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Section 9.1 Guass, Green and Stokes

In Subsection 6.2.3, I talked about the archetype of the fundamental theorem. The fundamental theorem of line integrals was a version of the fundemental theorem that applied to vector fields and parametric curves. In this section, I want to talk about versions of the fundamental theorem for parametric surfaces. Let me remind you of the most general idea of the fundamental theorem.

\begin{equation*} \int_{\del D} f = \int_D df \end{equation*}

Here, \(\del\) is a boundary operator for geoemtric object and \(df\) is some kind of derivative of \(f\text{.}\) On one side, there is a derivative; on the other side, there is a boundary. So far, the boundaries of curves or intervals are just points. The right side is the ‘integral’ of a function on two points, which is simply evaluation at the point. In this section, however, I'll be talking about surfaces. In general, the boundary of a surface will be a one dimensional object (a parametric curve, hopefully). There is no such thing as ‘evaluation’ along a parametric curve; instead, there is a line integral. Therefore, I really do expect integrals on the right side of the archetypical equation. The full understanding of the fundamental theorem is that there are integrations on both sides: integral of a derivative on the right and integration over a boundary on the left. Before I get to the major results, though, I have to talk about boundaries.

Subsection 9.1.1 Boundaries and Orientation

In Calculus III, I defined a topological definition of a boundary. With apologies for poor use of terminology, the notion of boundary in this section is not topological. You could think of this as a geometric boundary or a parametric boundary, if you wanted to give it a different label.

The boundary of a parametric object is the boundary in terms of the parameter. For a curve, the boundary is the endpoints, indicated by where the parameter starts and stops. Moreover, the curve has a direction. The boundary should reflect that direction. Therefore, boundaries of parametric objects are oriented boundaries. Orientation is going to mean something slightly different in each dimension. For curves, orientation assigns signs to the end points. The starting point is labelled as negative and the ending point as positive, so that the curve proceeds from negative to positive. This also fits with the fundamental theorem of line integrals: the line integral of a conservative fields is the value of the potential at the end point (positive) minus the value of the potential at the starting point (negative).

I'm not going to give a full, formal definition of the boundary of a parametric surface, because I don't want to get lost in the necessary technical details, Instead, for this course, I'll work with a less formal, more intuitive treatment of boundaries. Informally, the boundary of a parametric surface should be the one-dimensional edge of the surface, if such a thing exists. Let me give some examples to illustrate.

  • The boundary of a sphere is empty. There are no edges of the sphere, it just wraps around on itself. The same is true for many other objects The boundary of an ellipsoid is empty. The boundary of a cube is empty. The boundary of a cylinder which includes the top and bottom discs is empty.

  • A disc of radius \(R\) in the \(xy\) plane can be parametrized as a parametric surface. The boundary of the disc is the circle of the same radius. The circle is the edge of the disc: where the disc stops. Similarly, the boundary of a solid square is the edge of the square, and likewise for any two-dimensional shape which is treated as a parametric surface.

  • The boundary of a cylinder which does not include the top and bottom discs consists of two pieces: the circle at the top and the circle at the bottom. These circles are the edge of the cylinder, where is abruptly stops.

  • The boudary of a cone is a bit tricky. It certainly contains the circle at one end of the cone. What is less clear is whether the point at the other end of the cone is considered a boundary point. Informally, this could be argued several ways. For the purposes of this course, I will not treat such a point as a boundary point.

  • The boundary of a hemisphere is the circle at the edge of the hemisphere.

Hopefully, this is enough to get an informal sense of the boundary of a parametric surface. Now I need to talk about orientation. For curves, orientation came from the notion of direction: moving from the start to the end. For parametric surfaces, there isn't a direction of movement, but there is the direction of the normal. The normal defines a notion of above and below the surface, which is how I will treat orientation. By convention, the normal will point above, and the negative of the normal will point below. (Changing the order of the variables in the parametrization of a surface will change the order of the cross product, thus the sign of the normal.)

