanical device that traces the boundary of an irregular shape on a map or blueprint, computes area purely via the boundary-integral corollary A=12∮(x dy−y dx)A = \tfrac12\oint(x\,dy-y\,dx) — no direct area computation ever happens inside the machine. In computer graphics and computational geometry, the same "shoelace formula" (a discretized, polygonal version of the area corollary) is the standard algorithm for computing polygon areas from vertex coordinates. In complex analysis, Green's Theorem is used to derive the Cauchy–Riemann equations' consequences and to prove the Cauchy Integral Theorem for suitably smooth domains, linking Conservative Fields and Potentials in the plane to holomorphic function theory (a C1C^1 planar vector field is conservative on a simply connected domain exactly when its scalar curl vanishes identically — a direct corollary of Green's Theorem). In signal processing and PDE theory, Green's identities (integration-by-parts consequences of the same theorem in higher dimensions) underlie the weak/variational formulations used in finite element methods.

Ask the tutor: How does the winding-number version of Green's Theorem handle a figure-eight curve, and can you walk through computing ∮P dx+Q dy\oint P\,dx+Q\,dy on one explicitly?

DefinitionGreen's Theorem — statement of the setup

Let D⊂R2D \subset \mathbb{R}^2 be a bounded region whose boundary ∂D\partial D is a simple closed curve (a curve that is closed and does not self-intersect), piecewise-C1C^1, and given the positive (counterclockwise) orientation, meaning DD lies to the left as one traverses ∂D\partial D. Let P(x,y)P(x,y) and Q(x,y)Q(x,y) be real-valued functions with continuous first partial derivatives on an open set containing D‾=D∪∂D\overline{D} = D \cup \partial D. Define the vector field F=(P,Q)\mathbf{F} = (P, Q) and write the line integral

∮∂DP dx+Q dy=∮∂DF⋅dr.\oint_{\partial D} P\,dx + Q\,dy = \oint_{\partial D} \mathbf{F} \cdot d\mathbf{r}.

The quantity ∂Q∂x−∂P∂y\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} is called the scalar curl (or 2D curl) of F\mathbf{F}.

Intuition

Picture the region DD chopped into a fine grid of tiny rectangles. Around each small rectangle, walk counterclockwise and compute the tiny circulation of F\mathbf{F} — the line integral of F⋅dr\mathbf{F}\cdot d\mathbf{r} around that little square. Now sum all these tiny circulations over every rectangle in the grid. Here is the key cancellation: every interior edge of the grid is shared by two adjacent rectangles, and each rectangle traverses that shared edge in the opposite direction from its neighbor (one goes up, the other goes down along the same segment). So the contributions from every interior edge cancel exactly in pairs. All that survives is the sum of contributions along edges lying on the outer boundary ∂D\partial D — which reassembles into ∮∂DF⋅dr\oint_{\partial D}\mathbf{F}\cdot d\mathbf{r}. Meanwhile, the tiny circulation around a small rectangle of area ΔA\Delta A centered near a point is approximately (Qx−Py) ΔA(Q_x - P_y)\,\Delta A — this is essentially the definition of curl as a circulation density, and it is exactly what the proof below verifies rigorously for a single rectangle-like region via the Fundamental Theorem of Calculus. Summing (Qx−Py)ΔA(Q_x - P_y)\Delta A over the grid and refining gives the double integral ∬D(Qx−Py) dA\iint_D (Q_x - P_y)\,dA. This "telescoping of tiny loops" is the heart of why boundary information encodes interior information.

Figure
Cancellation of interior edges in a grid decomposition of D
opposite directions cancel ∂D (positively oriented, survives) D subdivided into small cells: interior edges cancel in pairs
TheoremGreen's Theorem

Let D⊂R2D \subset \mathbb{R}^2 be a region that is simultaneously xx-simple and yy-simple, meaning:

  • (yy-simple) D={(x,y):a≤x≤b, g1(x)≤y≤g2(x)}D = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\} for continuous g1≤g2g_1 \le g_2 on [a,b][a,b];
  • (xx-simple) D={(x,y):c≤y≤d, h1(y)≤x≤h2(y)}D = \{(x,y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\} for continuous h1≤h2h_1 \le h_2 on [c,d][c,d];

with ∂D\partial D piecewise-C1C^1 and positively oriented. Let P,QP, Q be C1C^1 on an open set containing D‾\overline{D}. Then

