Shear Flow: Curl Versus Boundary Circulation

Local Curl and Boundary Circulation

A shear flow can have a simple velocity pattern while still illustrating an important distinction: curl is measured locally, whereas circulation is accumulated around an entire loop. The two are connected, but they are not the same quantity. For a smooth velocity field, Stokes’ theorem says that the circulation around an oriented boundary equals the flux of curl through a surface spanning that boundary.

In planar flow, the curl points perpendicular to the plane. Its signed component tells us the local tendency of the flow to rotate. The circulation around a loop combines those local contributions over the enclosed region. If the curl varies from point to point, the total circulation depends on its integral across that region, not just on its value at one chosen point.

Two Basic Shear Fields

Use coordinates \((x,y,z)\), and let \(\mathbf e_z\) denote the unit vector in the positive \(z\)-direction. A horizontal shear has velocity that varies with \(y\), while a vertical shear has velocity that varies with \(x\). Write their velocity fields as follows, where \(U\) and \(W\) are differentiable functions:

\mathbf V_H(x,y,z)=(U(y),0,0), \qquad \mathbf V_V(x,y,z)=(0,W(x),0).

The curl of a field \((P,Q,R)\) is \((R_y-Q_z,\ P_z-R_x,\ Q_x-P_y)\). Applying this formula to the two shear fields gives:

\nabla\times\mathbf V_H=-U'(y)\mathbf e_z, \qquad \nabla\times\mathbf V_V=W'(x)\mathbf e_z.

The minus sign for horizontal shear matters. For example, if the horizontal speed increases as \(y\) increases, then \(U'(y)>0\), and the curl points in the negative \(z\)-direction.

Definition: For a velocity field \(\mathbf V\) and an oriented closed curve \(C\), the circulation is the line integral \(\displaystyle \oint_C \mathbf V\cdot d\mathbf r\). A counterclockwise traversal in the \(xy\)-plane, viewed from the positive \(z\)-axis, has the positive orientation associated with the normal \(\mathbf e_z\).

Circulation Around a Rectangle

Let \(R\) be the rectangle \(a\leq x\leq b\), \(c\leq y\leq d\), with \(a<b\) and \(c<d\). Its boundary is traversed counterclockwise. For horizontal shear, the velocity is parallel to the horizontal edges, and it is perpendicular to the vertical edges. The bottom edge contributes \((b-a)U(c)\), the top edge contributes \(-(b-a)U(d)\), and both vertical edges contribute zero. Therefore:

\oint_{\partial R}\mathbf V_H\cdot d\mathbf r =(b-a)\bigl(U(c)-U(d)\bigr).

The curl calculation gives the same result. The area integral is over the rectangle, and the fundamental theorem of calculus evaluates its inner integral:

\iint_R(\nabla\times\mathbf V_H)\cdot\mathbf e_z\,dA =\int_a^b\int_c^d -U'(y)\,dy\,dx =(b-a)\bigl(U(c)-U(d)\bigr).

This direct comparison is a rectangular case of Stokes’ theorem. The theorem applies when the field is continuously differentiable on an open region containing the oriented surface and its boundary; the boundary orientation must match the chosen surface normal. Here, the surface is the rectangle in the \(xy\)-plane with normal \(\mathbf e_z\).

Theorem: For a continuously differentiable field \(\mathbf V\) on a region containing an oriented surface \(S\) and its boundary \(C=\partial S\), Stokes’ theorem states that \(\displaystyle \oint_C\mathbf V\cdot d\mathbf r=\iint_S(\nabla\times\mathbf V)\cdot\mathbf n\,dS\), where \(\mathbf n\) is the chosen unit normal and the orientation of \(C\) is induced by that normal.

For vertical shear, the same rectangle has a complementary formula. Only the vertical edges contribute: the right edge contributes \((d-c)W(b)\), and the left edge contributes \(-(d-c)W(a)\). Thus:

\oint_{\partial R}\mathbf V_V\cdot d\mathbf r =(d-c)\bigl(W(b)-W(a)\bigr) =\int_a^b\int_c^d W'(x)\,dy\,dx.

Why Curl Is Local but Circulation Is Accumulated

For horizontal shear, the curl at a point has signed \(z\)-component \(-U'(y)\). The rectangle formula shows that circulation sums this component over the enclosed area. If the curl is constant throughout the rectangle, the circulation is that constant multiplied by the area. If curl varies, the area integral accounts for its variation.

The rectangle formula also shows how the local curl can be recovered from circulation density. Divide the circulation around a small rectangle by its area. For horizontal shear, with width \(b-a\) and height \(d-c\), this quotient is:

\frac{\oint_{\partial R}\mathbf V_H\cdot d\mathbf r}{(b-a)(d-c)} = -\frac{U(d)-U(c)}{d-c}.

As the height shrinks to zero around a fixed \(y\), this quotient approaches \(-U'(y)\), provided \(U\) is differentiable there. Thus curl describes circulation per unit area in the small-region limit. It is not itself the circulation around a finite loop.

Worked Examples

Worked Example: Linear Horizontal Shear on a Rectangle

Let \(\mathbf V(x,y,z)=(4y-1,0,0)\), and traverse the rectangle \(1\leq x\leq4\), \(0\leq y\leq2\) counterclockwise. First, the curl is:

\nabla\times\mathbf V =-4\mathbf e_z.

For a direct boundary calculation, the bottom edge has \(y=0\), so \(U(0)=-1\), and \(dx\) runs from \(1\) to \(4\). Its contribution is \((-1)(4-1)=-3\). The right and left edges contribute zero because \(dx=0\). On the top edge, \(y=2\), so \(U(2)=7\), and \(x\) runs from \(4\) to \(1\). Its contribution is \(7(1-4)=-21\). The total is:

$$ -3+0-21+0=-24. $$

The rectangle has area \(3\cdot2=6\). The curl integral is \((-4)(6)=-24\), agreeing with the direct calculation. This also checks the orientation and sign.

