Rotating a Field to Connect Green’s Theorem Forms

One Theorem, Two Boundary Measurements

Green’s theorem is commonly written in two forms. The circulation form relates a tangential boundary integral to the scalar curl over the enclosed region. The flux form relates an outward boundary flux to the divergence over that region. These formulas look different, but a quarter-turn of the vector field converts one into the other.

The key is to keep track of three changes at once: how the components of the field change, how its derivatives combine, and how the rotated field is measured along the same positively oriented boundary. A sign error in any one of these steps can make the connection appear to fail.

The Two Forms and Their Conditions

Let \(D\) be a bounded planar region whose boundary \(C\) is a positively oriented, piecewise-smooth simple closed curve. Positive orientation means that \(C\) is traversed counterclockwise, so the region remains on the left as the boundary is followed. Let \(\mathbf{F}(x,y)=(P(x,y),Q(x,y))\), where \(P\) and \(Q\) have continuous first partial derivatives on an open set containing \(D\).

For these hypotheses, the circulation form of Green’s theorem is

The quantity \(Q_x-P_y\) is the scalar curl of \(\mathbf{F}\) in the plane. The flux form is

The left side of the second equation is the outward flux. Indeed, if the boundary is parametrized positively by \((x(t),y(t))\), its outward normal multiplied by its length element is \((dy,-dx)\). Thus \(\mathbf{F}\cdot\mathbf{n}\,ds=P\,dy-Q\,dx\). The quantity \(P_x+Q_y\) is the divergence of \(\mathbf{F}\).

The Clockwise Quarter-Turn

Define the clockwise quarter-turn of a vector \((a,b)\) to be \((b,-a)\). Apply this rotation at every point to the field \(\mathbf{F}=(P,Q)\), producing a new field \(\mathbf{G}\):

This is a rotation of the field values, not a rotation of the region or its boundary. The coordinates \((x,y)\) and the curve \(C\) stay fixed; only the vector assigned at each point changes.

Two direct calculations reveal why this particular rotation is useful. First, the divergence of \(\mathbf{G}\) is

Second, use the outward normal element \((dy,-dx)\) on the same positively oriented curve. The flux of \(\mathbf{G}\) is

Consequently, the boundary flux of the rotated field is precisely the circulation of the original field. Both the area density and the boundary integrand match, with no extra minus sign.

Deriving the Circulation Form from the Flux Form

Suppose the flux form of Green’s theorem is available for continuously differentiable fields on regions satisfying the stated boundary conditions. Since \(P\) and \(Q\) have continuous first partial derivatives, the rotated field \(\mathbf{G}=(Q,-P)\) does too. Apply the flux form to \(\mathbf{G}\). The boundary and area calculations above give

This is the circulation form. The argument works because the quarter-turn makes the flux integrand of the new field equal the circulation integrand of the original field, while making the new divergence equal the original scalar curl.

Deriving the Flux Form from the Circulation Form

The reverse implication follows by applying the circulation form to a field chosen from an arbitrary field \(\mathbf{H}=(A,B)\). Define \(\mathbf{F}=(-B,A)\). Its clockwise quarter-turn is \((A,B)=\mathbf{H}\). The circulation integrand of \(\mathbf{F}\) is \(-B\,dx+A\,dy\), which is the outward flux integrand of \(\mathbf{H}\). Also,

Applying the circulation form to \(\mathbf{F}\) therefore yields

This is the flux form for \(\mathbf{H}\). Thus, under the same regularity and boundary hypotheses, either form implies the other. The equivalence does not remove the need to check those hypotheses: the rotated field must still have continuous first partial derivatives on an open set containing the region, and the boundary must still have the required positive orientation and regularity.

Worked Example: A Polynomial Field on a Rectangle

Let \(\mathbf{F}=(x^2-y,xy+3x)\) and let \(D=[0,2]\times[0,1]\), with its boundary traversed counterclockwise. We will compare the circulation of \(\mathbf{F}\) with the outward flux of its clockwise-rotated field.