∮∂DP dx+Q dy  =  ∬D(∂Q∂x−∂P∂y)dA.\oint_{\partial D} P\,dx + Q\,dy \;=\; \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.
Proof

Technique: direct reduction to the one-variable Fundamental Theorem of Calculus, applied once in each coordinate direction, exploiting the two simple-region descriptions separately. We prove the two "halves"

∮∂DP dx=−∬D∂P∂y dA,∮∂DQ dy=∬D∂Q∂x dA,\oint_{\partial D} P\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA, \qquad \oint_{\partial D} Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA,

and add them.

Step 1: the P dxP\,dx half, using the yy-simple description. Write D={a≤x≤b, g1(x)≤y≤g2(x)}D = \{a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Compute the double integral by iterated integration (justified since PyP_y is continuous on the compact set D‾\overline D, hence bounded and integrable, and Fubini's theorem for continuous integrands on such regions applies):

∬D∂P∂y dA=∫ab(∫g1(x)g2(x)∂P∂y(x,y) dy)dx.\iint_D \frac{\partial P}{\partial y}\,dA = \int_a^b \left(\int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}(x,y)\,dy\right) dx.

For fixed xx, the inner integral is computed by the Fundamental Theorem of Calculus (Part 2), since y↦P(x,y)y \mapsto P(x,y) is C1C^1:

∫g1(x)g2(x)∂P∂y(x,y) dy=P(x,g2(x))−P(x,g1(x)).\int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}(x,y)\,dy = P(x, g_2(x)) - P(x, g_1(x)).

Hence

∬D∂P∂y dA=∫ab[P(x,g2(x))−P(x,g1(x))] dx.(1)\iint_D \frac{\partial P}{\partial y}\,dA = \int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\,dx. \tag{1}

Now examine ∮∂DP dx\oint_{\partial D} P\,dx. Because DD is yy-simple, its positively oriented boundary decomposes into (at most) four pieces:

  • C1C_1: the graph y=g1(x)y = g_1(x), traversed left to right (x:a→bx: a \to b) — the "bottom";
  • C2C_2: the right vertical segment x=bx = b (if g1(b)≠g2(b)g_1(b) \ne g_2(b)), traversed bottom to top;
  • C3C_3: the graph y=g2(x)y = g_2(x), traversed right to left (x:b→ax: b \to a) — the "top", reversed;
  • C4C_4: the left vertical segment x=ax = a (if g1(a)≠g2(a)g_1(a) \ne g_2(a)), traversed top to bottom.

This is exactly the counterclockwise orientation: since DD lies above C1C_1 and below C3C_3, walking left-to-right along the bottom and right-to-left along the top keeps DD on the left, matching positive orientation. Along C2C_2 and C4C_4, xx is constant, so dx=0dx = 0 and these contribute nothing to ∮P dx\oint P\,dx. Along C1C_1, parametrize by x∈[a,b]x \in [a,b]: ∫C1P dx=∫abP(x,g1(x)) dx\int_{C_1} P\,dx = \int_a^b P(x, g_1(x))\,dx. Along C3C_3, parametrized by xx running from bb to aa (reversal reverses the sign of the standard left-to-right integral):

∫C3P dx=∫baP(x,g2(x)) dx=−∫abP(x,g2(x)) dx.\int_{C_3} P\,dx = \int_b^a P(x,g_2(x))\,dx = -\int_a^b P(x,g_2(x))\,dx.

