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Active Calculus - Multivariable

Steve Schlicker, Mitchel T. Keller, Nicholas Long

Section 12.11 Stokes’ Theorem

When we studied Green’s Theorem in Section 12.8, we saw how integrating the circulation density over a region in the plane bounded by a simple closed curve is equivalent to 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 Activity 12.11.1.

In this activity, we will look at how we can apply the ideas about circulation along overlapping curves from the beginning of Subsection 12.8.1 to curves in space.

(a)

For this part, consider the curves in Figure 12.11.1, where the yellow curve is \(Y\text{,}\) the blue curve is \(B\text{,}\) and the magenta curve is \(M\text{.}\)
Figure 12.11.1. Three curves in space
You should also go back and refamiliarize yourself with our notation for combining paths (as used in line integrals) from Convention 12.2.13. In our convention, \(Y+M\) would be closed loop, but \(Y-M\) would not make sense because the segment \(-M\) does not begin where \(Y\) begins.
(i)
Using the three segments in Figure 12.11.1, write out at least four different closed curves in terms of \(B\text{,}\) \(Y\text{,}\) and \(M\text{.}\) (Remember to consider orientation!)
(ii)
Let \(C_1=M+B\) and \(C_2=Y-B\text{.}\) Describe the curve given by \(C_1+C_2\text{.}\)
(iii)
Write a couple of sentences explaining how the circulation around \(C_1+C_2\) would compare to the circulation around \(C_1\) and the circulation around \(C_2\text{.}\) Write an equation in terms of \(\int_{C_1} \vF \cdot d\vr \text{,}\) \(\int_{C_2} \vF \cdot d\vr \text{,}\) and \(\int_{C_1+C_2} \vF \cdot d\vr \text{.}\)
(iv)
Explain how your arguments or equations from any of the parts above would or would not change if you considered the curves depicted in Figure 12.11.2.
Figure 12.11.2. Three slightly different curves in space

(b)

Let \(C\) be the simple closed curve consisting of the yellow and magenta curves in Figure 12.11.1. You can see \(C\) plotted in red in Figure 12.11.3. The drop-down allows you to select three different surfaces. You can 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.
Figure 12.11.3. Surfaces containing a common simple closed curve
The simple closed curve consisting of the yellow and magenta curves in Figure 12.11.1 can be parameterized by \(\langle \cos(t), \sin(t), \cos(2t)\rangle \) with \(0\leq t\leq 2 \pi\text{.}\) Let \(C_3 = Y+M\text{.}\) Use the given parameterization of \(C_3\) to show that \(C_3\) is on each of the following surfaces:
  • \(\displaystyle x^2-y^2=z\)
  • \(\displaystyle z=x^4-y^4\)
  • \(\displaystyle z=1-2y^2\)
  • \(\displaystyle z=-\cos(\pi \sqrt{x^2+y^2})(x^2-y^2)\)

Subsection 12.11.1 Circulation in three dimensions and Stokes’ Theorem

In part b of Preview Activity 12.11.1, 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 12.11.4, we show the curve \(C\) from Preview Activity 12.11.1 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 \(\vn\text{.}\)
Figure 12.11.4. 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 12.11.1, 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 12.7, the curl of a vector field in \(\R^3\) measures the rotation of the vector field. Theorem 12.7.18 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 \(D\) that gives a parameterization of the smooth surface \(S\text{.}\)
A rigorous proof of the following theorem is beyond the scope of this text. However, part a of Preview Activity 12.11.1 and our discussion of Green’s Theorem provide an intuitive description of why this theorem is true.

Subsection 12.11.2 Verifying and Applying Stokes’ Theorem

In this subsection, we will look at some examples and activities that will verify Stokes’ Theorem by calculation both side for a few different situations.

Example 12.11.6.

