Introduction to Line Integrals

Overview

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The definite integral \(∫_a^bf(x)dx\) that we're all familiar with gives us the area between the following two line segments: the curvilinear line segment \(f(x)\) along the interval \([f(a),f(b)]\) and the straight line segment along the interval \([a,b]\) on the x-axis. In other words, if you imagined building a "wall" whose base was along the straight line \(ab\) and whose ceiling was \(f(x)\) (over the interval \([f(a),f(b)]\), the area of that wall would represent the definite integral \(∫_a^bf(x)dx\). But what if that function popped into or out of the screen a little and the base of the wall, instead of just being a straight line, was actually a curved line? How would we find the area of that wall? The answer to that question is we would have to use something called a line integral which will be the subject of this lesson.

This video was produced by the Khan Academy.\(^{[1]}\)


 Figure 1: The width and height of each blue rectangle is given by the arc length \(ds\) and the function \(f(x,y)\), respectively. The line integral, \(\int_Cf(x,y)ds\), represents the infinite sum of the area of each blue rectangle along the curve \(C\) (the purple line on the \(xy\)-plane). This image is a derivative work. Original image: https://www.wikihow.com/User:Atheia

Figure 1: The width and height of each blue rectangle is given by the arc length \(ds\) and the function \(f(x,y)\), respectively. The line integral, \(\int_Cf(x,y)ds\), represents the infinite sum of the area of each blue rectangle along the curve \(C\) (the purple line on the \(xy\)-plane). This image is a derivative work. Original image: https://www.wikihow.com/User:Atheia

Defining and calculating line integrals

Let's suppose that we have some arbitrary curve \(C\) in the \(xy\)-plane whose endpoints are \(c_1\) and \(c_2\) as illustrated in Figure 1. We shall associate a value \(S\) known as the arc length (which could be measured either from \(c_1\) or \(c_2\)) with each point along the curve. Let's also say that \(f(x,y)\) can be any arbitrary surface as described in the video above. \(f(s)\) represents the \(z\)-values of the surface associated with each value of \(s\). Let's say that we wanted to find the area of the curvy, blue colored wall in Figure 1. How could we do that? We would essentially do something analogous to what we would have to do if we wanted to find the area of a "wall" formed by a single-variable function \(f(x)\) and the \(x\)-axis —we would have to take the infinite sum of the areas of infinitely many really skinny rectangles. The width of each rectangle in Figure 1 is \(dS\)—a very small change in arc length. The height of each rectangle is just \(f(s)\) (or, equivalently, \(f(x,y)\)). Thus, the area of each rectangle is \(f(s)ds\). If we take the infinite sum (\(∫\)) of the tiny areas of all of the infinitesimally skinny rectangles, we'll just get

$$∫_Cf(s)ds,\tag{1}$$

where \(∫_C\) denotes the infinite sum along the curve \(C\). Expression (1) is called a line integral and represents the area between a function and any curved line \(C\). What would happen if the curve \(C\) was a straight line segment lying on the \(x\)-axis as in Figure 3. Then the arc length (either measured from the \(x\)-value \(x_1\) or \(x_2\) associated with the point \(c_1\) or \(c_2\)) could be measured by \(x\) and a change in arc length would just be \(dx\). Thus, Expression (1) would simplify to

$$\int_{x_1}^{x_2}f(x)dx.$$

Thus, Expression (1) is a generalization of the definite integral. Expression (1), superficially, looks identical to a definite integral since swapping the variable \(x\) for another variable \(S\) doesn't seem to matter. But unlike the integrand of a definite integral of the form \(\int{f(x)dx}\), the integrand \(f(s)\) of Expression (1) is, clearly, a function of the independent variables \(x\) and \(y\) (see Notes 1); thus, we can write \(s(x,y)\). Substituting \(s(x,y)\) into Expression (1), we have

