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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 10.2 Line Integrals of Vector Fields

Subsection 10.2.1 Vectors Fields and Curves

Let me start this section with a vector field \(F\) on a set \(U \subset \RR^3\text{,}\) represented fluid flow or force. Now I’m going to consider a parametric curve \(\gamma(t)\) moving through the region \(U\text{.}\) I’m going to think of this as the path of an object moving through the situation caused by the field. If this field is, say, wind, this is an object moving through wind: perhaps with the wind, perhaps against it, perhaps across it. If the field is force, then the path is an object moving through the force; again, possibly along with the force and possibly against the force.
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Note that this path doesn’t need to be an integral curve. This can be a path which is simply influenced by the field, or it can a path of an object acting against the force. This could be the path of a leaf in the wind (which, with the right initial velocity, should be an integral curve). Or it could be the path of a person struggling to move upwind, acting against the force.
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In either case, there is an interaction of the field and the path. In the case of the leaf in the wind, the fields and the path work together. In the case of the person moving upwind, the field and the path are opposed. I’d like to measure this interaction. An integral curve was defined to be a curve where the tangent vector and the vector field coincided. For another curve, the tangent vector and the vector field may not coincide, but they exist and I can look at their interaction. I have a tool for this: the dot product. The dot product measures the similarity of two vectors, being largest when they are in the same direction, negative when the are in oppositve directions, and zero when they are perpendicular. Given an field \(F\) and a curve \(\gamma\text{,}\) the dot product \(F \cdot \gamma^\prime\) will measure the interaction.
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This interaction \(F \cdot \gamma^\prime\) tells me how the field and the direction of the curve interact at a point. However, I may want a more holistic understanding of what happens along the whole curve. In the process of moving along the curve, how does the field act? Does the field push the object along the curve (a positive influence) or does movement along the curve have to work hard to counteract the field? Or, perhaps, does the curve move perpendicularly to the field and there is no influence at all?
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To answer this question, I define the integral of the field along the parametric curve.
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Definition 10.2.1.

Let \(F\) be a vector field and \(\gamma(s): [0,L]\) be a parametric curve parametrized by arclength. The line integral of a vector field \(F\) along a path \(\gamma(s): [0,L] \rightarrow \RR^n\) is the integral of the scalar \(F(\gamma(s)) \cdot T(s)\) along the length of the curve.
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\begin{equation*} \int_{\gamma} F \cdot T ds = \int_{\gamma} F \cdot ds \defeq \int_0^L F(\gamma(s)) \cdot T(s) ds \end{equation*}
Both these notations, integrating \(F \cdot T\) where \(T\) is the unit tanget of the curve, and integrating \(F \cdot ds\) are conventional notations for line integrals.
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The term line integral is a bit annoying, since the curve does not need to actually be a striaght line. The name comes from historical usage, where ‘line’ and ‘curve’ could be used interchangable for things like parametric curves.
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The line integral is a measure of work or resitance of the field along the path. If the integral is positive then, on average, the field is working along with the curve. If the integral is negative then, on average, the curve is moving against the field. The integral curves of the field are the curves where the tangents of the curve and the field coincide: these are the curves that have the largest positive line integrals. As before, I should repeat a caution about the interpretation. Sometimes, curve are thought of as the paths objects take due to the action of the field (such as the integral curve, for objects starting at rest). However, sometimes curves are representing objects under some other volition, which may act with or against the fields; the line integral then measures the interaction.
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Subsection 10.2.2 Calculating Line Integrals

Using the arclength parameter for the definition is appropriate for a strong definition. It is a unique parametrizion and gives the line integral a single, clear definition. However, the arclength parmaeter is inconvenient for calculation. Instead, in this section, I’ll explain how to calculate line integrals for any parametrization. (Since I am starting from the arclength parameterization and going to any other parametrization, it will be automatically established that the calculation of the line integral will be independent of parametrization.)
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Let \(\gamma(t): [a,b] \rightarrow \RR^3\) be an arbitrary parametrization of a parametric curve. Then I can think of \(s(t)\) (the arclenth of the curve in terms of \(t\)) as a substitution for single variable integration. This substitution has \(ds = |\gamma^\prime(t)| dt\) and \(T(t) = \gamma^\prime(t) / |\gamma^\prime(t)|\text{.}\) I can apply this substitution to the definition of the line integral.
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\begin{align*} \int_{\gamma} F \cdot T ds \amp = \int_0^L F(\gamma(s)) \cdot T(s) ds \\ \amp = \int_a^b F(\gamma(t)) \cdot \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} |\gamma^\prime(t)| dt \\ \amp = \int_a^b F(\gamma(t)) \cdot \gamma^\prime(t) dt \end{align*}
Very conveniently, the length of the tangent \(|\gamma^\prime(t)|\) cancels off, so the calculation of the line integral is quite reasonable in any arbitrary parametrization. Let me move on to some examples.
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Example 10.2.2.

