What is the relationship between the circulation of a vector field \(\vF\) along a simple closed curve \(C\) in three-dimensional space and the flux integral of \(\curl(\vF)\) through a surface with \(C\) as its boundary?
Why does the flux integral of \(\curl(\vF)\) through a surface with boundary only depend on the boundary of the surface and not the shape of the surface’s interior?
When we studied Green’s 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 \(\R^3\text{,}\) the situation gets more complicated. However, there is an interesting, and perhaps surprising, generalization of Green’s Theorem for us to examine.
Preview Activity4.11.1.
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.
(a)
For this part, consider the curves in Figure 4.11.1, where the yellow curve is \(Y\text{,}\) the blue curve is \(B\text{,}\) and the magenta curve is \(M\text{.}\)
Figure4.11.1.Three curves in space
(i)
Write out the circulation line integrals of the closed curve consisting of the yellow curve \(Y\) and the and blue curve \(B\text{.}\) (Remember to consider orientation!)
(ii)
Write out the circulation line integrals of the closed curve consisting of the magenta curve \(M\) and the blue curve \(B\text{.}\) (Remember to consider orientation!)
(iii)
Write out the circulation line integrals of the closed curve consisting of the yellow curve \(Y\) and the magenta curve \(M\text{.}\)
(iv)
Why does it make sense that we can add the results of the first two parts and get the third?
(v)
Would your arguments from above change at all if you considered the curves depicted in Figure 4.11.2?
Figure4.11.2.Three slightly different curves in space
(b)
We will now let \(C\) be the simple closed curve consisting of the yellow and magenta curves in Figure 4.11.1. You can see \(C\) 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 \(C\text{.}\) Notice that the scale on the \(z\)-axis changes as you select different surfaces.
Figure4.11.3.Surfaces containing a common simple closed curve
Subsection4.11.1Circulation in three dimensions and Stokes’ Theorem
In Preview Activity 4.11.1, we saw that a simple closed curve in \(\R^3\) can bound many different surfaces. For now, however, we want to focus on a smooth surface \(S\) in \(\R^3\) that has a well-defined normal vector \(\vn\) at every point and a boundary curve \(C\text{.}\) We will use the normal vector to define an orientation of \(C\) so that if a person were to walk along \(C\) in the direction of the orientation with the top of their head pointing in the direction of \(\vn\text{,}\) their left arm would be over the surface \(S\text{.}\) Notice that this is the same convention that we used with Green’s Theorem if we assume that the normal vector being used is \(\vk\text{.}\)
In Figure 4.11.4, we show the curve \(C\) from Preview Activity 4.11.1 in magenta as well as a surface \(S\) that has \(C\) as its boundary. The chosen normal vector \(\vn\) to \(S\) is shown, as is the orientation of \(C\) that matches \(\vn\text{.}\)
Figure4.11.4.A surface \(S\) (and normal vector) bounded by an oriented simple closed curve \(C\)
Thinking back to Green’s Theorem, our main idea was that we could calculate the circulation around a simple closed curve in \(\R^2\) 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 \(\oint_C\vF\cdot d\vr\) into line integrals around other simple closed curves so that overlapping portions are oriented oppositely just as we did with the square grid for Green’s Theorem. To find a three-dimensional analog of Green’s Theorem, we require that a simple closed curve \(C\) in three dimensions bound a smooth surface \(S\) with a normal vector \(\vn\text{.}\) In doing this, we can choose our “smaller” curves similar to the squares we used in Green’s Theorem to lie on the surface \(S\text{.}\) This gives us almost all the ingredients used in Green’s 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 \(\R^3\) measures the rotation of the vector field. Theorem 4.5.17 says that for a unit vector \(\vv\text{,}\) the scalar \((\curl(\vF)(a,b,c))\cdot \vv\) measures the rotational strength of \(\vF\) at the point \((a,b,c)\) around the axis defined by \(\vv\text{.}\) When \(\vv\) is the normal vector to the surface \(S\) at the point \((a,b,c)\text{,}\) we have the appropriate analog for the circulation density of \(\vF\) on \(S\) at \((a,b,c)\text{.}\) 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 \(\iint_D
\curl(\vF)\cdot (\vr_s\times \vr_t)\, dA\text{,}\) where \(\vr(s,t)\) on the domain \(D\) gives a parametrization of the smooth surface \(S\text{.}\)
A rigorous proof of the following theorem is beyond the scope of this text. However, the previous subsection and our discussion of Green’s Theorem provide an intuitive description of why this theorem is true.