Summing,

∮∂DP dx=∫abP(x,g1(x)) dx−∫abP(x,g2(x)) dx=−∫ab[P(x,g2(x))−P(x,g1(x))] dx.(2)\oint_{\partial D} P\,dx = \int_a^b P(x,g_1(x))\,dx - \int_a^b P(x,g_2(x))\,dx = -\int_a^b\big[P(x,g_2(x)) - P(x,g_1(x))\big]\,dx. \tag{2}

Comparing (1)(1) and (2)(2):

∮∂DP dx=−∬D∂P∂y dA.(3)\oint_{\partial D} P\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA. \tag{3}

Step 2: the Q dyQ\,dy half, using the xx-simple description. By an entirely symmetric argument (swap the roles of xx and yy, using D={c≤y≤d, h1(y)≤x≤h2(y)}D = \{c \le y \le d,\ h_1(y) \le x \le h_2(y)\}), we get

∬D∂Q∂x dA=∫cd[Q(h2(y),y)−Q(h1(y),y)] dy,\iint_D \frac{\partial Q}{\partial x}\,dA = \int_c^d \big[Q(h_2(y),y) - Q(h_1(y),y)\big]\,dy,

and tracing ∂D\partial D via its xx-simple decomposition (now the "right" curve x=h2(y)x = h_2(y) traversed bottom-to-top, and the "left" curve x=h1(y)x = h_1(y) traversed top-to-bottom, which is again exactly the positive orientation since DD lies to the right of x=h1(y)x=h_1(y) and to the left of x=h2(y)x = h_2(y)) gives

∮∂DQ dy=∫cd[Q(h2(y),y)−Q(h1(y),y)] dy=∬D∂Q∂x dA.(4)\oint_{\partial D} Q\,dy = \int_c^d\big[Q(h_2(y),y) - Q(h_1(y),y)\big]\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA. \tag{4}

The crucial point is that the boundary decomposition in Step 1 (top/bottom via g1,g2g_1,g_2) and in Step 2 (left/right via h1,h2h_1,h_2) are two different parametrized descriptions of the same oriented curve ∂D\partial D; the line integral ∮∂DP dx+Q dy\oint_{\partial D} P\,dx + Q\,dy is independent of which piecewise parametrization is used, since a line integral is defined intrinsically on the oriented curve and reparametrization (preserving orientation) does not change its value. This independence of parametrization is inherited directly from the definition of Line Integrals as limits of Riemann sums along the curve, not from Green's Theorem itself, so no circularity arises.

Step 3: add. Adding (3)(3) and (4)(4):

∮∂DP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA.\oint_{\partial D} P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.

■\blacksquare

∎
CorollaryExtension to finite unions of simple regions

If DD is a finite union D=D1∪⋯∪DnD = D_1 \cup \cdots \cup D_n of regions each satisfying the hypotheses of the Theorem, with pairwise disjoint interiors, and each shared internal boundary arc traversed oppositely by the two adjacent sub-regions, then Green's Theorem still holds for DD: apply the theorem to each DiD_i and sum. Interior shared edges cancel in the sum of line integrals (as in the Intuition figure), leaving only the integral over the true outer boundary ∂D\partial D, while the double integrals simply add by additivity of the integral over the disjoint pieces. This is how Green's Theorem is extended to regions that are not literally both xx-simple and yy-simple — e.g. an annulus, handled correctly by cutting it into simple pieces (see the Counterexample below for what happens if orientation bookkeeping is done carelessly).

CorollaryArea via the boundary

Under the hypotheses of the Theorem, the area of DD satisfies

A(D)=∬DdA=∮∂Dx dy=−∮∂Dy dx=12∮∂D(x dy−y dx).A(D) = \iint_D dA = \oint_{\partial D} x\,dy = -\oint_{\partial D} y\,dx = \frac{1}{2}\oint_{\partial D}(x\,dy - y\,dx).
Proof

Apply Green's Theorem with P=0, Q=xP = 0,\ Q = x: then Qx−Py=1−0=1Q_x - P_y = 1 - 0 = 1, so ∮∂Dx dy=∬D1 dA=A(D)\oint_{\partial D} x\,dy = \iint_D 1\,dA = A(D). Apply it with P=−y, Q=0P = -y,\ Q = 0: then Qx−Py=0−(−1)=1Q_x - P_y = 0-(-1) = 1, so −∮∂Dy dx=A(D)-\oint_{\partial D} y\,dx = A(D). Averaging the two identities gives the symmetric formula A(D)=12∮∂D(x dy−y dx)A(D) = \frac12\oint_{\partial D}(x\,dy - y\,dx). ■\blacksquare

∎

Worked examples

Example 1 — direct verification on an asymmetric region. Let DD be the triangle with vertices (0,0)(0,0), (3,0)(3,0), (0,2)(0,2) (a right triangle, deliberately not isoceles, to avoid masking sign errors), oriented counterclockwise. Let

P(x,y)=x2y,Q(x,y)=xy2+sin⁡(y).P(x,y) = x^2 y, \qquad Q(x,y) = xy^2 + \sin(y).