For parametric curves, orientation always works: all curves have starting points, ending points, and movement from the start to the end. However, not all surfaces have reasonable orientation. Locally, orientation is always fine. In a small part of a surface, the normal will point to one side of the surface and a notion of above and below can be defined. But these local orientations may not fit together into an orientation that works for the entire surface. The classic example is the Möbius strip: a cylindrical ring with a single twist in it. This object does not have properly defined sides. Many other surfaces have similar problems. A surface that has a reasonable orientation over the whole surface is called orientable. Surfaces which don't, like the Möbius strip, are non-orientable. The results of this section are going to rely on orientability.

Now I have (informal) notions of orientation for both parametric curves and parametric surfaces. Typically, the boundary of an orientable parametric surface should be a parametric curve. How do the orientations interact? First, notice that the boundary of a parametric surface will always be a closed curve; that is, a curve with the same starting and endoing point. (It could be multiple closed curves, such as the two circles that bound an open cylinder, but each bounding curve can be dealt with separately as a closed curve.) Even more, I can expect for reasonable surfaces that the boundary is a simple, closed curve; that is, a closed curve with no self-intersection. Simple closed curves are cycles: they go around a loop. This loop could be smooth, like a circle or ellipse, or with sharp cornder, like a square or rectangle. I need a connection between the normal of the surface and the direction of the loop that forms its boundary.

The connection is given by a right-hand-rule. The normal defines a side which is above the surface. Looking down at the surface from this above side, the boundary loop must go counter-clockwise. This is called a right-hand-rule because if the thumb of a right hand points in the directin of the normal, the fingers curl in this counter-clockwise direction. Note that this is a completely arbitrary convention: using clockwise curves and a left-hand-rule would have also given consistent mathematics, as long as the same rule is used consistently.

In the theorems that follow in this section, I will be restricting to cases where all of this works: orientable surfaces; their simple, closed parametric curve boundarys; and compatible orientations according to this right-hand-rule. There are lots of ways that this setup can fail for certain surfaces. However, for most of the motivating applications, the surface are reasonable and can be parametrizated to fit these restriction.

The first surfaces in the list of examples above did not have boundaries (the sphere, the cube, etc.) Such surfaces are called closed surfaces. A closed surface will invariable define an inside and an outside. By convention, the outside is above and the inside is below; therefore, the normals to closed surfaces should always point outwards, not inward. These closed surfaces themselves form the boundary of solid region in \(\RR^3\text{.}\) What is a solid region in \(\RR^3\text{?}\) Again, a technical, formal definition is not what I want here. Instead, I'll simply treat a solid region in \(\RR^3\) as any set in \(\RR^3\) whose boundary (now this is the actual topological boundary!) is a closed parametric surface. The solid cube has the hollow cube as its boundary; the solid sphere has the hollow sphere as its boundary; the solid cone has the hollow cone as its boundary; and so on. The solid region itself doesn't have any orientation. However, the orientation on the closed surface boundary must point outward, matching the convention I already established.

I have one more note concerning the boundary of objects discussed in this section. Let \(\del\) be the operator that takes the boundary of a solid region or parametric object. Consider these two situation that I've describe above.

  • Start with a solid region \(D \subset \RR^3\text{.}\) Then \(\del D\) is a closed parametric surface (with outward pointing normal. The boundary of a closed parametric surface is empty, so I could write \(\del \del D = \emptyset\text{.}\)

  • Start with a orientable parametric surface \(\sigma\) (not necessarily closed). Then the boundary \(\del \sigma\) is a close parametric curve (with compartible orientation given by a right-hand-rule). Since the curve is closed, its has no boundary. So again I could write \(\del \del \sigma = \emptyset\text{.}\)

This is a general principle. For any reasonably defined geometric boundary operation \(\del\text{,}\) it should be true that \(\del\) satisfies \(\del^2 S = \emptyset\) for any reasonable geometric object \(S\text{.}\) By the end of the course, I want to use the archetype of the fundamental theorem to relate this fact to the composition of differential operators.

Finally, let me remind you about a term defined in previous courses. A curve is called piecewise differentiable if it is differentiable except for isolated points. A piecewise differentiable curve is allowed to have isolate places with sharp corners. The boundary of a rectangle is a good example: it is differentiable everywhere except for the four corners of the rectangle. Line integrals of piecewise differentiable curves are well defined: I just do the line integral over the pieces and add up all the results.