Worked Example: Polynomial Horizontal Shear on a Rectangle

Let \(\mathbf V(x,y,z)=(y^2-2y,0,0)\), and use the counterclockwise rectangle \(-1\leq x\leq2\), \(-1\leq y\leq2\). Its curl is:

\nabla\times\mathbf V=(2-2y)\mathbf e_z.

On the bottom edge, \(y=-1\), so \(U(-1)=(-1)^2-2(-1)=3\). The edge has width \(2-(-1)=3\), giving contribution \(3\cdot3=9\). On the top edge, \(y=2\), so \(U(2)=4-4=0\), giving zero. The vertical edges contribute zero. The total circulation is \(9\).

For the curl integral, integrate first over \(y\), then over the \(x\)-interval:

\begin{aligned} \iint_R(2-2y)\,dA &=3\int_{-1}^{2}(2-2y)\,dy\\ &=3\left[2y-y^2\right]_{-1}^{2}\\ &=3\bigl(0-(-3)\bigr)=9. \end{aligned}

The two calculations agree. Although the curl changes sign within the rectangle, its signed integral is \(9\); evaluating the curl at just one point would not give the total circulation.

Worked Example: Polynomial Vertical Shear on a Triangle

Let \(\mathbf V(x,y,z)=(0,x^2,0)\). Take the triangular region with vertices \((0,0)\), \((2,0)\), and \((0,1)\), traversed in that order and then back to the starting point. This order is counterclockwise. The curl is \((2x)\mathbf e_z\).

Along the bottom edge, \(dy=0\), so \(\mathbf V\cdot d\mathbf r=x^2dy=0\). Along the sloping edge, parameterize the path from \((2,0)\) to \((0,1)\) by \(x=2-2t\), \(y=t\), for \(0\leq t\leq1\). Since \(dy=dt\), its contribution is:

\int_0^1(2-2t)^2\,dt =4\int_0^1(1-2t+t^2)\,dt =4\left(1-1+\frac13\right)=\frac43.

The final edge has \(x=0\), so its velocity is zero and its contribution is zero. The direct circulation is therefore \(4/3\).

For the theorem-based calculation, the region is \(0\leq x\leq2\), \(0\leq y\leq1-x/2\). Hence:

\begin{aligned} \iint_S 2x\,dA &=\int_0^2\int_0^{1-x/2}2x\,dy\,dx\\ &=\int_0^2(2x-x^2)\,dx\\ &=\left[x^2-\frac{x^3}{3}\right]_0^2 =4-\frac83=\frac43. \end{aligned}

The direct boundary integral and the curl integral agree, with the counterclockwise orientation fixing the positive sign.

Worked Example: Linear Shear Around a Circle

Let \(\mathbf V(x,y,z)=(3y,0,0)\), and let \(C\) be the circle of radius \(2\) centered at the origin, traversed counterclockwise. Parameterize it by \(x=2\cos t\), \(y=2\sin t\), for \(0\leq t\leq2\pi\). Then \(dx=-2\sin t\,dt\), so the direct circulation is:

\oint_C\mathbf V\cdot d\mathbf r =\int_0^{2\pi}(6\sin t)(-2\sin t)\,dt =-12\int_0^{2\pi}\sin^2t\,dt =-12\pi.

The curl is \(-3\mathbf e_z\), constant over the disk. The disk area is \(\pi(2)^2=4\pi\), so the curl integral is \((-3)(4\pi)=-12\pi\), matching the direct computation.

A Common Sign Mistake

A frequent error is to use \(U'(y)\) instead of \(-U'(y)\) for horizontal shear. For the field \(\mathbf V=(4y-1,0,0)\), that mistake would give curl \(+4\mathbf e_z\) and predict circulation \(+4\cdot6=24\). But the counterclockwise boundary calculation gives \(-24\). The correction is to use the curl component \(Q_x-P_y=0-4=-4\). A sign disagreement can also arise from reversing the loop: clockwise traversal changes the sign of circulation and must be paired with the opposite surface normal.

Using the Comparison

These examples show why it is useful to compute circulation in two ways. A direct line integral tracks the velocity along each boundary segment and makes the orientation visible. A curl integral measures the local rotational contribution throughout a spanning surface and is often shorter when curl or the region is simple. Agreement between the two is also a strong check on signs, parameter limits, and area bounds.

In applications, curl and circulation carry different information: curl describes local rotation per unit area, while circulation is a boundary total. For instance, when a shear has constant curl, the circulation scales directly with the area enclosed. When its curl varies, regions with opposite signs can partially cancel. The same principle applies to other smooth loops, not only rectangles: the chosen boundary and surface orientations must match, and the curl must be integrated over the actual enclosed region.

Check Your Understanding

Use counterclockwise orientation for planar boundaries unless a question specifies otherwise.

For \(\mathbf V=(5y^2,0,0)\), what is the curl, including its direction?
For horizontal shear \(\mathbf V=(U(y),0,0)\), why does the top edge of a counterclockwise rectangle carry a minus sign in the direct circulation formula?
For \(\mathbf V=(0,3x,0)\), compute the circulation around the counterclockwise rectangle \(0\leq x\leq2\), \(0\leq y\leq1\) using the curl integral.
What does circulation divided by area approach as a rectangle shrinks to a point in a smooth planar flow?
How does the circulation change if the traversal direction around a loop is reversed while the velocity field stays fixed?