Timedora

09 Green’s Theorem and Planar Vector Calculus

A structured guide to planar vector fields, line integrals, circulation, flux, Green’s theorem, conservative fields, and the role of domain topology in vector calculus.

The language of planar vector calculus

A assigns a vector to each point in the plane. Write it as

F(x,y)=⟨P(x,y),Q(x,y)⟩.\mathbf F(x,y)=\langle P(x,y),Q(x,y)\rangle.

A parametrized curve has the form

r(t)=⟨x(t),y(t)⟩,a≤t≤b,\mathbf r(t)=\langle x(t),y(t)\rangle, \qquad a\le t\le b,

with differential dr=⟨dx,dy⟩d\mathbf r=\langle dx,dy\rangle. The vector line integral is therefore

∫CF⋅dr=∫CP dx+Q dy.\int_C\mathbf F\cdot d\mathbf r =\int_C P\,dx+Q\,dy.

This integral measures the work done by the field along the curve. When the curve is closed, use the notation ∮C\oint_C.

For a positively oriented closed curve, the enclosed region remains on the left as the curve is traversed. In the plane, positive orientation is usually counterclockwise.

Local rotation and expansion

Two derivatives summarize important local behavior:

  • The curl is

    curl⁡F=∂Q∂x−∂P∂y.\operatorname{curl}\mathbf F =\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}.

    It measures the local tendency to rotate counterclockwise. A positive value indicates counterclockwise rotational tendency, while a negative value indicates clockwise rotational tendency.

  • The is

    div⁡F=∂P∂x+∂Q∂y.\operatorname{div}\mathbf F =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}.

    It measures the local tendency to behave like a source or sink. Positive indicates net outward flow, and negative indicates net inward flow.

For example, if

F(x,y)=⟨−y,x⟩,\mathbf F(x,y)=\langle -y,x\rangle,

then

curl⁡F=2,div⁡F=0.\operatorname{curl}\mathbf F=2, \qquad \operatorname{div}\mathbf F=0.

The field has rotational behavior but no net source or sink behavior.

Takeaway: A line integral describes accumulated behavior along a curve, while curl and describe local behavior within the plane.

: and

connects a boundary line integral with a double integral over the enclosed region. Let CC be a positively oriented, piecewise smooth, simple closed curve enclosing a bounded region DD. Assume that PP and QQ have continuous first partial derivatives on an open region containing DD.

form

When the boundary quantity is , use

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

or, in vector notation,

∮CF⋅dr=∬Dcurl⁡F dA.\boxed{ \oint_C\mathbf F\cdot d\mathbf r = \iint_D\operatorname{curl}\mathbf F\,dA. }

The left side is total around the boundary. The right side accumulates local rotational tendency throughout the region.

For the unit circle and the field

F=⟨−y,x⟩,\mathbf F=\langle -y,x\rangle,

we have curl⁡F=2\operatorname{curl}\mathbf F=2. Thus

∮CF⋅dr=∬D2 dA=2Area⁡(D)=2π.\oint_C\mathbf F\cdot d\mathbf r = \iint_D2\,dA =2\operatorname{Area}(D) =2\pi.

This avoids a direct parametrization of the circle.

form

For a counterclockwise boundary, the outward normal satisfies

n ds=⟨dy,−dx⟩.\mathbf n\,ds=\langle dy,-dx\rangle.

Therefore,

F⋅n ds=P dy−Q dx.\mathbf F\cdot\mathbf n\,ds=P\,dy-Q\,dx.

The form is

∮CF⋅n ds=∬D(∂P∂x+∂Q∂y)dA\boxed{ \oint_C\mathbf F\cdot\mathbf n\,ds = \iint_D \left( \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y} \right)dA }

or

∮CF⋅n ds=∬Ddiv⁡F dA.\boxed{ \oint_C\mathbf F\cdot\mathbf n\,ds = \iint_D\operatorname{div}\mathbf F\,dA. }

For F(x,y)=⟨x,y⟩\mathbf F(x,y)=\langle x,y\rangle, the is 22. Across a circle of radius RR,

∮CF⋅n ds=∬D2 dA=2πR2.\oint_C\mathbf F\cdot\mathbf n\,ds = \iint_D2\,dA =2\pi R^2.

Takeaway: Match with curl and with before setting up the double integral.

Choosing and using the correct form

A reliable way to apply is to identify the boundary quantity before differentiating.

  1. Decide whether the problem asks for or outward .

  2. For , compute

    Qx−Py=∂Q∂x−∂P∂y.Q_x-P_y =\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}.

  3. For , compute

    Px+Qy=∂P∂x+∂Q∂y.P_x+Q_y =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}.