Here \(P=x^2-y\) and \(Q=xy+3x\), so the rotated field is \(\mathbf{G}=(xy+3x,-x^2+y)\). Its divergence is

The area integral is

For a direct boundary check, the outward flux integrand of \(\mathbf{G}=(A,B)\) is \(A\,dy-B\,dx\). On the bottom edge, \(y=0\) and \(x\) runs from \(0\) to \(2\). The contribution is

On the right edge, \(x=2\) and \(y\) runs from \(0\) to \(1\), giving \(\int_0^1(2y+6)\,dy=1+6=7\). On the top edge, \(y=1\) and \(x\) runs from \(2\) to \(0\), giving \(\int_2^0(x^2-1)\,dx=-\frac{2}{3}\). On the left edge, \(x=0\), so both \(A=0\) and \(B=0\), and the contribution is \(0\). The total is

As a separate check, the scalar curl of the original field is \(Q_x-P_y=(y+3)-(-1)=y+4\), exactly the divergence just computed for \(\mathbf{G}\). The circulation and the rotated-field flux both equal \(9\).

Worked Example: A Rotational Field on the Unit Circle

Let \(\mathbf{F}=(-y,x)\) on the unit disk \(D\), and let \(C\) be the unit circle traversed counterclockwise. Its rotated field is \(\mathbf{G}=(x,y)\). The scalar curl and the rotated divergence are

The area integral is \(2\) times the area \(\pi\) of the unit disk, so it is \(2\pi\). Check the boundary integral directly using \(x=\cos t\), \(y=\sin t\), for \(0\leq t\leq 2\pi\). Since \(dx=-\sin t\,dt\) and \(dy=\cos t\,dt\), the circulation is

For \(\mathbf{G}=(x,y)\), the outward unit normal on the unit circle is \((\cos t,\sin t)\), and \(ds=dt\). Its flux is therefore \(\int_0^{2\pi}(\cos^2t+\sin^2t)\,dt=2\pi\). The boundary and area checks agree.

Worked Example: Checking the Sign on a Square

Let \(\mathbf{F}=(x^2,xy)\) on the unit square \(D=[0,1]\times[0,1]\), with counterclockwise boundary. The rotated field is \(\mathbf{G}=(xy,-x^2)\). Its divergence, equal to the scalar curl of \(\mathbf{F}\), is

Thus \(\iint_D y\,dA=\int_0^1\int_0^1y\,dy\,dx=\frac{1}{2}\). To check the boundary flux, use \(A\,dy-B\,dx\) for \(\mathbf{G}=(A,B)\). The bottom edge contributes \(\int_0^1x^2\,dx=\frac{1}{3}\). The right edge contributes \(\int_0^1y\,dy=\frac{1}{2}\). The top edge, traversed from \(x=1\) to \(x=0\), contributes \(\int_1^0x^2\,dx=-\frac{1}{3}\). The left edge contributes \(0\). Hence

matching the area integral. In particular, the negative contribution from the top edge is required by the counterclockwise traversal; ignoring that direction would give an incorrect boundary total.

A Common Sign Error

A frequent mistake is to use a counterclockwise quarter-turn, \(\widetilde{\mathbf{G}}=(-Q,P)\), while claiming it converts circulation directly into outward flux and curl directly into divergence. In fact,

and its outward flux integrand is

Both quantities have the opposite sign. The counterclockwise rotation is not intrinsically wrong, but it must be accompanied by both minus signs. The clockwise choice \((Q,-P)\) is convenient because it matches the circulation and curl without changing their signs. Also keep the boundary orientation fixed when making these comparisons: reversing the direction of \(C\) reverses a boundary line integral.

Why the Connection Matters

The rotation explains structurally why Green’s theorem has two complementary forms. A field’s curl measures the area density associated with circulation; after a quarter-turn, that same quantity appears as divergence, the area density associated with outward flux. At the boundary, the rotation turns tangential measurement into normal measurement. The region and its orientation do not change, so the agreement follows from matching the integrands rather than from a separate geometric argument about the shape.

This is also a practical consistency check. If a calculation claims that a clockwise-rotated field has divergence \(-\operatorname{curl}\mathbf{F}\), or that its outward flux is the negative of the original circulation, revisit the rotation or the normal convention. With positive orientation and the outward normal element \((dy,-dx)\), the correct clockwise rotation is \((Q,-P)\), giving equal boundary integrands and equal area densities.