For ar example, consider the rotational flow \(F(x,y) = (-y,x)\) and \(\gamma(t)\) the counterclockwise circle \((r \cos t, r \sin t)\) for \(t \in [0, 2\pi]\text{.}\) I’ll calculate the line integral.
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\begin{align*} \gamma^\prime(t) \amp = (-r \sin t, r \cos t)\\ F(\gamma(t)) \amp = (-r\sin t, r \cos t)\\ F(\gamma(t)) \cdot \gamma^\prime(t) \amp = r^2 \sin^2 t + r^2 \cos^2 r = r^2\\ \int_{\gamma} F \cdot ds \amp = \int_0^{2\pi} r^2 dt = 2\pi r^2 \end{align*}
The work to move around the circle (with the vector field) of radius \(r\) is \(2\pi r^2\text{.}\) Since it is positive and follows the integral curve, I could think of it as the work the field accomplished to move the object along its integral curve.
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Example 10.2.3.

A very important example is gravity and potential energy. Here is the field describing the force (per unit mass) generated by a mass \(M\) at the origin.
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\begin{equation*} F = \frac{GM}{\sqrt{(x^2+y^2+z^2)^3}} (-x, -y, -z) \end{equation*}
Now consider an outward path \(\gamma(t) = (t,t,t)\) for \(t \in [a,b]\) and \(b>a>0\text{.}\) I’ll calculate the line integral.
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\begin{align*} \gamma^\prime \amp = (1,1,1)\\ F(\gamma(t)) \amp = \frac{GM}{\sqrt{(3t^2)^3}} (-t,-t,-t) = \frac{-GM}{t^3 3\sqrt{3}} (1,1,1)\\ F(\gamma(t)) \cdot \gamma^\prime(t) \amp = \frac{-3t GM}{t^3 3\sqrt{3}} = -\frac{GM}{t^2\sqrt{3}}\\ \int_{\gamma} F \cdot ds \amp = \int_b^a -\frac{GM}{t^2\sqrt{3}} dt = -\frac{GM}{\sqrt{3}} \int_a^b \frac{1}{t^2} dt\\ \amp = -\frac{GM}{\sqrt{3}} \left. \frac{-1}{t} \right|_a^b = - \frac{GM}{\sqrt{3}} \left( \frac{1}{a} - \frac{1}{b} \right)\\ \amp = \frac{GM(a-b)}{ab\sqrt{3}} \end{align*}
If the distances \(a\) and \(b\) are large, but \(a\) and \(b\) are close to each other, then the change \(a-b\) is much more significant than the change in the term \(ab\) in the denominator. If I set \(g = \frac{GM}{\sqrt{3}ab}\) and pretend this is locally constant (which is reasonable given the assumptions about the relative sizes of the distances), the line integral (approximately) evaluates to \(g(a-b)\text{.}\)
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Now recall that the field is defined as force per unit mass, so if it acts on a mass \(m\text{,}\) the output of the line integral will be potential of \(mg(a-b)\text{.}\) Finally, let me call the difference \(b-a\) by the letter \(h\text{,}\) indicating the ‘height’ gained in movement along this curve. The result then is \(-mgh\text{,}\) which is almost the high-school physics result of \(-mgh\text{,}\) the increase in potential energy, where \(h\) is the change in height and \(g\) the local acceleration due to gravity.
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The only remaining difference is the negative sign. This relates to my previous discussion about directions of gradients. The force of gravity is the negative gradient of the potential energy, which introduces a sign adjustment to all the interpretations for potential energy. Mutliplying by \((-1)\) to adjust for this sign discrepancy gives the desired result: the line integral represent the change in potential energey as the object moves along the curve. This example is archetypical for conservative force and will be expanded upon in the next section.
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Subsection 10.2.3 Fundamental Theorems