  4. Describe the enclosed region DD.

  5. Evaluate the resulting double integral.

  6. Check the orientation.

The essential correspondence is:

  • : The boundary integral is

    ∮CP dx+Q dy,\oint_C P\,dx+Q\,dy,

    and the corresponding interior quantity is

    curl⁡F=Qx−Py.\operatorname{curl}\mathbf F=Q_x-P_y.

  • Outward : The boundary integral is

    ∮CP dy−Q dx,\oint_C P\,dy-Q\,dx,

    and the corresponding interior quantity is

    div⁡F=Px+Qy.\operatorname{div}\mathbf F=P_x+Q_y.

Reversing the orientation changes the sign of both and . For a region with holes, the outer boundary is counterclockwise and each inner boundary is clockwise.

Computing area from a boundary integral

Choose

P=−y2,Q=x2.P=-\frac{y}{2}, \qquad Q=\frac{x}{2}.

Then

Qx−Py=12−(−12)=1.Q_x-P_y =\frac{1}{2}-\left(-\frac{1}{2}\right)=1.

gives the area formula

Area⁡(D)=12∮C(x dy−y dx)\boxed{ \operatorname{Area}(D) =\frac{1}{2}\oint_C(x\,dy-y\,dx) }

for a positively oriented simple closed curve.

Takeaway: The main difficulty is usually not integration; it is selecting the correct form, region, and orientation.

Conservative fields, , and holes

A has a scalar potential ϕ\phi satisfying

F=∇ϕ=⟨∂ϕ∂x,∂ϕ∂y⟩.\mathbf F=\nabla\phi =\left\langle \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y} \right\rangle.

For such a field, the line integral is determined by the endpoints:

∫CF⋅dr=ϕ(endpoint)−ϕ(starting point).\int_C\mathbf F\cdot d\mathbf r =\phi(\text{endpoint})- \phi(\text{starting point}).

This is . On a suitable , the following conditions are equivalent:

  1. F\mathbf F is conservative.

  2. The line integral is path independent.

  3. Every closed-curve integral is zero:

    ∮CF⋅dr=0.\oint_C\mathbf F\cdot d\mathbf r=0.

  4. The planar curl is zero:

    ∂Q∂x−∂P∂y=0.\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0.

For example, let

F(x,y)=⟨2x,2y⟩.\mathbf F(x,y)=\langle 2x,2y\rangle.

Its curl is zero, and it is the gradient of

ϕ(x,y)=x2+y2.\phi(x,y)=x^2+y^2.

For any path from (1,0)(1,0) to (0,2)(0,2),

∫CF⋅dr=ϕ(0,2)−ϕ(1,0)=4−1=3.\int_C\mathbf F\cdot d\mathbf r =\phi(0,2)-\phi(1,0) =4-1=3.

The condition on the region matters. A has no holes that obstruct the shrinking of closed curves. A disk, rectangle, or the entire plane is simply connected. A punctured plane and an annulus are not.

Consider

F(x,y)=⟨−yx2+y2,xx2+y2⟩,\mathbf F(x,y)= \left\langle -\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle,

which is defined on the plane with the origin removed. Its curl is zero wherever it is defined, but around the unit circle,

∮CF⋅dr=2π.\oint_C\mathbf F\cdot d\mathbf r=2\pi.

Thus it is not conservative on that domain. The hole at the origin prevents zero curl from guaranteeing .

Takeaway: Zero curl is sufficient for conservativeness on an appropriate simply connected domain, but not necessarily on a domain with holes.

Domains with holes and physical meaning

also applies to a region with holes when every boundary component is included with the correct orientation. The outer boundary is oriented counterclockwise, while each inner boundary is oriented clockwise. These orientations keep the region on the left as each boundary component is traversed.

If DD has multiple boundary components, then

∮∂DP dx+Q dy=∬D(Qx−Py) dA,\oint_{\partial D}P\,dx+Q\,dy = \iint_D(Q_x-P_y)\,dA,

where ∂D\partial D includes the outer boundary and all inner boundaries with their prescribed orientations.

For a curl-free field on a multiply connected region, the total over all boundary components is zero, but the around an individual hole may be nonzero. This is why checking curl alone is insufficient when the domain has holes.

The same topological issue explains the punctured-plane example: the field is not defined at the origin, so cannot be applied directly to the disk bounded by the unit circle. The hypotheses require the field to be defined throughout the region enclosed by the boundary.

Physical interpretation

For a fluid velocity field,

∮CF⋅dr\oint_C\mathbf F\cdot d\mathbf r

measures the fluid’s total tendency to move around the boundary. converts this quantity into the integral of local curl over the enclosed region.

Outward

∮CF⋅n ds\oint_C\mathbf F\cdot\mathbf n\,ds

measures the net amount of fluid crossing the boundary outward. converts it into the integral of , which measures local sources and sinks.

Takeaway: Topology and hypotheses are part of the theorem, not technical afterthoughts: boundary orientation, holes, and whether the field is defined on the full region can determine whether a conclusion is valid.