Similarly, I can define a piecewise differentiable parametric surface. This is a parametric surface that is differentiable everywhere except for some one-dimensional pieces. The hollow cube is a good example. I can parametrize the six faces of the cube with differentiable functions, but the six faces meet at sharp edges. Again, the flux integral of a piecewise differentiable parametric surface is well defined: I just do the flux integrals of the various pieces and add them up. Since the sharp edges are all one-dimensiona, they don't cause any problems for the two dimensional integral.

Subsection 9.1.2 Gauss-Green-Stokes

Having done a bunch or work on boundaries and orientation, I can finally state the two main theorems for vector calculus. These are both theorems that fit the archetype of the fundamental theorem. The first is called Stokes' Theorem.

This is the archetype of the fundamental theorem. There is a differential operator on the right, the curl in this case. There is a boundary on the left, the boundary of a parametric surface in this case.

The second major theorem is called Gauss' Theorem. In some texts, this is called the divergence theorem.

This theorem also has the desired form. There is a differential operator on the left, the divergence in this case. On the right there is a boundary operator, in this case the boundary of a solid region of \(\RR^3\text{.}\)

I've presented two main theorems here. In many texts, there are three important theorems presented in the matching section . Historically as well, there were three theorem in this development of this theory. I only presented two theorems since the third theorem is actually just a special case of Stokes' Theorem. It is called Green's theorem and it applies to \(\RR^2\text{;}\) it is a special case of Stokes' Theorem because if you calculate Stokes' theorem and set the \(z\) coordinates of everything to zero (thus essentially working in \(\RR^2\text{,}\) the result is exactly Green's Theorem. I will, however, state it here.

Since these three named theorems are of similar form, they are often referred to as Gauss-Green-Stokes (or even GGS). All three of these people were 19th century mathematics/physicists and all of them were involved with the development of the physics of electricity and magnetism. In Section 10.1, I'll give the details of that project. This mathematics was specifically invented to describe electricity and magnetism, since no previous mathematics could describe the experimental results observed in these 19th laboratories and workshops.

Subsection 9.1.3 Direct Uses of Gauss-Green-Stokes

The first use of GGS (what I will call the direct use) is using the equations of the theorem to change an integral into another integral. This can go in either directions: from left to right or right to left. There are many situations where the integral on one side is diffcult or impossible and the integral on the other side is reasonable. I'll some examples of these direct uses. Some of these examples generalize into useful proposition, which I will also state.

I'll start in \(\RR^2\) with Green's theorem. Let \(F(x,y) = (\sin x, x^2 y^3)\) and let \(\sigma\) be the triangle with vertices \((0,0)\text{,}\) \((2,0)\) and \((2,2)\text{.}\) Then \(\gamma\) is the path in three straight lines from \((0,0)\) to \((2,0)\) to \((2,2)\) and back to \((0,0)\text{.}\) (This shows why piece-wise differentiable conditions are useful: allowing these sharp corners on the boundary path is very reasonable.) The line integral of \(F\) over \(\gamma\) could be calculated in three steps, by parametrizing each line segments and calculating those line integrals. However, I've prefer not to have to figure out these three parametrization. Green's theorem gives me an alternate way to calcaulate the line integral by integrating over the triangle.

\begin{align*} \int_\gamma F \cdot ds \amp = \int_\sigma \left( \frac{\del F_2}{\del x} - \frac{\del F_1}{\del y} \right) \cdot dA = \int_\sigma ( 2xy^3 - -0) dA\\ \amp = \int_0^2 \int_0^x 2 x y^3 dy dx\\ \amp = 2 \int_0^2 \frac{xy^4}{4} \Bigg|_0^x dx\\ \amp = 2 \int_0^2 \frac{x^5}{4} dx = \frac{x^6}{12} \Bigg|_0^2 = \frac{2^6}{12} = \frac{16}{3} \end{align*}