Theorem4.11.5.Stokes’ Theorem.
Let \(S\) be a smooth surface in \(\R^3\) with a simple closed curve \(C\) as its boundary. Let \(\vF\) be a vector field that is smooth on \(S\) and \(C\text{.}\) Suppose that \(\vr(s,t)\) on the domain \(D\) gives a parametrization of \(S\) for which \(C\) is oriented so that a person walking along \(C\) in the direction of its orientation and head pointing in the direction of the normal vector \(\vr_s\times \vr_t\) would have their left hand above \(S\text{.}\) In this setting, the circulation of \(\vF\) along \(C\) is equal to the flux of \(\curl(\vF)\) through \(S\text{.}\) That is,
When we looked Green’s 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 Green’s Theorem to rewrite a double integral as a line integral because of the difficulty of finding a suitable vector field. The situation for Stokes’ Theorem will be similar, with the exception of Exercise 1 in this section. However, Stokes’ Theorem gives us an interesting additional piece of freedom: selecting the surface \(S\) through which we calculate the flux of \(\curl(\vF)\) from amongst possibly several reasonable surfaces with boundary \(C\text{.}\) The next two activities focus on the relationships between surfaces and their boundary.
Activity4.11.2.
Because Stokes’ 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 \(\R^3\text{.}\) 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 surface’s normal vector would hold their left hand over the surface.
(a)
The surface \(S\) is the portion of the sphere \(x^2+y^2+z^2=4\) described below. Assume the outward orientation on the sphere.
(i)
The points with \(z\geq 0\text{.}\)
(ii)
The points with \(z\geq x\text{.}\)
(b)
The surface \(S\) is the portion of the hyperbolic paraboloid \(z=x^2-y^2\) with \(x^2+y^2\leq 1\text{.}\) Assume the “upward” orientation, e.g., the normal vector at \((0,0,0)\) is \(\vk\text{.}\)
(c)
The surface \(S\) is the portion of the cylinder \(x^2+y^2=4\) for which \(-2\leq z\leq 2\text{,}\) assuming the outward orientation.
Hint.
It is fine for the boundary of a surface to be made up of more than one curve. Think carefully about how each piece is oriented!
Activity4.11.3.
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 \(C\) in \(\R^3\text{,}\) and your task is to find
a surface \(S\) so that \(C\) is the boundary of \(S\) and
a normal vector for the \(S\) so that a person walking along \(C\) in the direction of the given orientation with head pointing in the direction of your chosen normal vector would have their left hand over \(S\text{.}\)
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.
(a)
The curve \(C\) is the triangle with vertices \((1,0,0)\text{,}\)\((0,1,0)\text{,}\) and \((0,0,1)\) with orientation corresponding to the order the points are listed here.
(b)
The curve \(C\) is the circle parametrized as \(\vr(t) =
\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\) for \(0\leq t\leq 2\pi\text{.}\)
(c)
The curve \(C\) consists of (given in order of the orientation)
the quarter-circle \(C_1\)of radius \(2\) centered at the origin in the \(xy\)-plane from \((2,0,0)\) to \((0,2,0)\text{,}\)
the line segment \(C_2\) from \((0,2,0)\) to \((0,2,2)\text{,}\)
the quarter-circle \(C_3\) of radius \(2\) centered at \((0,0,2)\) in the plane \(z=2\) from \((0,2,2)\) to \((2,0,2)\text{,}\) and
the line segment \(C_4\) from \((2,0,2)\) to \((2,0,0)\text{.}\)
Subsection4.11.2Verifying and Applying Stokes’ Theorem
We close this section with a pair of activities. The first focuses on calculating both of the integrals in Stokes’ 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 Stokes’ Theorem).
Activity4.11.4.
In this activity, we will verify Stokes’ Theorem by calculating both a line integral and a flux integral.
(a)
Consider the vector field \(\vF = \langle x^2 ,y^2 ,z^2 \rangle\) and the circle \(C_1\) parametrized as \(\vr(t) =
\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\rangle\) for \(0\leq t\leq 2\pi\text{.}\)
(i)
Calculate \(\oint_{C_1} \vF\cdot d\vr\) directly using the given parametrization.
