Flux and Circulation

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Flux and Circulation

Based on “Flux” and “Circulation” in section 5.2 of the textbook.

Key Ideas

Flux and circulation are, respectively, integrals of components of a vector field perpendicular to, and tangent to, a curve.

Both have formulas for computing them, differing only in whether they involve a dot product with the curve’s tangent vector or its normal vector.

There are proofs for both computational formulas.

The orientation of the curve (i.e., the direction you move along it as the parameter value increases) matters. Changing orientation changes the signs of the integrals.

Curves may be closed (i.e., start and end at the same place), particularly with circulation integrals.

Examples

One of the things I sometimes do in my spare time is brew beer. One of the steps in brewing is to steep bags of grain in hot water to give the beer its flavor. I do this by swirling the bag around a kettle of water; relative to the bag, the water is moving in a circular flow. In 2 dimensions, the edge of the bag is a circle of radius 1, centered at (let’s say) 2 inches from the center of the kettle. The water’s velocity relative to the bag can be modeled by the vector field F(x,y) = ⟨ -y, x ⟩:

What’s the flux of water (i.e., amount of water per unit of time) crossing the edge of the bag?

Start by coming up with a parametric description of the bag. This is mostly the parametric form for a circle of radius 1, except that it has to be translated to have its center at (2,0):

Now you can find the derivative of r(t), use it to calculate the normal vector n(t), and plug into the calculation formula for a flux line integral:

So the flux is 0. That’s disappointing, it seems to say no water gets into the grain to flavor the beer? Not really, it could just be saying that all the water that goes into the grain eventually comes out again. To check this, lets calculate flux across just the edge of the bag that faces the flow, i.e., the edge between t = π and t = 2π. We’ve already done most of the work for this calculation, we just have to evaluate the integral between different bounds:

Now we have a negative amount of flux. The negative sign simply means the flow is directed opposite to the direction of the normal vectors, and in this problem the normals point out of the bag. So negative flux is what we’d expect from water flowing in.

What’s the circulation of water around the bag?

This requires integrating the field dotted with the tangent to the curve, which is the original sort of line integral we did in vector fields. Once again, we’ve already done a lot of the work for solving this problem:

Next

Line integrals in conservative vector fields.

Please read “Curves and Regions,” “Fundamental Theorem for Line Integrals,” and “Conservative Vector Fields and Potential Functions” in section 5.3 on the textbook.

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