Let \(F = (\sin x^2, e^{y^2} + x^2, z^2 + 2x^2)\) and let \(\gamma\) be the counterclockwise path around the triangle in \(\RR^3\) with vertices \((3,0,0)\text{,}\) \((0,2,0)\) and \((0,0,1)\text{.}\) I could parametrize the three pieces of the triangle, but I will use Stokes' theorem with the triangle as the surface. The triangle is flat, so it may be parametrized with a constant normal. The equation of the plane containing the triangle is \(\frac{x}{3} + \frac{y}{2} + z = 1\text{,}\) with normal \(\left( \frac{1}{3}, \frac{1}{2}, 1 \right)\text{.}\) The curl of \(F\text{,}\) after some algebra is \((0, -4x, 2x)\text{.}\) (Here the curl is much easier to work with than the original \(F\text{!}\)) Then \((\nabla \times F) \cdot N = (0, -4x, 2x) \cdot \left( \frac{1}{3}, \frac{1}{2}, 1 \right) = 0\text{.}\) This lets me treat the original (very complicated) line integral as a very straightforward surface integral.

\begin{equation*} \int_\gamma F \cdot ds = \int_\sigma (\nabla \times F) \cdot N dA = \int_{\sigma} 0 dA = 0 \end{equation*}

Let \(F\) be any \(C^1\) vector field and let \(\sigma = S^2\text{,}\) the unit sphere in \(\RR^3\text{.}\) I'll apply Stokes' theorem to the integral of the curl of \(F\) over the sphere.

\begin{equation*} \int_{\sigma} (\nabla \times F) \cdot N dA = = \int_{\emptyset} F \cdot dV \end{equation*}

The integal on the right is the integral over the empty set. There is no set to integrate over, so the value of the integral is zero. Therefore, the integral on the left is also zero.

I can generalize the previous example into a useful proposition. Nothing about the sphere was important to the example; it only needed to have an empty boundary.

Let \(\sigma(u,v) = (u,v,4-u^2-v^2)\) for \((u,v)\) in the disc about the origin of radius \(2\text{,}\) (\(u^2 + v^2 \leq 4\)). This \(\sigma\) is a parabaloid above the \(xy\) plane. Its bounding curve is the circle of radius \(2\) about the origin in the \(xy\) plane, going counterclockwise (to fit the upward normal of the surface with a right-hand-rule). The boundary curve is parametrized as \(\gamma(t) = (2 \cos t, 2 \sin t, 0)\text{.}\) Let \(F(x,y,z) = (z^2, x^2, y^2)\text{.}\) The curl of \(F\) is \(\nabla \times F = (2y, 2z, 2x)\text{.}\) Instead of calculating the flux of this curl over the whole paraboloid, Stokes' theorem lets me calculate the line integral instead.

\begin{align*} \int_{\sigma} (\nabla \times F) \cdot dA \amp = \int_{\sigma} (2y, 2z, 2x) \cdot dA = \int_\gamma F \cdot ds \\ \amp = \int_0^{2\pi} F(\gamma(t)) \cdot \gamma^\prime(t) dt\\ \amp = \int_0^{2\pi} (0, 4\cos^2 t, 4 \sin^2 t) \cdot (-2\sin t, 2\cos t, 0) dt\\ \amp = \int_0^{2\pi} 8 \cos^3 t dt = \frac{8}{3} (2 + \cos^2 2t) \sin t \Bigg|_0^{2\pi} = 0 \end{align*}

It I had decided to calculate the flux directly, this is how I would proceed.

\begin{align*} \sigma_u \amp = (1,0, -2u)\\ \sigma_v \amp = (0,1, -2v)\\ \sigma_u \times \sigma_v \amp = (2u, 2v, 1)\\ \nabla \times F \amp = (8v, 8-2u^2-2v^2, 2u)\\ \int_{\sigma} \nabla \times F \cdot dA \amp = \int_D (4uv - 16 v - 4vu^2 - 4v^3 + 2u) du dv \end{align*}

I could continue and verity that this integral does evaluate to zero if I wanted to double-check Stokes' theorem in this case.

So far, I've just done examples with Stokes' theorem. Let me move on to uses of Gauss' theorem.