In the previous section, I showed how to calculate line integral buy doing a single-variable integral where the variable of integration was the curve parameter. The new kind of integral, a line integral, was reduced to a single-variable integral which we already understand from single-variable calculus. This mirrors the approach from earlier sections in this course. For integration over two, three and higher dimensional region, the method of calculation was iterated integras, which reduce the problem to a series of understood single-variable integrals.
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Reduction to single variable integrals is an excellent strategy. It means that I don’t have to invent entirely new techniques and methods for multiple integrals; I can, instead, rely on all the previous knowledge I already have for integral. It is also very typical for mathematics: many, many mathematical problems are ‘solved’ by simply reducing them to another type of problem that is already considered solved. However, as good as these reduction methods are, they miss something important. To understand what they are missing, let me return to the Fundamental Theorm of Calculus. To remind you of one of the formulations of the fundamental theorem, let \(f(x)\) be a differentiable function.
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\begin{equation*} \int_a^b \frac{df}{dx} dx = f(b) - f(a) \end{equation*}
The fundamental theorem say that integrating a derivative is the same as evaluating the original function on the boundary (endpoints) of the interval. My question now is: are there versions of this theorem for our new kinds of integrals? To understand the question, let me talk about the main pieces of the fundamental theorem.
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I want to make an archetype out of the fundamental theorem. To that end, I’m going to boil down the idea to this basic: the fundamental theorem relates integrals where there is a derivative on the right and a boundary on the left. In this context, I can ask for other similar results: what theorem relate integrals where there is some kind of differential operator on the left and some kind of set boundary on the right? In symbols, if \(Df\) is some differential opreator and \(\del S\) is the boundary of some set, I am looking for theorems that fit this archetype:
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\begin{equation*} \int_S Df = \int_{\del S} f \end{equation*}
It turns out there are many such theorems. A major part of the rest of this course will be understanding these theorems. What kind of differential operators can show up on the left side, and for what kinds of integrals? What kinds of set and boundary match with the differential operators on the right? I can start answering this question with line integrals.
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Subsection 10.2.4 The Fundamental Theorem of Line Integrals

The first new ‘fundamental theorem’ is about line integrals. The differential operator on the leff will be the gradient of a scalar field. The boundary operator on the right will be the end poitn of a parametric curve.
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Proof.

I calculate the integral in steps, working in \(\RR^3\) to demonstrate (though the proof works in any dimension).
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\begin{align*} \int_{\gamma} F \cdot ds \amp = \int_a^b \nabla f(\gamma(t)) \cdot \gamma^\prime(t) dt\\ \amp = \int_a^b \left( \frac{\del f}{\del x} (\gamma(t)), \frac{\del f}{\del y} (\gamma(t)), \frac{\del f}{\del z} (\gamma(t)) \right) \cdot \left( \gamma_1^\prime(t), \gamma_2^\prime(t), \gamma_3^\prime(t) \right) dt\\ \amp = \int_a^b \frac{\del f}{\del x} (\gamma(t)) \gamma_1^\prime(t) + \frac{\del f}{\del y} (\gamma(t)) \gamma_2^\prime(t) + \frac{\del f}{\del z} (\gamma(t)) \gamma_3^\prime(t) dt \end{align*}
This is the most important step of the proof. So far, I’ve just expanded the notation for calculating an arbitrary line integral of a gradient in \(\RR^3\text{,}\) doing the dot product and getting a general expression for the resulting single variable integral. However, the expression I get is precisely the definition of the chain rule for the derivative of a scalar field evaluated along a parametric curve. Therefore, I can write this single variable integral as a derivative, going backwards with the chain rule. After doing so, I use the ordinary single-variable fundamental theorem applied to the function \(f(\gamma(t))\text{.}\)
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\begin{align*} \amp = \int_a^b \frac{d}{dt} \left( f(\gamma(t)) \right) dt\\ \amp = f(\gamma(b)) - f(\gamma(a)) \end{align*}
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Side by side, let me clearly show the similarity of the theorem with the original fundamental theorem: derivatives on one side and (evaluation on) boundaries on the other side.
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\begin{align*} \int_a^b \frac{df}{dx} dx \amp = f(b) - f(a)\\ \int_{\gamma} \nabla f \cdot ds \amp = f(\gamma(b)) - f(\gamma(a)) \end{align*}
The result of the fundamental theorem of line integrals says that only the endpoints determine the value of the line integral. There is a term for this fact.
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Definition 10.2.5.