In this example, we will verify Stoke’s Theorem for the curve used in Preview Activity 12.11.1 using the parameterization and surfaces from part b of Preview Activity 12.11.1. We will use the vector field \(\vF=\langle z-x, y+z, xy \rangle \) throughout this problem.
(a)
We first calculate the circulation of \(\vF=\langle z-x, y+z, xy \rangle \) around the curve \(C\) with parameterization given by \(\vr(t)= \langle \cos(t), \sin(t), \cos(2t)\rangle \) with \(0\leq t\leq 2 \pi\text{.}\) Note here that \(\vr'(t)= \langle -\sin(t), \cos(t), -2 \sin(2t) \rangle \text{.}\) Applying Theorem 12.3.6 to calculate the circulation, we have
\begin{align*} \int_C \vF \cdot d\vr = \amp \int_0^{2\pi} \langle \cos(2t)-\cos(t),\sin(t)+\cos(2t),\cos(t)\sin(t) \rangle\\ \amp \quad \cdot \langle -\sin(t), \cos(t), -2 \sin(2t) \rangle \, dt \end{align*}
If you were to write out this dot product and combine like terms, then you would be left with four terms. Each of these can be evaluated using a few trig identities and substitutions. In fact, three of the four terms will correspond to functions that have as much area below the axis as above and thus will have integrals of zero.
\begin{align*} \int_C \vF \cdot d\vr = \amp \int_0^{2\pi} 2\cos(t)\sin(t) \, dt \\ \amp - \int_0^{2\pi} \cos(2t)\sin(t) \, dt \\ \amp + \int_0^{2\pi} \cos(2t)\cos(t) \, dt \\ \amp - \int_0^{2\pi} 2\cos(t)\sin(t)\sin(2t) \, dt \\ \amp = 0 -0 + 0 -\pi = -\pi \end{align*}
The circulation of \(\vF\) around \(C_3\) is \(-\pi\text{.}\) This result being negative means that more of the vector field moves opposite the direction of travel given by the parameterization.
(b)
In this part, we will calculate the flux of \(\curl(\vF) \) through a surface with boundary \(C\text{.}\) As demonstrated by part b of Preview Activity 12.11.1, there are several different surfaces that we can use in this problem. For our first case, we will use the part of the surface \(z=1-2y^2\) that is bounded by \(C\) as shown in Figure 12.11.7. This surface can be parameterized by \(\vr(s,t)=\langle s\cos(t),s\sin(t), 1-2 s^2 \sin(t)^2 \rangle \) with \(0\leq s\leq 1\) and \(0\leq t\leq 2\pi\text{.}\)
Figure 12.11.7. A portion of the surface \(z=1-2y^2\) bounded by \(C\)
Using our parameterization, we have the following for the partial derivative functions and the corresponding normal vector:
\begin{align*} \vr_s(s,t) \amp =\langle \cos(t), \sin(t),-4s\sin(t)^2\rangle \\ \vr_r(s,t) \amp =\langle -s\sin(t), s\cos(t),-4s^2\sin(t)\cos(t)\rangle \\ \vw=\vr_s \times \vr_t \amp =\langle -4s^2\sin(t)^2\cos(t)+4s^2\sin(t)^2\cos(t) , \\ \amp \quad \quad 4s^2\cos(t)^2\sin(t)+4s^2\sin(t)^2\sin(t), s \cos(t)^2 + s \sin(t)^2 \rangle \\ \amp = \langle 0, 4s^2\sin(t), s \rangle \end{align*}
In particular, our parameterization yields \(\vr_s \times \vr_t \, dA = \langle 0,4s\sin(t), 1 \rangle s \, ds dt\text{.}\) Since we used polar coordinates as our parameter, \(dA = s ds dt\text{.}\)
The next step in setting up the flux integral on the right side of Stokes’ Theorem is calculating the curl of \(\vF\text{.}\) Since \(\vF=\langle z-x, y+z, xy \rangle \text{,}\) \(\curl(\vF)=\langle x-1,1-y,0 \rangle \) and converting to \(s\) and \(t\) gives \(\curl(\vF) (s,t)=\langle s\cos(t)-1,1-s\sin(t),0 \rangle \text{.}\) Now we are able to apply Theorem 12.9.7, which give the following iterated integral:
