What is the relationship between the circulation of a vector field along a simple closed curve in three-dimensional space and the flux integral of through a surface with as its boundary?
Why does the flux integral of through a surface with boundary only depend on the boundary of the surface and not the shape of the surfaceโs interior?
When we studied Greenโs Theorem in Section 4.6, we saw how integrating the circulation density over a region in the plane bounded by a simple closed curve gave an alternative way of calculating the circulation along the boundary curve. When we consider simple closed curves in , the situation gets more complicated. However, there is an interesting, and perhaps surprising, generalization of Greenโs Theorem for us to examine.
In this activity, we will look at how we can apply the ideas about circulation along overlapping curves from the beginning of Subsection 4.6.1 to curves in space.
Write out the circulation line integrals of the closed curve consisting of the yellow curve and the and blue curve . (Remember to consider orientation!)
We will now let be the simple closed curve consisting of the yellow and magenta curves in Figure 4.11.1. You can see plotted in red in Figure 4.11.3. The drop-down allows you to select three different surfaces. Visually verify that each of the three surfaces contains . Notice that the scale on the -axis changes as you select different surfaces.
Figure4.11.3.Surfaces containing a common simple closed curve
In Preview Activity 4.11.1, we saw that a simple closed curve in can bound many different surfaces. For now, however, we want to focus on a smooth surface in that has a well-defined normal vector at every point and a boundary curve . We will use the normal vector to define an orientation of so that if a person were to walk along in the direction of the orientation with the top of their head pointing in the direction of , their left arm would be over the surface . Notice that this is the same convention that we used with Greenโs Theorem if we assume that the normal vector being used is .
In Figure 4.11.4, we show the curve from Preview Activity 4.11.1 in magenta as well as a surface that has as its boundary. The chosen normal vector to is shown, as is the orientation of that matches .
Thinking back to Greenโs Theorem, our main idea was that we could calculate the circulation around a simple closed curve in by taking the double integral of the circulation density over the region bounded by the curve. As we saw in Preview Activity 4.11.1, we can break up into line integrals around other simple closed curves so that overlapping portions are oriented oppositely just as we did with the square grid for Greenโs Theorem. To find a three-dimensional analog of Greenโs Theorem, we require that a simple closed curve in three dimensions bound a smooth surface with a normal vector . In doing this, we can choose our โsmallerโ curves similar to the squares we used in Greenโs Theorem to lie on the surface . This gives us almost all the ingredients used in Greenโs Theorem, but we still need to find a suitable replacement for the circulation density.
As we saw in Section 4.5, the curl of a vector field in measures the rotation of the vector field. Theorem 4.5.17 says that for a unit vector , the scalar measures the rotational strength of at the point around the axis defined by . When is the normal vector to the surface at the point , we have the appropriate analog for the circulation density of on at . Thus, the equivalent idea to integrating the circulation density of a two-dimensional vector field over a region in the plane is calculating the flux integral , where on the domain gives a parametrization of the smooth surface .
A rigorous proof of the following theorem is beyond the scope of this text. However, the previous subsection and our discussion of Greenโs Theorem provide an intuitive description of why this theorem is true.
Let be a smooth surface in with a simple closed curve as its boundary. Let be a vector field that is smooth on and . Suppose that on the domain gives a parametrization of for which is oriented so that a person walking along in the direction of its orientation and head pointing in the direction of the normal vector would have their left hand above . In this setting, the circulation of along is equal to the flux of through . That is,
When we looked Greenโs Theorem, it was generally most useful when we were given a line integral and we calculated it using a double integral. In fact, except in the circumstances described in exercises 4.6.6.1 and 4.6.6.3, we did not use Greenโs Theorem to rewrite a double integral as a line integral because of the difficulty of finding a suitable vector field. The situation for Stokesโ Theorem will be similar, with the exception of Exercise 1 in this section. However, Stokesโ Theorem gives us an interesting additional piece of freedom: selecting the surface through which we calculate the flux of from amongst possibly several reasonable surfaces with boundary . The next two activities focus on the relationships between surfaces and their boundary.
Because Stokesโ Theorem requires us to consider a surface (with normal vector) and the boundary of the surface, this activity will give you a chance to practice identifying the boundary of some surfaces in . For each surface below:
Describe the boundary in words.
Find a parametrization for the boundary.
Ensure that a person walking along the boundary in the direction of your parametrization with head pointing in the direction of the surfaceโs normal vector would hold their left hand over the surface.
In some sense, this activity considers the reverse problem of that considered in Activity 4.11.2. Here, each part of the activity gives you an oriented simple closed curve in , and your task is to find
a surface so that is the boundary of and
a normal vector for the so that a person walking along in the direction of the given orientation with head pointing in the direction of your chosen normal vector would have their left hand over .
You are encouraged to think about multiple possible answers, since as we saw in Preview Activity 4.11.1, there may be more than one reasonable choice of a surface with a particular boundary.
We close this section with a pair of activities. The first focuses on calculating both of the integrals in Stokesโ Theorem. The second asks you to calculate some line integrals along simple closed curves and gives you the discretion to choose the best method to use for this (as well as the best surface to use, if you choose Stokesโ Theorem).
Consider the vector field and the curve , which is the triangle with vertices ,, and with orientation corresponding to the order the points are listed here.
Find the circulation of along the curve consisting of (given in order of the orientation) the quarter-circle of radius centered at in the plane from to , the line segment from to , the quarter-circle of radius centered at in the plane from to , and the line segment from to .
In part part b of Activity 4.11.5, there are two โreasonableโ choices for the surface bounded by the circle. If you did not do so while doing the activity, we encourage you to identify both of them and compare which one makes doing the flux integral easier. In general, this will vary depending on the (curl of the) vector field in question, so we cannot give a rule for determining what surface or coordinate system to use. However, we do encourage you to think about which surface will make evaluating the flux integral easiest.
Stokesโ Theorem tells us that we can calculate the circulation of a smooth vector field along a simple closed curve in that bounds a surface (with normal vector) on which the vector field is also smooth by calculating the flux of the curl of the vector field through the surface.
Given two surfaces and with the same boundary (and assuming normal vectors that give the same orientation on ), the flux of through and through is the same because Stokesโ Theorem tells us that this flux is equal to the circulation of along .
Stokesโ Theorem is generally used to turn a line integral into a flux integral. Sometimes it is possible to be given a flux integral and recognize that the given vector field is for some vector field , however. When this is the case, we call a vector potential for the vector field , much like a function so that is called a potential function for .
When finding an anti-derivative of a function of a single variable, you know that there is an infinite family of anti-derivatives, but that any two anti-derivatives differ by a constant. This is why we write expressions such as . A similar phenomenon occurs with (scalar) potential functions for gradient vector fields. Find a second vector field with the same curl as in part a, and do so in a way that is not a constant vector. That is, after simplifying fully, must contain at least one of the variables .
Verify that for the vector fields you found above, is a gradient vector field. Explain why for every pair of vector potentials for a vector field , you must have that is a gradient vector field.