If \(\sigma\) is the sphere of radius \(4\) in \(\RR^3\) and \(F = (2x^3, 2y^3, 2z^3)\text{,}\) then let \(D\) be the solid ball of radius \(4\text{.}\) Instead of parametrizing teh sphere and calculating the flux of the field over that surface, I can use Gauss' theorem to integrate over the solid sphere instead. (That solid sphere integral is a triple integral in \(\RR^3\text{.}\) I'll switch to spherical coordinates to do that integral.)

\begin{align*} \int_\sigma F \cdot dA \amp = \int_D \nabla \cdot F dV = \int_D 6 (x^2 + y^2 + z^2) dV\\ \amp = \int_0^4 \int_0^{\pi} \int_0^{2\pi} 6 r^2 r^2 \sin \phi d\theta d \phi dr\\ \amp = 12\pi ( -\cos \phi ) \Bigg|_0^{\pi} \frac{r^5}{2} \Bigg|_0^4 = \frac{12288\pi}{5} \end{align*}

Let \(D\) be the region inside the cylinder \(x^2 + y^2 = 4\) bounded below by the \(xy\) plane and above by the plane \(x+z=6\text{.}\) Let \(F = (x^2 + \sin z, xy + \cos z, e^y)\text{.}\) It would be challenging to calculate the flux of this field over the surface \(\del D\text{;}\) I'd have to parametrize the surface, with its strange diagonal upper bound and then calculate the flux integral of this complicated field. Gauss' theorem lets me calculate a triple integral instead. Conveniently, the divergence of this field is a much simpler expression that the original field. (I do still need to deal with this strange diagonal upper bounds. I do this by switching to cylindrical coordinates, where this upper bounding plane becomes \(z = 6 - r \cos \theta\text{;}\) I can use this equation as the upper bound for \(z\) in the triple integral.)

\begin{align*} \int_{\sigma} F \cdot dA \amp = \int_D \nabla \cdot F dV = \int_D 3x dV\\ \amp = \int_0^{2\pi} \int_0^2 \int_0^{6-r\cos \theta} 3 r \cos \theta r dz dr d\theta\\ \amp = \int_0^{2\pi} \int_0^2 3 \cos \theta r^2 (6-r\cos \theta) dr d\theta\\ \amp = \int_0^{2\pi} \left( 16 \frac{r^3}{3} \cos \theta - \frac{3r^4}{4} \cos^2 \theta \right) \Bigg|_0^2 d \theta\\ \amp = \frac{128}{3} ( -\sin \theta) \Bigg|_0^{2\pi} - 12 \left( \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right) \Bigg|_0^{2\pi} = -12\pi \end{align*}

If \(F = (z^2 + xy^2, \cos(x+z), e^{-y} + zy^2)\text{,}\) then \(\nabla \cdot F = 0\text{,}\) so \(F\) is incompressible. Then, if \(D\) is any solid region in \(\RR^3\) with boundary \(\sigma\text{,}\) Gauss' theorem gives us a vanishing result.

\begin{equation*} \int_{\sigma} F \cdot dA = \int_D \nabla \cdot F dV = 0 \end{equation*}

The previous example can be generalized into a proposition. The specifics of the field were not important to the argument; all that was needed was the fact that the divergence was zero (that the field was incompressible).

Subsection 9.1.4 Indirect Uses of Gauss-Green-Stokes

The examples in the previous section used Gauss-Green-Stokes directly, changing the integral on one side of the equation to a (hopefully easier) integral on the other side of the equation. In this section, I want to demonstrate some trickier and more subtle uses of the theorems to calculate integrals. Let me start with an example.

