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Flux and Circulation Applications · Tutorial 994 of 999

Circulation Around Holes and Multiple Boundaries

Track the induced direction of every boundary component and use Stokes’ theorem carefully on multiply connected regions and punctured domains.

Advanced 10 min read

What You'll Learn

  • State the hypotheses for applying Stokes’ theorem to a surface with multiple boundary components
  • Determine each boundary direction from the chosen surface normal
  • Explain why inner boundary curves have the opposite orientation to the outer boundary
  • Compute circulation around annuli and regions with several holes
  • Apply Stokes’ theorem when a field is defined on a punctured domain
  • Recognize the error caused by assigning every boundary component the same direction

Why Holes Change the Boundary Calculation

A region with a hole has more than one boundary curve. Stokes’ theorem still relates the circulation around the entire oriented boundary to the curl across the surface, but “the entire boundary” includes every component, with its induced orientation. For a planar region viewed from above, the outer boundary runs counterclockwise while each hole boundary runs clockwise.

This reversal is not an optional sign convention. It is required for the boundary orientation to agree with the chosen surface normal. It also matters when the vector field is defined everywhere on the region but not at a point inside a hole: Stokes’ theorem can apply to the region around the excluded point without applying to a surface that includes it.

Definition: An oriented surface with holes is a surface whose boundary consists of an outer curve and one or more inner curves surrounding omitted regions. Each boundary component is oriented by the surface orientation. For a planar region with upward normal, the outer curve is counterclockwise and every inner curve is clockwise.

Stokes’ Theorem with Multiple Boundary Components

Let \(S\) be a suitably smooth oriented surface, with a finite number of piecewise-smooth boundary curves \(C_0,C_1,\ldots,C_m\). Let \(\mathbf F\) have continuous first partial derivatives on an open set containing \(S\). Orient the boundary curves consistently with the chosen orientation of \(S\). Stokes’ theorem says that the circulation around all the boundary components, with those orientations, equals the flux of the curl through the surface:

$$ \oint_{\partial S}\mathbf F\cdot d\mathbf r = \sum_{j=0}^{m}\oint_{C_j}\mathbf F\cdot d\mathbf r = \iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS. $$

Here \(\mathbf n\) is the chosen unit normal. The notation \(\partial S\) means the full boundary, not just its outermost curve. If the direction of \(\mathbf n\) is reversed, every induced boundary direction must also be reversed.

For an upward-oriented planar region, the induced direction has a practical description: as you travel along any boundary component, the surface region remains on your left. This places the region to the left along the outer boundary when it runs counterclockwise. Along a hole boundary, keeping the region to your left requires traveling clockwise.

Orientation rule: If \(\boldsymbol{\nu}\) is the outward-pointing unit conormal to the boundary within the surface, and \(\mathbf n\) is the unit normal to the surface, the induced unit tangent is \(\mathbf t=\mathbf n\times\boldsymbol{\nu}\). This rule reverses all boundary directions if the surface normal is reversed.

Why the Inner Boundary Runs Backward

The orientation rule can also be understood by cutting a narrow path from each hole to the outer edge. The cut turns the surface into pieces with simpler boundaries. Each side of a cut is traversed once in one direction and once in the opposite direction, so the two line integrals along that cut cancel. What remains is the original outer boundary and the boundaries around the holes.

Along each remaining boundary, the surface lies to the left as it is traversed. On the outside edge this gives counterclockwise motion for an upward normal. Around a hole, the surface lies outside the hole, so this same left-side rule gives clockwise motion. Thus the inner direction is a consequence of the orientation and the geometry, not a separate adjustment made to the integral.

Worked Example: An Annulus and a Smooth Rotational Field

Worked Example: An Annulus and a Smooth Rotational Field

Let \(S\) be the annulus \(1\leq x^2+y^2\leq 9\) in the plane \(z=0\), oriented upward. Let \(\mathbf F=\langle -y,x,0\rangle\). Compute the total boundary circulation and verify Stokes’ theorem.

The boundary consists of an outer circle of radius \(3\), oriented counterclockwise, and an inner circle of radius \(1\), oriented clockwise. For a counterclockwise circle of radius \(R\), use \(\mathbf r(t)=\langle R\cos t,R\sin t,0\rangle\), where \(0\leq t\leq 2\pi\). Then \(d\mathbf r=\langle -R\sin t,R\cos t,0\rangle\,dt\) and \(\mathbf F(\mathbf r(t))=\langle -R\sin t,R\cos t,0\rangle\). Therefore,

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =R^2\sin^2t+R^2\cos^2t =R^2, \qquad \oint\mathbf F\cdot d\mathbf r =\int_0^{2\pi}R^2\,dt =2\pi R^2. $$

The outer integral is \(2\pi(3^2)=18\pi\). The same calculation in the clockwise direction changes the sign, so the inner integral is \(-2\pi(1^2)=-2\pi\). The total boundary circulation is \(18\pi-2\pi=16\pi\).

For the surface calculation, \(\nabla\times\mathbf F=\langle 0,0,2\rangle\). The annulus has area \(\pi(3^2-1^2)=8\pi\), so its curl flux is \(2(8\pi)=16\pi\). This agrees with the sum of the two induced boundary integrals.

Worked Example: A Field Defined Only Away from a Puncture

Worked Example: A Field Defined Only Away from a Puncture

Consider the vector field \(\mathbf F=\left\langle -y/(x^2+y^2),x/(x^2+y^2),0\right\rangle\), defined for \(x^2+y^2>0\). Let \(S\) be the annulus \(1\leq x^2+y^2\leq 4\), oriented upward. Compute the boundary circulation and compare it with the curl integral.

