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Stokes’ Theorem · Tutorial 925 of 999

Stokes' Theorem as Circulation Equals Curl Flux

Learn how Stokes’ theorem equates a boundary circulation with the flux of curl through an oriented surface, and how to identify each quantity correctly.

Advanced 10 min read

What You'll Learn

  • State Stokes’ theorem and its smoothness and orientation requirements.
  • Distinguish the line integral of a field around a boundary from the flux of its curl through a surface.
  • Compute curl and identify the relevant normal component in concrete examples.
  • Verify Stokes’ theorem by calculating both integrals.
  • Explain, using a small rectangle, why curl measures local circulation per unit area.

From Boundary Motion to Surface Rotation

A vector field assigns a vector to each point in space. If a particle travels along a closed curve in that field, the line integral of the field along the curve measures its circulation. The field may also rotate locally at points across a surface. Curl describes that local rotation, and its component normal to the surface contributes to the flux of curl through the surface.

Stokes’ theorem connects these two measurements: the total circulation around the boundary equals the total flux of curl through the oriented surface. The theorem is useful because it relates an integral along a curve to an integral across a surface, while keeping track of orientation and the required smoothness.

Central idea: Stokes’ theorem says that circulation around a positively oriented boundary is the flux of curl through the surface that boundary encloses.

The Three Quantities in the Theorem

Let \(\mathbf{F}=(P,Q,R)\) be a vector field in three-dimensional space, and let \(S\) be an oriented surface. The orientation specifies a consistent choice of unit normal vector \(\mathbf{n}\) on the surface. The boundary curve is denoted by \(C=\partial S\).

The circulation integral of \(\mathbf{F}\) around an oriented, piecewise-smooth curve \(C\) is the line integral

$$ \oint_C \mathbf{F}\cdot d\mathbf{r}. $$

Here \(d\mathbf{r}\) is the vector displacement along the curve. The dot product means that only the component of the field in the direction of travel contributes. Reversing the direction of travel changes the sign of the integral.

In three dimensions, the curl of \(\mathbf{F}\) is the vector field

$$ \nabla\times\mathbf{F} = \left( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right). $$

The flux integrand for Stokes’ theorem is the component of this curl in the chosen unit-normal direction: \((\nabla\times\mathbf{F})\cdot\mathbf{n}\). The curl flux through \(S\) is the surface integral of that scalar quantity.

Definition: The circulation of \(\mathbf{F}\) around \(C\) is \(\oint_C\mathbf{F}\cdot d\mathbf{r}\). The flux of curl through an oriented surface \(S\) is \(\iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS\), where \(\mathbf{n}\) is the chosen unit normal and \(dS\) is the surface-area element.

Stokes’ Theorem

The theorem requires a smooth oriented surface \(S\), with a positively oriented, piecewise-smooth boundary \(C=\partial S\). The vector field \(\mathbf{F}\) must have continuous first partial derivatives on an open region containing \(S\), so its curl is defined and continuous there. The direction of \(C\) must be the one induced by the orientation of \(S\); the right-hand rule expresses this compatibility.

Theorem: If \(\mathbf{F}\) has continuous first partial derivatives on an open region containing a smooth oriented surface \(S\), and \(C=\partial S\) is given its positive orientation, then $$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS. $$

The two sides are different kinds of integrals. The left side follows the boundary curve and uses the component of \(\mathbf{F}\) tangent to that curve. The right side covers the surface and uses the component of \(\nabla\times\mathbf{F}\) normal to it. Stokes’ theorem asserts that, under its hypotheses and with compatible orientations, these totals agree.

Why Curl Measures Circulation

A small planar rectangle gives a direct explanation of the curl component normal to a surface. Consider a rectangle in the \(xy\)-plane with corners \((x,y)\), \((x+\Delta x,y)\), \((x+\Delta x,y+\Delta y)\), and \((x,y+\Delta y)\), traversed counterclockwise as viewed from above. Write \(\mathbf{F}=(P,Q,R)\). Because the displacement is horizontal, only \(P\) and \(Q\) contribute to the circulation around this rectangle.