Let \(F\) be a vector field and \(\gamma(t)\) an parametric curve. If the line integral \(\int_{\gamma} F \cdot dx\) only depends on the endpoint of \(\gamma\text{,}\) then the line integral is called path independent.
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We can succincly rephrase the fundamental theorem of line integrals by just saying that line integrals of conservative fields are always path independent.
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Subsection 10.2.5 Implications for Conservative Vector Fields

The fundamental theorem for line integral applies to conservative vector fields: it says that the integral of a conservative vector field can be calculated simply by evaluation on the end points. That means all the points on the path inbetween are irrelevant!
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No matter how strange the path is, no matter how many times it loops around and goes elsewhere, only the endpoints matter. Let me go back to conservative fields as forces which come from a potential energy scalar field. If \(f\) is this potential energy, then \(f(\gamma(b)) - f(\gamma(a))\) is the potential energy at the end less the potential energy at the start. This difference is the change in potential energy. The work to move through the force field is equal to the change in potential energy. The work can be defined in terms of kinetic energy lost or gained, so this says that the kinetic energy lost or gained is equal to the change in potential energy. Therefore, energy is conserved! This explains the term ‘conservative vector field’: these are vector fields where conservation of energy makes sense. (As before, there are sign issues for both gravity and electromagnetic force, since the gradient point in directions of increasing potential energy, but the model of physics is based on the principle of decreasing potential energy).
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I said that the change in potential energy (the result of the line integral using the fundamental theorem) should be the negative of the change in kinetic energy. I can actually be very explicit with this calculation. Let \(F = \nabla f\) be a conservative force. Then \(F = ma\text{,}\) where \(a\) is acceration. Also, if \(\gamma\) is a curve representing movement, then \(\gamma^{\prime \prime}\) is the acceleration. Then look carefully at the change in energy using the fundamental theorem of line integrals. I will replace the force \(F\) with \(ma = m \gamma^{\prime \prime}\text{.}\) (The reult of the line integral is the negative change in potential energy due to the already-frequently-mention sign reversal in these physics application.)
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\begin{align*} -\Delta PE \amp = f(\gamma(b)) - f(\gamma(a))\\ \amp = \int_{\gamma} F \cdot ds\\ \amp = \int_a^b m \gamma^{\prime \prime}(t) \cdot \gamma^\prime(t) dt \end{align*}
This dot product can be written as a derivative of a different dot product (using the product rule). After doing so, I use the vector identity that a dot product of a vector with itself is its length squared.
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\begin{align*} \amp = \frac{m}{2} \int_a^b \frac{d}{dt} (\gamma^\prime(t) \cdot \gamma^\prime(t)) dt\\ \amp = \frac{m}{2} \int_a^b \frac{d}{dt} |\gamma^\prime(t)|^2 dt\\ \amp = \frac{m}{2} |\gamma^\prime(t)|^2 \bigg|_a^b \end{align*}
The magnitude of \(\gamma^\prime\) is just the scalar speed. Evaluated at the ends points gives the starting speed and the ending speed. I’ll write these as \(v_a\) and \(v_b\text{,}\) respectively,
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\begin{equation*} = \frac{mv_b^2}{2} - \frac{mv_a^2}{2} = \Delta KE \end{equation*}
The expression \(\frac{1}{2} m v^2\) is the definition of kinetic energy. Therefore, the result of this calculation is simply the ending kinetic energy minus the starting kinetic energy: this is the change in kinetic energy. Therefore, the calculation shows that the change in potential energey is the negative of the change in potential energy.
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Subsection 10.2.6 Properties of Conversative Vector Fields

In Section 10.1 I showed that conservative vector fiels are irrotational and, conversely, an irrotation field defined on a simply connected set was guaranteed to be conservative. I now have a new characterization of conversative fields: their line integrals are path independent. Again, I can ask if the converse holds and again, there is a topological condition.
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Putting together the various statements on conservative vector fields, I can summarize their definiting properties in the following proposition.
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