$$∫_Cf(s(x,y))dS=∫_Cf(x,y)dS.\tag{2}$$

Equation (2) shows that the integrand is a multi-variable fucntion of \(x\) and \(y\); also, we can see that the variables inside of the integrand (\(x\) and \(y\)) are in terms of the limits of integration with respect to \(S\). Thus, the integral in Equation (2), at least initially, looks pretty complicated to solve. But we'll see that by making two simplifications, the integrals \(∫_Cf(x,y)dS\) will simplify to something that your already familiar with—namely, it'll simplify to a definite integral as we'll see shortly. The first simplification that we'll have to make is to express each \((x,y)\) coordinate on the curve \(C\) as parametric functions of some variable \(t\). Doing so, we have \(x=x(t)\) and \(y=y(t)\). This allows us to write \(S(x,y)\) as \(S(x(t),y(t))=S(t)\). Substituting \(S(x,y)\) with \(S(x(t),y(t)\) into Equation (2), we have

$$∫_Cf(x,y)dS=∫_Cf(x(t),y(t))dS.\tag{3}$$

 Figure 2: By the Pythagorean theorem, each infinitesimal arc length \(ds\) can be represented in terms of \(x\) and \(y\) as \(\sqrt{x+2+y^2}\).

Figure 2: By the Pythagorean theorem, each infinitesimal arc length \(ds\) can be represented in terms of \(x\) and \(y\) as \(\sqrt{x+2+y^2}\).

Now the question that you might be asking yourself is: what was the point of expressing the integrand in terms of \(t\). This brings us to the second simplification that we'll want to do to Equation (3): express \(dS\) as a function of \(t\). To do this, let's start out by writing \(dS\) in terms of \(x\) and \(y\). If you imagined looking at Figure 1 from a "birds eye" view, you would see the curve \(C\) lying along the flat \(xy\)-plane as illustrated in Figure 2. As you can see from Figure 2, the length \(dS\) forms the hypotenuse of a right triangle where the lengths of the two other sides of the triangle are given by \(dx\) and \(dy\). Using the Pythagorean theorem, we can rewrite \(dS\) as

$$dS=\sqrt{dx^2+dy^2}.$$

Now, let's multiply the right-hand side of the above equation by \(dt/dt\) (\(=1\)) and so some algebraic simplifications to get

$$dS=\frac{1}{dt}\sqrt{dx^2+dy^2}dt=\sqrt{\frac{1}{dt^2}(dx^2+dy^2)}dt=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt=\sqrt{(x'(t))^2+(y'(t))^2}dt.$$

Substituting the above expression into Equation (3), we have

$$∫_Cf(s)dS=∫_{t_0}^tf(x(t),y(t))\sqrt{(x'(t))^2+(y'(t))^2}dt.\tag{4}$$

As you can see on the right-hand side of Equation (4), both the integrand and the limits of integration are expressed in terms of the single variable \(t\). Equation (4) is nice since it means that all line integrals can be evaluated as a definite integral of a single variable. Expression (1) represents, conceptuaally, what a line integral actually is (the area between a function and a curved line \(C\)); but Equation (4) is what we use to the line integral.


This article is licensed under a CC BY-NC-SA 4.0 license.

Notes

1. For example, if the curve \(C_1C_{1m}\) in Figure 2 has an arc length of \(S=1m\), then \(S(x_{1m},y_{1m})=1m\); whereas, if the curve \(arcC_1C_{2m}\) in Figure 2 has an arc length of \(s=2m\), then \(s(x_{2m},y_{2m})=2m\). Thus, the value of \(s\) depends upon the values of both \(x\) and \(y\) and to capture this fact we write the function \(S(x,y)\).


References

1. Khan Academy. "Introduction to the line integral | Multivariable Calculus | Khan Academy". Online video clip. YouTube. YouTube, 16 December 2016. Web. 27 March 2017.

2. Atheia. How to Calculate Line Integrals. Retrieved from https://www.wikihow.com/Calculate-Line-Integrals