Adding the contributions from the four edges gives an exact identity. The bottom and top edges contribute the difference in \(P\), while the right and left edges contribute the difference in \(Q\):

$$ \begin{aligned} \oint_{\partial R}\mathbf{F}\cdot d\mathbf{r} &= \int_x^{x+\Delta x}P(s,y)\,ds -\int_x^{x+\Delta x}P(s,y+\Delta y)\,ds\\ &\quad+\int_y^{y+\Delta y}Q(x+\Delta x,t)\,dt -\int_y^{y+\Delta y}Q(x,t)\,dt. \end{aligned} $$

Apply the one-variable fundamental theorem of calculus to each difference. Assuming the partial derivatives are continuous, the result is

$$ \oint_{\partial R}\mathbf{F}\cdot d\mathbf{r} = \int_x^{x+\Delta x}\int_y^{y+\Delta y} \left( \frac{\partial Q}{\partial x}(s,t) - \frac{\partial P}{\partial y}(s,t) \right)\,dt\,ds. $$

The integrand is the \(z\)-component of curl. Dividing by the rectangle’s area \(\Delta x\,\Delta y\) shows that the circulation per unit area is the average value of that curl component over the rectangle. As the rectangle shrinks to a point, continuity makes this average approach the curl component at that point. Thus \(\partial Q/\partial x-\partial P/\partial y\) measures local circulation per unit area for a small horizontal loop. The surface flux in Stokes’ theorem adds the corresponding normal component over the whole oriented surface.

Worked Examples

Worked Example: Circulation Around a Circle

Let \(\mathbf{F}=(-y/2,x/2,0)\). Take \(S\) to be the disk \(x^2+y^2\leq 4\) in the \(xy\)-plane, with upward normal, and let \(C\) be its positively oriented boundary circle.

First compute the curl. Here \(P=-y/2\), \(Q=x/2\), and \(R=0\), so

$$ \nabla\times\mathbf{F} = \left(0,0,\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) = \left(0,0,\frac12-\left(-\frac12\right)\right) = (0,0,1). $$

The upward unit normal is \(\mathbf{n}=(0,0,1)\). Therefore the flux integrand is \((\nabla\times\mathbf{F})\cdot\mathbf{n}=1\), and the disk has area \(\pi(2)^2=4\pi\). The curl flux is \(1\cdot4\pi=4\pi\).

For the circulation, parametrize the positively oriented circle by \(\mathbf{r}(t)=(2\cos t,2\sin t,0)\), for \(0\leq t\leq2\pi\). Then \(\mathbf{r}'(t)=(-2\sin t,2\cos t,0)\) and \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t,0)\). Hence

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \left[(-\sin t)(-2\sin t)+(\cos t)(2\cos t)\right]dt = \int_0^{2\pi}2\,dt = 4\pi. $$

Both integrals equal \(4\pi\), as Stokes’ theorem requires.

Worked Example: A Rectangular Surface

Let \(\mathbf{F}=(-y,x,0)\), and take \(S\) to be the rectangle \(0\leq x\leq2,\ 0\leq y\leq1\) in the \(xy\)-plane with upward normal. Its positively oriented boundary runs counterclockwise when viewed from above.

The curl is \(\nabla\times\mathbf{F}=(0,0,2)\). Since the unit normal is \((0,0,1)\), the flux integrand equals \(2\). The rectangle has area \(2\cdot1=2\), so the curl flux is \(2\cdot2=4\).

To check the circulation, sum the integrals over the four edges. On the bottom edge, \(y=0\), \(x\) runs from \(0\) to \(2\), and \(d\mathbf{r}=(dx,0,0)\). The field is \((0,x,0)\), so the dot product is \(0\), giving an integral of \(0\). On the right edge, \(x=2\), \(y\) runs from \(0\) to \(1\), and \(d\mathbf{r}=(0,dy,0)\). The field is \((-y,2,0)\), so the contribution is \(\int_0^1 2\,dy=2\).

