Vector calculus. Stokes' theorem. Circulation

Statement

The circulation of a field around a closed contour equals the flux of its curl through any surface spanning that contour:

Γ \oint a \cdot d r = S \iint rot a \cdot d S,

where

rot a = \nabla \times a = i \frac{\partial}{\partial x} P j \frac{\partial}{\partial y} Q k \frac{\partial}{\partial z} R .

The orientations of the contour and the surface must be matched: seen from the tip of the normal, $Γ$ is traversed anticlockwise (the right-hand rule). A mismatch gives the answer with the opposite sign.

Green's theorem is a special case: a plane domain, a field with no $z$ -component, the normal along $O z$ .

What it is for

Any surface will do — and that is what gets used. A contour of three segments gives three line integrals of the second kind; the flat triangle spanning it gives one surface integral, often of a constant. The reverse also pays off: if the curl is complicated and the contour is a circle, it is easier to compute the circulation directly.

The steps

Find $rot a$ from the determinant.
Choose a surface spanning the contour — usually a flat one.
Write out the normal consistently with the direction of traversal.
Compute the flux of the curl.

If a problem asks for it "two ways", both answers must agree; that is the built-in check.

Common mistakes

The sign of the middle component of the curl. In the determinant it is taken with a minus: $(\partial P / \partial z - \partial R / \partial x)$ , not the other way round.
A normal inconsistent with the traversal — the answer is right in absolute value and wrong in sign.
Forgetting that the normal to an inclined plane is not a unit vector. If $N = (A, B, C)$ is taken, the integration is over the projection with $d x d y$ ; if $\hat{n}$ is taken, it is over the area with $d S$ . The two must not be mixed.

Problems

Problem 1 (Circulation around a circle, two ways). Given the vector field $a = z i + x^{2} j + y k$ and the closed contour $Γ$ : $x = 2 cos t$ , $y = 2 sin t$ , $z = 0$ , $t \in [0; 2 π]$ .

Find the circulation of the field $a$ along $Γ$ in two ways: directly and by Stokes' theorem.

Solution.

Method 1: a line integral of the second kind

The circulation is

Γ \oint a \cdot d r = Γ \oint P d x + Q d y + R d z,

where $P = z$ , $Q = x^{2}$ , $R = y$ .

From the parametrisation: $d x = - 2 sin t d t$ , $d y = 2 cos t d t$ , $d z = 0$ .

On the contour $z \equiv 0$ , so the term $P d x$ vanishes; the term $R d z$ disappears with $d z$ . Only $Q d y$ remains:

Γ \oint a \cdot d r = \int_{0}^{2 π} (2 cos t)^{2} \cdot 2 cos t d t = 8 \int_{0}^{2 π} cos^{3} t d t .

The integral of an odd power of the cosine over a full period is zero. Formally: $cos^{3} t = \frac{3 cos t + cos 3 t}{4}$ , and both terms integrate to zero on $[0; 2 π]$ .

Γ \oint a \cdot d r = 0.

Method 2: Stokes' theorem

The curl.

rot a = (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) i + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) j + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) k .

Compute: $\partial R / \partial y = 1$ , $\partial Q / \partial z = 0$ ; $\partial P / \partial z = 1$ , $\partial R / \partial x = 0$ ; $\partial Q / \partial x = 2 x$ , $\partial P / \partial y = 0$ . So $rot a = i + j + 2 x k$ .

The middle component is non-zero here, and this is the case where it is easy to lose: in the determinant it is taken with a minus sign, that is, it equals $\partial P / \partial z - \partial R / \partial x$ , not the other way round.

The surface. Span $Γ$ with the disc $S$ of radius $2$ in the plane $z = 0$ . The traversal with $t$ from $0$ to $2 π$ is anticlockwise, so the matching normal is $n = k$ .

The flux of the curl. The dot product picks out the third component only: $rot a \cdot n = 2 x$ .

S \iint 2 x d S = \int_{0}^{2 π} \int_{0}^{2} 2 r cos θ \cdot r d r d θ = 2 \int_{0}^{2} r^{2} d r \int_{0}^{2 π} cos θ d θ = 0,

because $\int_{0}^{2 π} cos θ d θ = 0$ . The same conclusion without computation: the disc is symmetric about the $O y$ axis, and the function $2 x$ is odd in $x$ .