Consider the interal of \(F = (z^2, y, xz)\) over a closed curve \(\gamma\text{.}\) There isn't explicitly a surface here, but I can still try to apply Stokes' theorem. The curve is a closed curve, so it will be the boundary of some surface (in fact, inifinitely many surfaces). Let's say it is the boundary of some surface \(\sigma\) (I won't actually care about the other details of the surface). Then I apply Stokes' theorem.

\begin{equation*} \int_{\gamma} F \cdot ds = \int_{\sigma} \nabla \times F \cdot dA = \int_{\gamma} \left( 0, z, 0 \right) \cdot dA \end{equation*}

Now consider the vector field \(G(x,y,z) = \left( \frac{z^2}{2}, 0, 0 \right)\text{.}\) The curl of this vector field is also \((0,z,0)\text{.}\) I'll apply Stokes' theorem for the integral of the curl of this field over \(\sigma\text{.}\)

\begin{equation*} \int_{\gamma} \left( 0, z, 0 \right) \cdot dA = \int_{\sigma} \nabla \times G \cdot dA = \int_{\gamma} G \cdot ds \end{equation*}

Now let me chain the two equalities together.

\begin{equation*} \int_{\gamma} F \cdot ds = \int_{\gamma} \left( 0, z, 0 \right) \cdot dA = \int_{\gamma} G \cdot ds \end{equation*}

What I've done here is replace the field \(F\text{,}\) which is quite complicated, with the field \(G\text{,}\) which is much easier to deal with.

To make a general statement based on the previous example, I need a new definition.

Definition 9.1.14.

Let \(G\) and \(F\) be \(C^1\) vectors fields with \(G = \nabla \times F\text{.}\) The field \(F\) is called a vector potential for \(G\text{.}\) (If I want to be careful with previous terminology, I'll use the term scalar potential to describe the potentials of conservative vector fields, in order to make clear which kind of potential is under discussion.)

Now the previous example can be generalized into a proposition. Succinctly, for a line integral over a closed curve, I can replace the field with any other field which has the same curl. Here is the statement with all the necessary assumptions.

I'd like to say a bit more about vector potential before moving on. For scalar fields, I stated conditions to check if a field was conservative. I'd like to do the same here. For scalar fields, I checked the curl, because \(\nabla \times \nabla f = 0\text{.}\) Now, I also have the equation \(\nabla \cdot (\nabla \times F) = 0\text{.}\) So all fields which have a vector potential must have zero divergence. Is the inverse true? Again, it is true up to a topological condition.

So, on a simply connected open set, the divergence checks for the existence of a vector potential. If the check works and a vector potential exists, how is it calculated? If \(F\) is the vector potential for a field \(G\text{,}\) that means \(\nabla \times F = G\text{.}\) If I write this in components, I get a system of three differential equations.

\begin{align*} \frac{\del F_3}{\del y} - \frac{\del F^2}{\del z} \amp = G_1\\ \frac{\del F_1}{\del z} - \frac{\del F^3}{\del x} \amp = G_2\\ \frac{\del F_2}{\del x} - \frac{\del F^1}{\del y} \amp = G_3 \end{align*}

This system is usually prohibitively difficult. Moreover, this is a very underdetermined system. A field can have a huge variety of vector potentials. The system is much less determined than scalar potentials -- those potentials only differed by a constant. How much option is there for a vector potential? I'll give a proposition for this.

I'll just calculate the curl of \(F + \nabla f\)

\begin{equation*} \nabla \times (F + \nabla f) = \nabla \times F + \nabla \times \nabla f = \nabla \times F + (0,0,0) = \nabla \times F = G \end{equation*}

If \(H\) is some other vector potential for \(G\text{,}\) let me calculate \(\nabla times (F - H)\text{.}\)

\begin{equation*} \nabla \times (F - H) = \nabla \times F - \nabla \times H = G - G = 0 \end{equation*}

The difference \((F-H)\) has zero curl on the simply connected open set, so it is conservative. Therefore \(F - H = \nabla f\) for some scalar potential, which rearrnages to \(H = F + \nabla f\text{.}\) This proves the second part of the proposition, that any other vector potential has the desired form.