\begin{equation*} \int_0^1 \int_0^{2\pi} \langle s\cos(t)-1,1-s\sin(t),0 \rangle \cdot \langle 0,4s\sin(t), 1 \rangle s \, dt\, ds\text{.} \end{equation*}
Evaluating this dot product and computing each integral gives
\begin{equation*} \int_0^1 \int_0^{2\pi} -4s^3 \sin(t)^2 + 4s^2 \sin(t) \, dt ds = -\pi\text{,} \end{equation*}
which matches our result for the calculation of the circulation around \(C\text{.}\)
(c)
As we saw in part b of Preview Activity 12.11.1 there is more than one surface that has \(C\) as a boundary. We will calculate the flux integral (the right side of Stokes’ Theorem) for a different surface to help motivate why it will not matter which surface we use (as long as we have the correct orientation and boundary). For this part we will use the surface \(z=x^2-y^2\text{,}\) which will be parameterized by \(\vr(s,t)=\langle s\cos(t),s\sin(t), s^2 (\cos(t)^2-\sin(t)^2) \rangle \) with \(0\leq s\leq 1\) and \(0\leq t\leq 2\pi\text{.}\) Additionally, we will use the trig identity \(\cos(2t)=\cos(t)^2-\sin(t)^2 \) to write the last component of our parameterization as \(s^2 \cos(2t)\text{.}\)
Using our parameterization, we have the following for the partial derivative functions and the corresponding normal vector:
\begin{align*} \vr_s(s,t) \amp =\langle \cos(t), \sin(t),2s\cos(2t)\rangle \\ \vr_r(s,t) \amp =\langle -s\sin(t), s\cos(t),-2s^2\sin(2t)\rangle \\ \vw=\vr_s \times \vr_t \amp =\langle -2s^2\sin(2t)\sin(t)-2s^2\cos(2t)\cos(t) , \\ \amp \quad \quad 2s^2\cos(t)\sin(2t)-2s^2\sin(t)\cos(2t), s \cos(t)^2 + s \sin(t)^2 \rangle \\ \amp = \langle -2s^2 \cos(t), 2s^2 \sin(t), s \rangle \end{align*}
(There are a number of equivalent algebraic simpmlifications that can be done here by choosing different trigonometric identities.)
Now we are ready to set up the flux integral as the following iterated integral:
\begin{equation*} \int_0^1 \int_0^{2\pi} \langle s\cos(t)-1,1-s\sin(t),0 \rangle \cdot \langle -2s^2 \cos(t), 2s^2 \sin(t), s \rangle \, dt\, ds \end{equation*}
You probably noticed that this integral looks slightly more complicated than our work in the previous part, but if we are consistent, we should get the same result. Evaluating this dot product and computing each integral gives
\begin{align*} \amp \int_0^1 \int_0^{2\pi} (-2s^2 \cos(t))(s\cos(t)-1) + (2s^2 \sin(t))(1-s\sin(t)) \, dt\, ds \\ =\amp \int_0^1 \int_0^{2\pi} -2s^3 \cos(t)^2 + 2s^2 \cos(t) + 2s^2 \sin(t) -2s^3\sin(t)^2 \, dt\, ds \\ =\amp \int_0^1 \int_0^{2\pi} -2s^3 + 2s^2 \cos(t) + 2s^2 \sin(t) \, dt\, ds \text{.} \end{align*}
Because both sine and cosine will integrate to zero over the interval from \(0\) to \(2\pi\text{,}\) we only need to evaluate
\begin{equation*} 2 \pi \int_0^1 -2s^3 \, ds= -\pi\text{,} \end{equation*}
which gives exactly the same result as our circulation integral and our flux integral with the other surface.
We close this subsection 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).