We compute ∮∂DP dx+Q dy\oint_{\partial D} P\,dx + Q\,dy two ways. Via the double integral. Here Qx=y2Q_x = y^2 and Py=x2P_y = x^2, so

Qx−Py=y2−x2.Q_x - P_y = y^2 - x^2.

Describe DD as yy-simple: x∈[0,3]x \in [0,3], and for fixed xx, yy ranges from 00 up to the hypotenuse. The hypotenuse joins (3,0)(3,0) to (0,2)(0,2), equation x3+y2=1\frac{x}{3}+\frac{y}{2}=1, i.e. y=2−2x3y = 2 - \frac{2x}{3}. So

∬D(y2−x2) dA=∫03∫02−2x/3(y2−x2) dy dx.\iint_D (y^2-x^2)\,dA = \int_0^3\int_0^{2-2x/3} (y^2-x^2)\,dy\,dx.

Inner integral: with g(x)=2−2x3g(x) = 2-\tfrac{2x}{3},

∫0g(x)(y2−x2) dy=g(x)33−x2g(x).\int_0^{g(x)}(y^2-x^2)\,dy = \frac{g(x)^3}{3} - x^2 g(x).

Now integrate over x∈[0,3]x\in[0,3]. Substitute u=2−2x3u = 2-\tfrac{2x}{3} where convenient, or expand directly. Expand g(x)=2−23xg(x) = 2 - \tfrac23 x:

g(x)3=8−3⋅4⋅23x+3⋅2⋅49x2−827x3=8−8x+83x2−827x3.g(x)^3 = 8 - 3\cdot4\cdot\tfrac23 x + 3\cdot 2\cdot \tfrac49x^2 - \tfrac{8}{27}x^3 = 8 - 8x + \tfrac{8}{3}x^2 - \tfrac{8}{27}x^3.

So g(x)33=83−83x+89x2−881x3\dfrac{g(x)^3}{3} = \tfrac83 - \tfrac83 x + \tfrac89 x^2 - \tfrac{8}{81}x^3. Also x2g(x)=2x2−23x3x^2 g(x) = 2x^2 - \tfrac23 x^3. Thus the inner integral, as a function of xx, is

g(x)33−x2g(x)=(83−83x+89x2−881x3)−(2x2−23x3)\frac{g(x)^3}{3} - x^2g(x) = \left(\tfrac83 - \tfrac83x + \tfrac89x^2 - \tfrac{8}{81}x^3\right) - \left(2x^2 - \tfrac23x^3\right)
=83−83x+(89−2)x2+(−881+23)x3=83−83x−109x2+4681x3.= \tfrac83 - \tfrac83 x + \left(\tfrac89-2\right)x^2 + \left(-\tfrac{8}{81}+\tfrac23\right)x^3 = \tfrac83 - \tfrac83x - \tfrac{10}{9}x^2 + \tfrac{46}{81}x^3.

Integrate over x∈[0,3]x \in [0,3] term by term:

∫0383 dx=8,∫03−83x dx=−83⋅92=−12,\int_0^3 \tfrac83\,dx = 8,\qquad \int_0^3 -\tfrac83 x\,dx = -\tfrac83\cdot\tfrac92 = -12,
∫03−109x2 dx=−109⋅9=−10,∫034681x3 dx=4681⋅814=464=11.5.\int_0^3 -\tfrac{10}{9}x^2\,dx = -\tfrac{10}{9}\cdot 9 = -10, \qquad \int_0^3 \tfrac{46}{81}x^3\,dx = \tfrac{46}{81}\cdot\tfrac{81}{4} = \tfrac{46}{4} = 11.5.

Sum: 8−12−10+11.5=−2.58 - 12 - 10 + 11.5 = -2.5. So

∬D(y2−x2) dA=−52.\iint_D (y^2 - x^2)\,dA = -\frac{5}{2}.