In the previous example, the two fields had the same curl. Let me actually prove that their difference is conservative. Again, let \(G = (0,z,0)\text{.}\) \(F_a = (z^2, y, xz)\text{,}\) \(F_b = \left( \frac{z^2}{2}, 0, 0 \right)\) are both vector potentials for \(G\text{.}\) Their difference must be a conservative field.

\begin{equation*} F_a - F_b = \left( \frac{z^2}{x}, y, xz \right) = \nabla \left( \frac{xz^2}{2} + \frac{y^2}{2} \right) \end{equation*}

If \(\gamma\) is any closed curve entirely in the \(xy\) plane, then since \(F_b = 0\) on the \(xy\) plane, we have a general vanish result about line integrals along \(\gamma\text{.}\)

\begin{equation*} \int_\gamma F_a \cdot ds = \int_\gamma F_b \cdot ds = \int_\gamma 0 \cdot ds = 0 \end{equation*}

The following example illustrates another indirect method of using Gauss-Green-Stokes.

Consider a windsock which is the surface of revolution of the function \(y = \sqrt{1 - \frac{x}{3}}\) for \(x \in [0,3]\text{,}\) with outward oriented normal. Let \(F = (7, -z, y)\) be a field describing the wind near the windsock. What is the flux through the windsock?

First, I can check \(\nabla \cdot F = 0\text{,}\) so \(F\) is incompressible and has a vector potential. Assume \(F = \nabla \times G\) for some vector potential \(G\text{.}\) If we call \(D\) the solid interior of the windsock, then the boundary of \(D\) is the windsock and the circle at its mouth. I will label the circle \(\mu\) and consider it a parametric surface with normal in the negative \(x\) direction. Then I apply Gauss twice, once in each direction. The first integral in the middle piece of the equation is zero since \(\nabla \times F = 0\text{;}\) this is just the integral of zero over a solid region in \(\RR^3\text{.}\)

\begin{equation*} \int_{\sigma} F \cdot dA = \int_{D} \nabla \cdot F dV - \int_\mu F \cdot dA = -\int_\mu F \cdot dA \end{equation*}

This tells me that I can actually change the surface of integration via Gauss' theorem, at least when the field is incompressible. I'm not going to give a second argument for the same result, this time using Stokes' theorem. Let \(\gamma\) be the boundary of the circle, which is the same as the boundary of the windsock excet that the the curve must go in a different direction to match the orientation. (The negative sign reflects that change in direction and orientation.)

\begin{equation*} \int_\sigma F \cdot dA = \int_\gamma G \cdot ds = -\int_\mu F \cdot dA \end{equation*}

The result of either Guass' or Stokes' here is the same. Since the field is incompressible, I can with a much easier surface: a disc of radius \(1\) in the \(yz\) plane, parametrized as \(\mu(r,\theta) = (0, r\cos \theta, r \sin \theta)\) for \(r \in [0,1]\) and \(\theta \in [0, 2\pi]\text{.}\) We calculate the flux, reversing the conventional order of the cross product to account for the normal in the negative \(x\) direction.

\begin{align*} \sigma_r \amp = (0, \cos \theta, \sin \theta)\\ \sigma_\theta \amp = (0, -r\sin \theta, r \cos \theta)\\ \sigma_\theta \times \sigma_l \amp = (-r\sin^2 \theta - r \cos^2 \theta, 0 ,0) = (-r,0,0)\\ \int_\mu F \cdot dA \amp = -\int_0^{2\pi} \int_0^1 (7, -r\sin \theta, r \cos \theta) \cdot (-r, 0, 0) dr d\theta\\ \amp = -\int_0^{2\pi} \int_0^1 -7rdr d\theta = 2\pi 7 \frac{r^2}{2} \Bigg|_0^1 = 7\pi \end{align*}

This is substantially easier than working with the windsock directly.

I can generalize the previous example into another proposition. This roughly says that for flux integrals of incompressible fields, the surface can be changed as long as the boundary is preserved.

I have one more example of an indirect use of GGS.

Let \(F = (y, z e^{y^2}, x + z^2 e^{z^2})\) and let \(\gamma\) be the semicirlce from \((-3,0,0)\) to \((3,0,0)\) which lies in the positive \(y\) part of the \(xy\)plane. What is the line integral?

First, I can check that \(\nabla \times F = (-1,1,0)\text{,}\) so the field is not conservative. I might seem that no theorem can help here. The field is not conservative, so the fundamental theorem of line integrals doesn't apply. The curve is not closed, so Stokes' theorem doesn't apply.