Activity 12.11.2.

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 \(C_1\) parameterized 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)
Write a sentence to explain why the sign of your answer to the previous two parts makes sense.

Activity 12.11.3.

(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 12.11.3, 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.

Subsection 12.11.3 Practice with Surfaces and their Boundaries

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 Exercise 6 and Exercise 8 of Section 12.8, 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 4 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.

Activity 12.11.4.

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:
  1. Describe the boundary in words.
  2. Find a parametrization for the boundary.
  3. 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 \(S_1\) is the portion of the sphere \(x^2+y^2+z^2=4\) with \(z\geq x\text{.}\) Assume the outward orientation on the sphere.
(b)
The surface \(S_2\) is the portion of the sphere \(x^2+y^2+z^2=4\) with \(z\geq 0\text{.}\) Assume the outward orientation on the sphere.
(c)
The surface \(S_3\) 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 \(\vk\text{.}\)
(d)
The surface \(S_4\) 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!

Activity 12.11.5.

In some sense, this activity considers the reverse problem of that considered in Activity 12.11.4. 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 12.11.1, 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 parameterized as \(\vr(t) =\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\rangle\) 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 \((2,0,0)\text{.}\)

Subsection 12.11.4 Summary

  • 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 \(C\text{.}\)

Exercises 12.11.5 Exercises

1.

Find \(\int_C\vec F\cdot d\vec r\) where \(C\) is a circle of radius \(1\) in the plane \(x+y+z=3\text{,}\) centered at \((4,4,-5)\) and oriented clockwise when viewed from the origin, if \(\vec F = 4y\vec i - 3x\vec j + 4\!\left(y-x\right)\vec k\)
\(\int_C\vec F\cdot d\vec r =\)

2.

Use Stokes’ Theorem to evaluate \(\displaystyle \int_{C} \mathbf{F} \cdot d\mathbf{r}\) where \(\mathbf{F}(x, y, z) = x\mathbf{i} + y\mathbf{j} + 8\!(x^{2} + y^{2})\mathbf{k}\) and \(C\) is the boundary of the part of the paraboloid where \(z = 16 - x^{2} - y^{2}\) which lies above the xy-plane and \(C\) is oriented counterclockwise when viewed from above.

3.

Verify Stokes’ theorem for the helicoid \(\Psi(r,\theta) = \langle r\cos \theta, r\sin \theta, \theta \rangle\) where \((r,\theta)\) lies in the rectangle \([0,1] \times [0,\pi/2]\text{,}\) and \(\mathbf{F}\) is the vector field \(\mathbf{F} = \langle 6 z, 6 x, 6 y \rangle\text{.}\)
First, compute the surface integral:
\(\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta\text{,}\) where
\(a =\), \(b =\), \(c =\), \(d =\), and
\(f(r, \theta) =\) (use "t" for theta).
Finally, the value of the surface integral is .
Next compute the line integral on that part of the boundary from \((1,0,0)\) to \((0,1,\pi/2)\text{.}\)
\(\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b g(\theta)\,d\theta\text{,}\) where
\(a =\), \(b =\), and
\(g(\theta) =\) (use "t" for theta).

4.

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.

5.

For each of the following vector fields, determine whether a vector potential exists. If so, find one.
For this problem, enter your vectors with angle-bracket notation: \(\lt a, b, c>\text{,}\) not in \(ijk\)-notation.
(a) \(\vec F = 6x\,\mathit{\vec i}+\left(2y-z^{2}\right)\,\mathit{\vec j}+\left(x+8z\right)\,\mathit{\vec k}\)
\(\vec F\)
  • has a vector potential
  • does not have a vector potential
\(\vec H\text{,}\) \(\vec H =\)
(If there is no potential function, enter none for the function.)
(b) \(\vec F = 6x\,\mathit{\vec i}+\left(2y-z^{2}\right)\,\mathit{\vec j}+\left(x-8z\right)\,\mathit{\vec k}\)
\(\vec F\)
  • has a vector potential
  • does not have a vector potential
\(\vec H\text{,}\) \(\vec H =\)
(If there is no potential function, enter none for the function.)

Subsection 12.11.6 Notes to Instructors and Dependencies

This section relies heavily on understanding flux integrals Section 12.9 as well as the calculation of circulation around a closed curve (from Section 12.8 and Section 12.2). Subsection 12.11.3 gives some reminders about the different ways to parameterize surfaces, which was first introduced in Section 11.6.
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