Via the line integral (spot-check on one edge, to confirm consistency without redoing the whole boundary computation from scratch would be circular reassurance — instead we verify the area-type piece independently). As a genuine independent check, note ∬D(y2−x2) dA=∬Dy2 dA−∬Dx2 dA\iint_D(y^2-x^2)\,dA = \iint_D y^2\,dA - \iint_D x^2\,dA. By symmetry of the triangle under swapping (x,y)↔(23y,32x)(x,y)\leftrightarrow(\tfrac23 y, \tfrac32 x) scaled appropriately... instead we directly verify ∬Dx2 dA\iint_D x^2\,dA and ∬Dy2 dA\iint_D y^2\,dA numerically as a sanity check:

∬Dx2 dA=∫03x2(2−23x)dx=∫03(2x2−23x3)dx=2⋅9−23⋅814=18−13.5=4.5.\iint_D x^2\,dA = \int_0^3 x^2\left(2-\tfrac23x\right)dx = \int_0^3\left(2x^2 - \tfrac23x^3\right)dx = 2\cdot 9 - \tfrac23\cdot\tfrac{81}{4} = 18 - 13.5 = 4.5.
∬Dy2 dA=∫03g(x)33 dx=∫03(83−83x+89x2−881x3)dx=8−12+89⋅9−881⋅814=8−12+8−2=2.\iint_D y^2\,dA = \int_0^3\frac{g(x)^3}{3}\,dx = \int_0^3\left(\tfrac83-\tfrac83x+\tfrac89x^2-\tfrac{8}{81}x^3\right)dx = 8 - 12 + \tfrac89\cdot 9 - \tfrac{8}{81}\cdot\tfrac{81}{4} = 8-12+8-2 = 2.

Then ∬Dy2 dA−∬Dx2 dA=2−4.5=−2.5\iint_D y^2\,dA - \iint_D x^2\,dA = 2 - 4.5 = -2.5, matching −52-\tfrac52 exactly. This confirms the double-integral computation is internally consistent. By Green's Theorem, the original line integral equals this double integral:

∮∂Dx2y dx+(xy2+sin⁡y) dy=−52.\oint_{\partial D} x^2y\,dx + (xy^2+\sin y)\,dy = -\frac{5}{2}.

Note the sin⁡(y)\sin(y) term contributed nothing to the curl computation, since ∂x(sin⁡y)=0\partial_x(\sin y) = 0 — it is present precisely to illustrate that Green's Theorem handles transcendental terms in QQ as long as they don't depend on xx in a way that alters QxQ_x; here it silently drops out, which is itself a useful sanity check (the term is a gradient-like piece with no xx-dependence contributing to curl).

Example 2 — using Green's Theorem to compute area of a boundary given only parametrically (astroid-type curve). Let DD be the region enclosed by the curve

x(t)=3cos⁡3t,y(t)=3sin⁡3t,t∈[0,2π],x(t) = 3\cos^3 t, \qquad y(t) = 3\sin^3 t, \qquad t \in [0, 2\pi],

(an astroid of "radius" 33), traversed counterclockwise as tt increases (one checks the standard astroid parametrization is counterclockwise for t:0→2πt: 0 \to 2\pi). We compute the enclosed area using the Area Corollary, A=12∮(x dy−y dx)A = \tfrac12\oint (x\,dy - y\,dx), rather than converting to Cartesian — illustrating why the boundary formula is often easier than a direct double integral over a region with a complicated non-simple-looking boundary. Compute differentials:

dx=−9cos⁡2tsin⁡t dt,dy=9sin⁡2tcos⁡t dt.dx = -9\cos^2t\sin t\,dt, \qquad dy = 9\sin^2t\cos t\,dt.

Then

x dy−y dx=3cos⁡3t⋅9sin⁡2tcos⁡t dt  −  3sin⁡3t⋅(−9cos⁡2tsin⁡t) dtx\,dy - y\,dx = 3\cos^3t \cdot 9\sin^2t\cos t\,dt \;-\; 3\sin^3t\cdot(-9\cos^2t\sin t)\,dt
=27cos⁡4tsin⁡2t dt+27sin⁡4tcos⁡2t dt=27cos⁡2tsin⁡2t(cos⁡2t+sin⁡2t) dt=27cos⁡2tsin⁡2t dt.= 27\cos^4t\sin^2t\,dt + 27\sin^4t\cos^2t\,dt = 27\cos^2t\sin^2t(\cos^2t+\sin^2t)\,dt = 27\cos^2t\sin^2t\,dt.