However, I can do something clever to make sure one of the theorems applies. Consider the path \(\delta\) from \((3,0,0)\) to \((-3,0,0)\) along the \(x\) axis. Then \(\gamma\) followed by \(\delta\) is a closed path which is the boundary of the half-circle of radius \(3\) in the \(xy\) plane. Call the half circle \(\sigma\text{.}\) Using this as the surface, I can apply Stokes' theorem. F

\begin{equation*} \int_{\del \sigma} F \cdot ds = \int_\sigma (\nabla \times F) \cdot dA \end{equation*}

The line integral over \(\del \sigma\) is the line integral over \(\delta\) plus the line integral over \(\gamma\text{.}\) Writing the two line integral, I can solve for the line integral over \(\gamma\text{.}\)

\begin{equation*} \int_\gamma F \cdot ds = \int_\sigma (\nabla \times F) \cdot dA - \int_\delta F \cdot ds \end{equation*}

This clever use of Stokes' theorem has changes the original integral into two integrals. This might seem counterproductive, but these two integrals on the right are notably easier than the first integral. \(\delta\) is entirely on the \(x\) axis, so \(F = (0,0,x)\) when \(y = z = 0\text{.}\) The tangent to \(\delta\) is \((1,0,0)\) so \(F \cdot N = 0\text{,}\) which means the second integral on the right vanishes.

The normal to \(\sigma\) is \((0,0,-1)\text{,}\) pointing in the negative \(z\) direction to match orientation. \((\nabla \times F) \cdot N = (-1,1,0) \cdot (0,0,-1) = 0\text{,}\) so the first integral on the right also vanishes. Together, I can conclude that the original line integral is also zero.

Note that the field is not conservative, so a different path from \((-3,0,0)\) to \((3,0,0)\) might have a different value. If I worked in the \(xz\) plane instead of the \(xy\) plane with the same setup, the second integral would still vanish but the normal to \(\sigma\) would be \((0,-1,0)\text{,}\) so \((\nabla \times F) \cdot N = (-1)\) and the value of the first integral would be non-zero.

\begin{equation*} \int_\sigma -1 dA = - \frac{9\pi}{2} \implies \int_\gamma F \cdot ds = \frac{-9\pi}{2} \end{equation*}

Subsection 9.1.5 Strategies for Line and Surface Integrals

Through the many examples in this section, a variety of uses of the theorems have been presented. Let me give you a convenient list to describe the various strategies that I've introduced. I'll state the most important conditions for applying these strategies, but I won't state all the conditions each time. (Basically, assume that the fields are \(C^1\) on the appropriate sets and the parametric objects are piecewise differentiable and have compatible orientations when required.)

  • For line integrals of conservative fields, the line fundamental theorem of line integrals applies and the line integral can be calculed by finding a (scalar) potential and evaluating the potential on the endpoints.

  • For line integrals over closed curves, Stokes' theorem can be applied to change the line integral into the flux integral of the curl over some surface whose boundary is the closed curve.

  • For flux integrals over any surface, if the field has a vector potential, then Stokes' theorem can be applied to change the flux integral into a line integral of the vector potential over the boundary of the surface.

  • For flux integrals over closed surfaces, Guass' theorem can be applied to change the flux integral into a triple interated integral of the divergence of the field over the solid region contain within the surface.

  • For triple integrals over a solid region in \(\RR^3\text{,}\) if the scalar field being integrated is the divergence of some vector field, Guass' theorem can be applied to change the triple integral into a flux integral of the vector field over the boundary of the region.

  • For a line integral of a field \(F\) over a closed curve, the field \(F\) can be replaced with the field \(G\) as long as they have the same curl.

  • For a flux integral of a field \(F\) over a closed surface, the field \(F\) can be replace with the field \(G\) as long as they have the same divergence.

  • For a line integral of a conservative field over a curve \(\gamma\text{,}\) the curve can be changed to any curve with the same endpoint.

  • For a flux integral of an irrotation field over a surface \(\sigma\text{,}\) the surface can be changed to any surface with the same boundary.