On the top edge, \(y=1\), \(x\) runs from \(2\) to \(0\). Here \(d\mathbf{r}=(dx,0,0)\) and the field is \((-1,x,0)\), so the contribution is \(\int_2^0(-1)\,dx=2\). On the left edge, \(x=0\), \(y\) runs from \(1\) to \(0\); the field is \((-y,0,0)\) and \(d\mathbf{r}=(0,dy,0)\), so this contribution is \(0\). The total circulation is \(0+2+2+0=4\), matching the curl flux.

Worked Example: A Surface with Normal in the \(y\)-Direction

Let \(\mathbf{F}=(0,0,x)\). Take \(S\) to be the square \(y=0,\ 0\leq x\leq1,\ 0\leq z\leq1\), oriented with normal \(\mathbf{n}=(0,1,0)\). Use the boundary direction that starts at \((0,0,0)\), proceeds to \((0,0,1)\), then to \((1,0,1)\), then to \((1,0,0)\), and returns to its starting point.

The curl is

$$ \nabla\times\mathbf{F} = \left( 0,\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, 0 \right) = (0,0-1,0) = (0,-1,0). $$

Thus the flux integrand is \((0,-1,0)\cdot(0,1,0)=-1\). The square has area \(1\), so the curl flux is \(-1\).

For the circulation, the integrand is \(\mathbf{F}\cdot d\mathbf{r}=x\,dz\), because only the \(z\)-component of the field is nonzero. Along the first edge, \(x=0\), so the integral is \(0\). Along the second edge, \(z=1\) is constant, so \(dz=0\) and the integral is \(0\). Along the third edge, \(x=1\) and \(z\) decreases from \(1\) to \(0\), giving \(\int_1^0 1\,dz=-1\). Along the final edge, \(x=0\), so the integral is \(0\). The total is \(-1\), equal to the curl flux.

A Common Sign Mistake

In the square example, the magnitude of the curl is \(1\), and the surface area is \(1\). Multiplying these gives \(1\), but that is not the flux: it ignores the direction of curl relative to the oriented surface. The correct integrand is the dot product with the chosen normal, \((0,-1,0)\cdot(0,1,0)=-1\), so the flux is \(-1\). A surface integral of curl uses the signed normal component, not the magnitude of curl.

When the Theorem Applies

The equality is not a substitute for checking hypotheses. The surface must have an orientation, its boundary must be piecewise smooth and positively oriented with respect to that surface, and the field must have continuous first partial derivatives on an open region containing the surface. In particular, the field cannot have an undefined or singular point on the surface while still satisfying these assumptions.

These conditions explain why both sides are well-defined and why the orientations matter. Reversing the boundary direction reverses the circulation’s sign. Reversing the surface normal reverses the sign of the curl flux. The theorem equates the signed quantities when the boundary and surface orientations are compatible.

Why the Connection Matters

A circulation integral can be difficult to evaluate directly along a complicated curve. Stokes’ theorem shows that the same quantity can be understood as curl flux over a surface with that curve as its positively oriented boundary. Conversely, a surface curl integral is tied to the circulation around its edge. The theorem is therefore more than a computational shortcut: it reveals that total boundary circulation records the accumulated local rotation across the surface.

Check Your Understanding

For each question, identify the relevant quantity before attempting a calculation.

  1. In Stokes’ theorem, what does the line integral measure, and what does the surface integral measure?
  2. What regularity condition on \(\mathbf{F}\) is stated in the theorem?
  3. For \(\mathbf{F}=(0,0,2x)\), what is \(\nabla\times\mathbf{F}\), and what is its dot product with the upward unit normal?
  4. Why is the magnitude of curl not generally the correct flux integrand?
  5. If the direction of the boundary curve is reversed while the surface orientation is fixed, what happens to the circulation’s sign?

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