On a circle of radius \(R\), parameterized counterclockwise by \(\mathbf r(t)=\langle R\cos t,R\sin t,0\rangle\), the field becomes \(\mathbf F(\mathbf r(t))=\langle-\sin t/R,\cos t/R,0\rangle\). Since \(\mathbf r'(t)=\langle-R\sin t,R\cos t,0\rangle\),

$$ \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) =\sin^2t+\cos^2t =1, \qquad \oint\mathbf F\cdot d\mathbf r =\int_0^{2\pi}1\,dt =2\pi. $$

The outer circle has radius \(2\), so its counterclockwise circulation is \(2\pi\). The inner circle has induced clockwise orientation, so its circulation is \(-2\pi\). Their sum is zero.

At every point where the field is defined, its curl is zero. For example, with \(P=-y/(x^2+y^2)\) and \(Q=x/(x^2+y^2)\),

$$ \frac{\partial Q}{\partial x} =\frac{y^2-x^2}{(x^2+y^2)^2}, \qquad \frac{\partial P}{\partial y} =\frac{y^2-x^2}{(x^2+y^2)^2}, \qquad (\nabla\times\mathbf F)_z =\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0. $$

Thus the curl flux through the annulus is also zero. Stokes’ theorem applies because the annulus does not contain the origin, where the field is undefined. It would not be valid to apply the theorem to the full disk, since the field is not defined throughout a surface containing the origin.

Worked Example: A Square with Two Circular Holes

Worked Example: A Square with Two Circular Holes

Take the square \(-3\leq x\leq 3\), \(-3\leq y\leq 3\), and remove two disjoint circular holes: one of radius \(1\) centered at \((-1.5,0)\), and one of radius \(0.5\) centered at \((1.5,0)\). Orient the remaining planar region upward, and use \(\mathbf F=\langle-y/2,x/2,0\rangle\). Find its total boundary circulation.

First compute the outer square integral directly. On the bottom edge, \(y=-3\), \(x\) increases from \(-3\) to \(3\), and \(F_x=3/2\), giving \((3/2)(6)=9\). On the right edge, \(x=3\), \(y\) increases from \(-3\) to \(3\), and \(F_y=3/2\), giving \(9\). On the top edge, \(y=3\), \(x\) decreases from \(3\) to \(-3\), and \(F_x=-3/2\), giving \((-3/2)(-6)=9\). On the left edge, \(x=-3\), \(y\) decreases from \(3\) to \(-3\), and \(F_y=-3/2\), giving \((-3/2)(-6)=9\). The counterclockwise outer integral is therefore \(9+9+9+9=36\).

For a counterclockwise circle of radius \(r\) centered at \((a,b)\), set \(x=a+r\cos t\) and \(y=b+r\sin t\). Then \(dx=-r\sin t\,dt\) and \(dy=r\cos t\,dt\). Substitution into \(\mathbf F\cdot d\mathbf r=(-y/2)\,dx+(x/2)\,dy\) gives

$$ \mathbf F\cdot d\mathbf r = \frac r2\bigl(b\sin t+a\cos t+r\bigr)\,dt. $$

Over \(0\leq t\leq 2\pi\), the sine and cosine terms integrate to zero, leaving \(\pi r^2\). The induced direction around each hole is clockwise, so the hole contributions are \(-\pi(1^2)=-\pi\) and \(-\pi(0.5^2)=-\pi/4\). The total circulation is \(36-\pi-\pi/4=36-5\pi/4\).

As a check, the curl is \(\langle 0,0,1\rangle\). The remaining area is \(36-\pi-\pi/4\), so the curl flux equals \(36-5\pi/4\), matching the boundary calculation.

A Common Orientation Mistake

A frequent error is to orient every boundary curve counterclockwise when the surface normal points upward. For the annulus in the first example, that would give \(18\pi+2\pi=20\pi\). This is wrong: the inner circle is clockwise as part of the induced boundary, so its contribution is \(-2\pi\), not \(+2\pi\). The correct total is \(18\pi-2\pi=16\pi\), which agrees with the curl flux.

Before evaluating any line integrals, identify the surface normal and list the induced direction of every boundary component. This simple check helps prevent sign errors and clarifies whether a singular point lies on the surface or only inside an excluded hole. Stokes’ theorem requires the field and its first partial derivatives to be defined on an open set containing the surface; a singularity inside a hole is allowed only when it is outside that surface and the theorem’s other conditions hold.

Key takeaway: Stokes’ theorem uses the sum of all boundary integrals with induced orientations. For an upward-oriented planar region, the outer boundary is counterclockwise and every hole boundary is clockwise. A field may be singular inside an excluded hole, but it must be smooth on an open set containing the surface itself.

Check Your Understanding

Use the orientation and hypotheses of Stokes’ theorem to answer these questions.

  1. For a planar annulus with upward normal, what is the induced direction of the outer circle and of the inner circle?
  2. If the surface normal is reversed, what happens to each induced boundary direction?
  3. For a field undefined at the origin, when can Stokes’ theorem still be applied to a planar annulus centered at the origin?
  4. For \(\mathbf F=\langle-y,x,0\rangle\), what is the circulation on a counterclockwise circle of radius \(R\)?
  5. Why is it incorrect to add the circulation of every boundary component using counterclockwise orientation?

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