Both methods gave the same result, which is the check.

Answer: the circulation is $0$ .

Problem 2 (Circulation around a triangular contour). The vector field $a = y i + z j + x k$ is given.

The contour $Γ$ is the triangle cut out of the first octant by the plane $x + y + z = 1$ , with vertices $A (1; 0; 0)$ , $B (0; 1; 0)$ , $C (0; 0; 1)$ , traversed in the order $A \to B \to C \to A$ .

Find the circulation of the field $a$ along $Γ$ .

Solution.

Direct computation would take three integrals along the three sides. Stokes' theorem reduces the problem to one flux through the triangle $S$ spanning the contour — and here the curl turns out to be constant, so the integral can be done in one's head.

The curl. For $P = y$ , $Q = z$ , $R = x$ :

rot a = (\frac{\partial x}{\partial y} - \frac{\partial z}{\partial z}) i + (\frac{\partial y}{\partial z} - \frac{\partial x}{\partial x}) j + (\frac{\partial z}{\partial x} - \frac{\partial y}{\partial y}) k = - i - j - k .

Orientation. Seen from the side of positive coordinates, the traversal $A \to B \to C \to A$ is anticlockwise, so the matching normal points "outwards", away from the origin:

\hat{n} = \frac{( 1 ; 1 ; 1 )}{3} .

This can be checked with the right-hand rule: going from $A$ to $B$ and then to $C$ , the thumb of the right hand points towards $(1; 1; 1)$ .

The flux of the curl.

rot a \cdot \hat{n} = \frac{- 1 - 1 - 1}{3} = - 3 .

The quantity is constant, so the flux is its product with the area:

Γ \oint a \cdot d r = - 3 \cdot S (△ A B C) .

The triangle is equilateral with side $∣ A B ∣ = 2$ , so

S = \frac{3}{4} \cdot (2)^{2} = \frac{3}{2} .

Γ \oint a \cdot d r = - 3 \cdot \frac{3}{2} = - \frac{3}{2} .

A check by direct computation. On the side $A B$ : $r (t) = (1 - t; t; 0)$ , $d r = (- 1; 1; 0) d t$ , the field is $a = (t; 0; 1 - t)$ , the dot product is $- t$ , and the integral is $- 1/2$ . By the symmetry of the field under the cyclic substitution $x \to y \to z \to x$ , the other two sides give the same, $- 3/2$ in total. It agrees.

Answer: $- \frac{3}{2}$ .

Problem 3 (Circulation around a triangular contour (variant)). The vector field $a = y i + z j + x k$ is given.

The contour $Γ$ is the triangle cut out of the first octant by the plane $x + y + z = 2$ , with vertices $A (2; 0; 0)$ , $B (0; 2; 0)$ , $C (0; 0; 2)$ , traversed in the order $A \to B \to C \to A$ .

Find the circulation of the field $a$ along $Γ$ .

Solution.

The field and the direction of the normal are the same as in the previous problem; only the size of the triangle has changed. The curl does not depend on it:

rot a = - i - j - k, \hat{n} = \frac{( 1 ; 1 ; 1 )}{3}, rot a \cdot \hat{n} = - 3 .

The area changes. The side is now $∣ A B ∣ = 22$ , so

S = \frac{3}{4} \cdot (22)^{2} = \frac{3}{4} \cdot 8 = 23 .

Γ \oint a \cdot d r = - 3 \cdot 23 = - 6.

The answer grew fourfold, not twofold: the contour was stretched by a factor of two along each axis, and area is a quadratic quantity. Here the curl is constant, so the scaling is predictable; with a variable curl such an estimate cannot be trusted, and the integral must be computed again.

A check by direct computation. On the side $A B$ : $r (t) = (2 - 2 t; 2 t; 0)$ , $d r = (- 2; 2; 0) d t$ , the field is $a = (2 t; 0; 2 - 2 t)$ , the dot product is $- 4 t$ , and the integral is $- 2$ . By symmetry the three sides give $- 6$ .

Answer: $- 6$ .

Statement

What it is for

The steps

Common mistakes

Problems

stokes.ipynb