(ii)
Let \(S_1\) be the hemisphere of the sphere of radius \(2\) centered at the origin with \(y\leq x\text{.}\) Calculate the flux of \(\curl(\vF)\) through \(S_1\text{.}\)
(iii)
What could you have observed about \(\vF\) that would have gotten you the same answer without doing either of the above calculations?
(b)
Consider the vector field \(\vG = x\vi + y^2z\vj +
x^2\vk\) and the curve \(C_2\text{,}\) which is the triangle with vertices \((1,0,0)\text{,}\)\((0,1,0)\text{,}\) and \((0,0,1)\) with orientation corresponding to the order the points are listed here.
(i)
Find the circulation of \(\vG\) along \(C_2\) by calculating the appropriate line integrals.
(ii)
The vertices of \(C_2\) lie in a plane. Let \(S_2\) be the portion of this plane lying in the first octant, i.e., the portion with \(x,y,z\geq 0\text{.}\) Find the flux of \(\curl(\vG)\) through \(S_2\text{.}\)
(iii)
Explain why the sign of your answer to the previous two parts makes sense.
Activity4.11.5.
(a)
Find the circulation of \(\vF = \langle 3yz, xz,
-xy\rangle\) along the curve \(C\) consisting of (given in order of the orientation) the quarter-circle of radius \(1\) centered at \((0,-2,0)\) in the plane \(y=-2\) from \((0,-2,1)\) to \((1,-2,0)\text{,}\) the line segment from \((1,-2,0)\) to \((1,5,0)\text{,}\) the quarter-circle of radius \(1\) centered at \((0,5,0)\) in the plane \(y=5\) from \((1,5,0)\) to \((0,5,1)\text{,}\) and the line segment from \((0,5,1)\) to \((0,-2,1)\text{.}\)
(b)
Find the circulation of \(\vG = 3z^2\vi -z^2 + 2x\vj+zy\vk\) along the circle in the \(xy\)-plane of radius \(3\) centered at the origin. Assume the counterclockwise orientation of the circle.
In part part b of Activity 4.11.5, there are two “reasonable” 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.
Subsection4.11.3Summary
Stokes’ Theorem tells us that we can calculate the circulation of a smooth vector field along a simple closed curve in \(\R^3\) 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 \(S_1\) and \(S_2\) with the same boundary \(C\) (and assuming normal vectors that give the same orientation on \(C\)), the flux of \(\curl(\vF)\) through \(S_1\) and through \(S_2\) is the same because Stokes’ Theorem tells us that this flux is equal to the circulation of \(\vF\) along \(C\text{.}\)
Exercises4.11.4Exercises
1.
Stokes’ 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 \(\vF\) is \(\curl(\vG)\) for some vector field \(\vG\text{,}\) however. When this is the case, we call \(\vG\) a vector potential for the vector field \(\vF\text{,}\) much like a function \(f\) so that \(\vF =
\grad(f)\) is called a potential function for \(\vF\text{.}\)
(a)
Find a vector field \(\vF\) so that \(\curl(\vF) =
\langle x-y^2,4xy-y-4,-4xz\rangle\text{.}\)
Hint.
Your experience in finding potential functions for gradient vector fields will be useful to you here, although you will have more flexibility.
(b)
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 \(\displaystyle\int\cos(x)\, dx = \sin(x) + C\text{.}\) A similar phenomenon occurs with (scalar) potential functions for gradient vector fields. Find a second vector field \(\vG\) with the same curl as in part a, and do so in a way that \(\vF-\vG\) is not a constant vector. That is, after simplifying fully, \(\vF-\vG\) must contain at least one of the variables \(x,y,z\text{.}\)
(c)
Verify that for the vector fields you found above, \(\vF-\vG\) is a gradient vector field. Explain why for every pair \(\vF, \vG\) of vector potentials for a vector field \(\vH\text{,}\) you must have that \(\vF-\vG\) is a gradient vector field.
(d)
Explain why if \(\vH\) is a vector field with a vector potential \(\vF\text{,}\)\(\divg(\vH) = 0\text{.}\) Such a vector field is called a solenoidal vector field or divergence-free vector field.