Using cos⁡2tsin⁡2t=14sin⁡2(2t)=18(1−cos⁡4t)\cos^2t\sin^2t = \tfrac14\sin^2(2t) = \tfrac18(1-\cos4t):

x dy−y dx=27⋅18(1−cos⁡4t) dt=278(1−cos⁡4t) dt.x\,dy-y\,dx = 27\cdot\frac18(1-\cos 4t)\,dt = \frac{27}{8}(1-\cos4t)\,dt.

Integrate over t∈[0,2π]t\in[0,2\pi]: the cos⁡4t\cos 4t term integrates to 00 over a full period (∫02πcos⁡4t dt=0\int_0^{2\pi}\cos4t\,dt = 0), leaving

∮(x dy−y dx)=278⋅2π=27π4.\oint(x\,dy-y\,dx) = \frac{27}{8}\cdot 2\pi = \frac{27\pi}{4}.

Hence

A(D)=12⋅27π4=27π8.A(D) = \frac12\cdot\frac{27\pi}{4} = \frac{27\pi}{8}.

This matches the known general formula for an astroid x=acos⁡3t,y=asin⁡3tx=a\cos^3t, y=a\sin^3t, namely A=3πa28A = \tfrac{3\pi a^2}{8}, evaluated at a=3a=3: 3π⋅98=27π8\tfrac{3\pi\cdot9}{8} = \tfrac{27\pi}{8}. ✓ Consistent.

CounterexampleWhy simple-connectedness / correct boundary bookkeeping is needed

Green's Theorem as stated applies to a region bounded by a single simple closed curve. If DD has a hole — say DD is the annulus 1≤x2+y2≤41 \le x^2+y^2 \le 4 — then ∂D\partial D consists of two curves: the outer circle C2:x2+y2=4C_2: x^2+y^2=4 and the inner circle C1:x2+y2=1C_1: x^2+y^2=1. The correctly oriented boundary ∂D\partial D requires the outer circle traversed counterclockwise and the inner circle traversed clockwise (so that DD is always on the left) — this is the correct extension of the Theorem (via the Corollary on finite unions, cutting the annulus into two simple half-annuli and canceling the cut edges). Here is the concrete failure if one instead (incorrectly) orients both circles counterclockwise, ignoring the hole, and naively tries to apply the single-curve version of Green's Theorem to the vector field

P=−yx2+y2,Q=xx2+y2,P = \frac{-y}{x^2+y^2}, \qquad Q = \frac{x}{x^2+y^2},

which is C1C^1 everywhere on DD (the origin, where it's undefined, is excluded from DD since DD starts at radius 11). A direct computation gives Qx−Py=0Q_x - P_y = 0 at every point of DD (a standard identity, checked by quotient rule: Qx=(x2+y2)−x(2x)(x2+y2)2=y2−x2(x2+y2)2Q_x = \frac{(x^2+y^2)-x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, and Py=−(x2+y2)+y(2y)(x2+y2)2=y2−x2(x2+y2)2P_y = \frac{-(x^2+y^2)+y(2y)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, so Qx−Py=0Q_x - P_y = 0). So the double integral over DD is ∬D0 dA=0\iint_D 0\,dA = 0. If one (wrongly) applied the naive single-curve conclusion "∮∂D=0\oint_{\partial D} = 0" treating ∂D\partial D as just the outer circle C2C_2 alone (forgetting the inner boundary entirely, as if DD had no hole), one would predict ∮C2P dx+Q dy=0\oint_{C_2} P\,dx+Q\,dy = 0. But direct computation on C2C_2, parametrized x=2cos⁡t,y=2sin⁡tx=2\cos t, y=2\sin t, t∈[0,2π]t\in[0,2\pi], gives

