The circulation density or curl of a two dimensional field F = M i + N j at a point (x, y) is described by the scalar quantity

In three-dimensions, the circulation around a point P in a plane is described by a vector. This vector is normal to the plane of circulation and points in the direction that gives it a right-hand relation to the circulation line. Also, the length of this vector gives the rate of the fluid's rotation. In fact, the vector of greatest circulation in a flow with velocity field F = M i + N j + P k is

The equation for the curl F is usually written using the symbolic operator del

Stoke's theorem states that, under conditions normally met in practice, the circulation of a vector field around the boundary of an oriented surface in space in the directions counterclockwise with respect to the surface's unit normal vector field n equals the integral of the normal component of the curl of the field over the surface.

The circulation of F = M i + N j + P k around the boundary of C of an oriented surface S in the direction counterclockwise with respect to the surface's unit normal vector n equals the integral of Ñ ´ F · n over S.

NOTE:

If two different oriented surfaces S₁ and S₂ have the same boundary D, then their curl integrals are equal:

NOTE:

If C is a curve in the xy-plane, oriented counterclockwise, and R is the region in the xy-plane bounded by C, then ds = dx dy an

and Stoke's theorem becomes

Notice that this is the circulation-curl form of Green's theorem.

EXAMPLE 1:

Calculate the circulation of the field F = x ² i + 2x j + z ² k around the curve C: the ellipse 4x ² + y ² = 4 in the xy-plane, counterclockwise when viewed from above.

SOLUTION:

Since it is in the xy-plane, then n = k and (Ñ ´ F) · n = 2.

We are working with the ellipse 4x ² + y ² = 4 or x ² + y ²/4 = 1, so I will use the transformation x = r cos q and y = 2r sin q to transform this ellipse into a circle. I will also have to use the Jacobian to find the integrating factor for this integral.

EXAMPLE 2:

Calculate the circulation of the field F = y i + xz j + x ² k around the curve C: the boundary of the triangle cut from the plane x + y + z = 1 by the first octant, counterclockwise when viewed from above.

SOLUTION: Using the shortcut formula

where M = y, N = xz, and P = x ², I will find Ñ ´ F.

Ñ ´ F = (0 - x) i + (0 - 2x) j + (z - 1) k = -x i -2x j + (z - 1) k

The triangle that we are looking at from above is in the plane x + y + z = 1, and the vector perpendicular to the plane is p = i + j + k.

Let f = x + y + z - 1, and since the shadow is in the xy-plane, let p = k.

When z = 0, then x + y = 1 or y = 1 - x. When y = 0, then x = 1. Finally, we have to get rid of the z in the integrand, so solve x + y + z = 1 for z. z = 1 - x- y

EXAMPLE 3:

Use the surface integral in Stoke's theorem to calculate the flux of the curl of the field F = 2z i + 3x j + 5y k across the surface r (r, q ) = (r cos q ) i + (r sin q ) j + (4 - r ²) k, 0 £ r £ 2, 0 £ q £ 2p in the direction of the outward unit normal n.

SOLUTION: Before we start to solve this problem, we need a fact from integration of parametric surfaces, and here is the fact.

FACT:

Now apply this to Ñ ´ F · n ds .

EXAMPLE 4:

Use the surface integral in Stoke's theorem to calculate the flux of the curl of the field F = 2y i + (5 - 2x) j + (z ² - 2) k across the surface r (j , q ) = (2sin j cos q ) i + (2sin j sin q ) j + (2cos j ) k, 0 £ j £ p /2, 0 £ q £ 2p in the direction of the outward unit normal n.

SOLUTION: Apply the fact stated in the previous example to do this problem.

Stoke's theorem can be extended to an oriented surface S that has one or more holes in a way that is analogous to the extensions of Green's theorem. Here is how it can be extended.

The surface integral over S of the normal component Ñ ´ F equals the sum of the line integrals around all of the boundary curves of the tangential component of F where the curves are to be traced in the direction induced by the orientation of S.

FACT:

curl grad f = 0 or Ñ ´ Ñ f = 0.

Recall that the field F is conservative in an open region D in space is equivalent to saying that the integral of F around every closed loop in D is zero. This is also equivalent in simply connected open regions to saying Ñ ´ F = 0. But what does it mean to be simply connected?

DEFINITION:

A region D is simply connected if every closed path in D can be contracted to a point in D without leaving D. (See figure 1)

figure 1

THEOREM:

If Ñ ´ F = 0 at every point of a simply connected open region D in space, then on any piecewise smooth closed path C in D,