Stokes’ Theorem

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Circulation from local rotation

For an oriented piecewise smooth surface $S$ with boundary $C$ , and a continuously differentiable field on a neighborhood of $S$ ,

\oint_C\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS.

The right-hand rule links the two orientations: looking from the tip of the chosen normal toward the surface, the positive boundary runs counterclockwise. For holes, the induced direction on an inner boundary is reversed.

Visual guide

VISUAL GUIDEBoundary orientation agrees with the normal

Viewed from above, the disk boundary is counterclockwise and its normal points toward you. For F = (−y/2, x/2, 0), curl is a uniform upward unit vector. Circulation around a radius-2 boundary therefore equals disk area 4π.

Choose a convenient spanning surface

When several surfaces share the same oriented boundary and the field is smooth near them, each gives the same curl flux. A flat disk may be easier than a curved cap. This statement concerns flux of curl, not arbitrary flux of the original field.

Worked example: a circular boundary

For $\mathbf F=(-y/2,x/2,0)$ , curl is $(0,0,1)$ . Let $C$ be the radius- $2$ circle in $z=3$ , counterclockwise from above. Choose its flat disk with upward normal. Curl flux equals disk area, $4\pi$ , so circulation is $4\pi$ . Translating the disk vertically does not alter this field or the result.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Stokes’ theorem relates circulation around a boundary to…

Flux of curl through a spanning surface

Flux of divergence through a surface

Volume integral of the vector field

Hint 1 · Find a starting point

Curl describes local circulation.

Hint 2 · Take the next step

Match the curve direction and surface normal using the right-hand rule.

Show the reasoning

Answer: Flux of curl through a spanning surface

∮F·dr=∫∫(∇×F)·n dS for a compatible orientation and the theorem’s regularity assumptions.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a triangular surface

Let $\mathbf F=(0,x,0)$ , again with curl $(0,0,1)$ . The triangle $x+y+z=1$ in the first octant has upward vector area $(-z_x,-z_y,1)dx\,dy=(1,1,1)dx\,dy$ . Its projection is a right triangle of area $1/2$ , so circulation around its induced boundary is $1/2$ .

Avoiding theorem mix-ups

Stokes relates a line integral to a surface integral of curl. The divergence theorem relates a closed-surface integral of the field to a volume integral of divergence. Green’s circulation theorem is the planar case of Stokes. Write the type of boundary and integral before choosing one.

Practice

Reverse the orientation of the radius- $2$ circle above.
Find circulation of a smooth gradient around a closed loop bounding a suitable surface.
Use Stokes for $(-y,x,0)$ on the unit circle counterclockwise.

Show worked solutions

The answer becomes $-4\pi$ ; the compatible normal is downward.
Curl of a smooth gradient is zero, giving zero circulation.
Curl is $(0,0,2)$ , so the answer is $2\pi$ .

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

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