P dx+Q dy=−2sin⁡t⋅(−2sin⁡t)+2cos⁡t⋅2cos⁡t4 dt=4sin⁡2t+4cos⁡2t4 dt=dt,P\,dx+Q\,dy = \frac{-2\sin t \cdot(-2\sin t) + 2\cos t\cdot 2\cos t}{4}\,dt = \frac{4\sin^2t+4\cos^2t}{4}\,dt = dt,

so ∮C2F⋅dr=∫02πdt=2π≠0\oint_{C_2}\mathbf F\cdot d\mathbf r = \int_0^{2\pi}dt = 2\pi \ne 0. The resolution: the correct statement requires both boundary circles, oriented as prescribed (outer counterclockwise, inner clockwise). One computes similarly ∮C1,ccw=2π\oint_{C_1,\text{ccw}} = 2\pi, so with C1C_1 traversed clockwise (the correct orientation) its contribution is −2π-2\pi. Then the correctly-oriented total is

∮C2,ccw+∮C1,cw=2π+(−2π)=0=∬D0 dA,\oint_{C_2,\text{ccw}} + \oint_{C_1,\text{cw}} = 2\pi + (-2\pi) = 0 = \iint_D 0\,dA,

which matches perfectly. The lesson: dropping the hypothesis that the region is bounded by a single simple closed curve (i.e., dropping simple-connectedness) does not make Green's Theorem false — it makes the naive single-curve statement false, and the theorem must be applied in its correctly generalized multi-boundary form, with every component of ∂D\partial D oriented so DD stays on the left.

Scope, edge cases & common traps

SituationWhat happens
DD simply connected, ∂D\partial D single simple closed C1C^1 curve, P,Q∈C1(D‾)P,Q \in C^1(\overline D)Theorem applies exactly as stated.
DD has holes (multiply connected)Must include all boundary components, each oriented so DD is on the left; naive single-curve version fails (see Counterexample).
P,QP,Q fail to be C1C^1 at an interior point of DD (e.g. origin for F=(−y,x)/(x2+y2)\mathbf F = (-y,x)/(x^2+y^2) when DD contains the origin)Theorem does not apply directly; must exclude a small disk around the singularity and treat as multiply connected.
∂D\partial D only piecewise-C1C^1 (corners)Fine — the proof only used the Fundamental Theorem of Calculus on finitely many smooth arcs; corners contribute no extra terms.
Orientation reversed (clockwise)Formula picks up an overall minus sign: ∮−∂D=−∬D(Qx−Py) dA\oint_{-\partial D} = -\iint_D(Q_x-P_y)\,dA.
Region not xx-simple and yy-simple simultaneouslyCut into finitely many pieces that are (apply the Corollary on unions); true for any "reasonable" region (e.g. any region bounded by a piecewise-smooth simple closed curve).

The mathematician behind it

Origins · 1 minGeorge Green and how this was discovered
GG
George Green
1793–1841 · Nottingham, England — self-taught miller-mathematician

George Green worked for most of his life in his father's windmill in Nottingham, with almost no formal schooling beyond about one year of arithmetic classes. He taught himself mathematics from library books and, in 1828, at his own expense, privately published a 72-page essay, An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, that introduced what we now call Green's functions, Green's identities, and the theorem bearing his name — though in a form oriented toward potential theory in three dimensions rather than the plane. The pamphlet sold to only about 100 subscribers and was largely unnoticed until William Thomson (later Lord Kelvin) rediscovered it years after Green's death and arranged for its republication, recognizing its foundational importance for electromagnetism.

History
- 1793 — George Green born in Sneinton, near Nottingham, England.
- 1828 — Green privately publishes his Essay, containing the identities and boundary-to-interior conversion now called Green's Theorem and Green's identities.
- 1833 — Green, at age 40, enrolls as an undergraduate at Gonville and Caius College, Cambridge, one of the few formal academic credentials he ever obtained.
- 1841 — Green dies in Nottingham; his work remains obscure outside a small circle.
- 1846–1850 — William Thomson (Lord Kelvin) discovers copies of Green's essay, republishes it, and popularizes the results, cementing "Green's Theorem," "Green's function," and "Green's identities" as standard names in analysis and mathematical physics.
Why does Green's Theorem require P and Q to be C¹, not just continuous?

The proof directly integrates ∂P/∂y\partial P/\partial y and ∂Q/∂x\partial Q/\partial x, so these partial derivatives must exist; continuity of the partials (making P,Q∈C1P,Q \in C^1) ensures the Fundamental Theorem of Calculus applies cleanly to the inner iterated integrals and that Fubini-type interchange of integration order is valid. If PP or QQ is merely continuous but not differentiable, the right-hand double integral of Qx−PyQ_x - P_y may not even be defined. Weaker hypotheses (e.g. Lipschitz functions with derivatives existing almost everywhere) allow generalized versions of the theorem, but these require substantially more measure-theoretic machinery and are not part of the classical statement.

Is Green's Theorem just a special case of the Divergence Theorem in 2D?

They are closely related but conceptually distinct: Green's Theorem (circulation form) relates a line integral of the tangential component of F\mathbf F to the curl integral, while the 2D Divergence Theorem relates a line integral of the normal component of F\mathbf F to ∬Ddiv⁡F dA\iint_D \operatorname{div}\mathbf F\, dA. You can derive one from the other by rotating the vector field 90°90°: replacing (P,Q)(P,Q) with (−Q,P)(-Q,P) turns the circulation form into the flux form. Both are 2D shadows of the same generalized Stokes' Theorem in the language of differential forms, but as classically stated they emphasize different physical quantities (circulation vs. flux).

Can Green's Theorem be used to prove a vector field is conservative?

Indirectly, yes, but with care. If F=(P,Q)\mathbf F = (P,Q) is C1C^1 on a simply connected open set and Qx=PyQ_x = P_y everywhere there, then Green's Theorem shows ∮CF⋅dr=∬D(Qx−Py) dA=0\oint_{C}\mathbf F\cdot d\mathbf r = \iint_D(Q_x-P_y)\,dA = 0 for every simple closed curve CC bounding a region DD inside the domain, and vanishing circulation around every closed loop is equivalent to path-independence, hence to the existence of a potential function (see Conservative Fields and Potentials). Simple connectedness is essential here: the field (−y,x)/(x2+y2)(-y,x)/(x^2+y^2) has Qx=PyQ_x=P_y everywhere it's defined but is not conservative on any annulus surrounding the origin, precisely because that domain is not simply connected.

Does Green's Theorem work for non-convex or oddly shaped regions?

Yes, as long as the boundary is a piecewise-C1C^1 simple closed curve. The proof given here assumes DD is both xx-simple and yy-simple for concreteness, but any region bounded by a single piecewise-smooth simple closed curve — however non-convex, star-shaped-or-not, with corners, spikes, or concavities — can be cut by finitely many auxiliary line segments into finitely many pieces that are each simultaneously xx-simple and yy-simple. Applying the theorem to each piece and summing (with interior cut-edges canceling, as in the Corollary on unions) extends the result to essentially any "reasonable" bounded planar region.

How is Green's Theorem related to the Fundamental Theorem of Calculus?

Directly: the proof consists of applying the one-variable Fundamental Theorem of Calculus twice, once integrating ∂P/∂y\partial P/\partial y in yy and once integrating ∂Q/∂x\partial Q/\partial x in xx, then reassembling the boundary terms produced by each application into the line integral over ∂D\partial D. Conceptually, Green's Theorem is exactly the statement "the integral of a derivative over a region equals the values (integrated appropriately) on the boundary" — the same principle as the ordinary Fundamental Theorem of Calculus, just applied on a two-dimensional domain instead of an interval.

What goes wrong if the curve is not simple, i.e., it crosses itself?

If ∂D\partial D self-intersects, the "region enclosed" is ambiguous or may have a winding number greater than 11 on some sub-regions, and the naive double integral ∬D(Qx−Py) dA\iint_D(Q_x-P_y)\,dA no longer matches the line integral term-by-term. The correct generalization uses the winding number n(γ,x)n(\gamma,\mathbf x) of the curve around each point x\mathbf x, giving the weighted formula ∮γP dx+Q dy=∬(Qx−Py) n(γ,x) dA\oint_\gamma P\,dx+Q\,dy = \iint (Q_x-P_y)\,n(\gamma,\mathbf x)\,dA; the classical statement of Green's Theorem sidesteps this entirely by assuming simplicity (winding number exactly 11 inside, 00 outside).

Before this