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Advanced Engineering Mathematics Lecture Notes: West Virginia Wesleyan College

Jesse Oldroyd

Section 4.1 Line Integrals

An integral is fundamentally determined by two objects: the integrand and the domain of integration. In this section, we study integrals that involve one-dimensional domains of integration. Such integrals are known as line integrals. We will have a different type of line integral depending on whether our integrand is a scalar field or a vector field.

Subsection Scalar Line Integrals

Consider the problem of finding the mass of a wire that follows a path \(C\) in either \(\RR^2\) or \(\RR^3\) given a density function \(\rho\text{.}\) As seen in calculus, the mass can be found by integrating the density. This leads to integrals of the form \(\int_C \rho\dd{s}\) which are known as (scalar) line integrals. The fundamental domain of integration for a line integral is a smooth curve.

Definition 4.1.1. Smooth Curves.

Let \(C\) denote a curve in \(\RR^n\) parameterized by a vector function \(\rr(t)\) defined on an interval \(I\text{.}\) We say that \(C\) is smooth if \(\rr(t)\) is differentiable and \(\rrp(t)\neq\vb{0}\) anywhere on \(I\text{.}\)
We can rephrase Definition 4.1.1 as follows: \(C\) is smooth if and only if it has a smooth parameterization \(\rr(t)\text{.}\) With this in mind, we can now define line integrals over smooth curves.

Definition 4.1.2. Scalar Line Integrals.

Let \(C\) be a smooth curve in \(\RR^2\text{.}\) Let \(f(x, y)\) denote a function that is continuous on some open set containing \(C\text{.}\) The scalar line integral of \(f\) over \(C\) is given by the limit
\begin{equation*} \int_C f(x, y)\dd{s} = \lim_{\Delta s\to0}\sum_{k=1}^{n}f(x_k, y_k)\Delta s \end{equation*}
where the points \((x_k, y_k)\) are lie on \(C\text{,}\) and \(\Delta s\) gives the length of the line segment from one such point to the next.
Geometrically, we view the line integral \(\int_C f(x, y)\dd{s}\) as giving the area between the curve \(C\) in the \(xy\)-plane and the surface \(z = f(x, y)\text{.}\) Although Definition 4.1.2 assumes that \(C\) lies in \(\RR^2\text{,}\) the definition can be extended to higher dimensions.
The definition given in Definition 4.1.2, while important, will not typically be used for computing line integrals. Instead, we will take another approach using arc length. The differential \(\dd{s}\) in the line integral formula corresponds to the arc length function along \(C\text{.}\) Therefore we may write
\begin{equation*} \dd{s} = \norm{\rrp(t)}\dd{t} \end{equation*}
and
\begin{equation} \int_C f(x, y)\dd{s} = \int_a^b f(\rr(t))\norm{\rrp(t)}\dd{t}\tag{1} \end{equation}
for some suitable smooth parameterization \(\rr(t), a\leq t\leq b\) of \(C\text{.}\)

Example 4.1.3. Computing a Scalar Line Integral.

Compute \(\int_C (xy)^{\frac{1}{3}}\dd{s}\) where \(C\) is the parabolic segment \(y = x^2\) from \(x = 0\) to \(x = 1\text{.}\)
Solution.
First, we need to parameterize \(C\text{.}\) This can be done using \(\rr(t) = \mqty[t \amp t^2]\) for \(0\leq t\leq 1\text{.}\) (1) then gives
\begin{align*} \int_C (xy)^{\frac{1}{3}}\dd{s} \amp = \int_0^1 t\sqrt{1 + 4t^2}\dd{t} \\ \amp = \int_1^5\frac{1}{8}\sqrt{u}\dd{u} \\ \amp = \frac{5^{\frac{3}{2}} - 1}{12} \text{.} \end{align*}
Recall that Definition 4.1.2 defines scalar line integrals over smooth curves only. If a curve \(C\) is piecewise smooth, which means that it can be written as the union of finitely many smooth curves \(C_1\) through \(C_n\text{,}\) then we can extend Definition 4.1.2 to \(C\) as well. All we need to do is compute the line integral along the smooth components \(C_i\) and add them up:
\begin{equation*} \int_C f(x, y)\dd{s} = \sum_{i=1}^n \int_{C_i} f(x, y)\dd{s}\text{.} \end{equation*}

Example 4.1.4. Computing a Line Integral Along a Piecewise Smooth Curve.

Let \(C\) denote the curve consisting of the line segment from \((-1, 0)\) to \((0, 1)\) followed by the line segment from \((0, 1)\) to \((1,0)\text{.}\) Find \(\int_C (2x - 3y)\dd{s}\text{.}\)
Solution.
We need to decompose \(C\) into smooth components. The natural choice is to pick \(C_1\) to be the first segment and \(C_2\) to be the second segment. These curves are smooth and have the following smooth parameterizations:
\begin{align*} C_1 : \rr_1(t) \amp = \mqty[t - 1 \amp t], 0\leq t\leq 1 \\ C_2 : \rr_2(t) \amp = \mqty[t \amp 1-t], 0\leq t\leq 1 \text{.} \end{align*}
Now we can compute the resulting line integrals:
\begin{align*} \int_{C_1} (2x - 3y)\dd{s} \amp = \int_0^1 [2(t-1) - 3t]\sqrt{2}\dd{t} \\ \amp = \int_0^1 (-2 - t)\sqrt{2}\dd{t} \\ \amp = -\frac{5}{2}\sqrt{2} \text{.} \end{align*}
Likewise,
\begin{equation*} \int_{C_1} (2x - 3y)\dd{s} = \int_0^1 (5t - 3)\sqrt{2}\dd{t} = -\frac{1}{2}\sqrt{2}\text{.} \end{equation*}
Adding these integrals together, we get
\begin{equation*} \int_C (2x - 3y)\dd{s} = -3\sqrt{2}\text{.} \end{equation*}

Subsection Line Integrals of Vector Fields

Using the scalar line integral in Definition 4.1.2, we can also define line integrals of vector fields along smooth and piecewise-smooth curves.

Definition 4.1.5. Vector Line Integrals.

Let \(C\) be a (piecewise) smooth curve and let \(\FF\) be a vector field that is continuous on \(C\text{.}\) The vector line integral of \(\FF\) over \(C\) is defined to be
\begin{equation*} \int_{C}\FF\cdot\TT\dd{s}\text{,} \end{equation*}
where \(\TT\) denotes the unit tangent to \(C\) with respect to a previously chosen (piecwise) smooth parameterization of \(C\text{.}\)
We interpret the vector line integral as measuring how well the vector field \(\FF\) and the curve \(C\) align. We also call it the circulation of \(\FF\) along \(C\text{.}\) In physics, Definition 4.1.5 can also be used to find the work done by a force field \(\FF\) on a particle that moves along the curve \(C\text{.}\)
The formula in Definition 4.1.5 can be simplified in much the same way as the original definition of a scalar line integral was in (1). First, suppose that \(C\) is parameterized by the smooth vector function \(\rr(t):[a,b]\to\RR^n\text{.}\) Then
\begin{align*} \FF\cdot\TT\dd{s} \amp = \FF(\rr(t))\cdot\frac{\rrp(t)}{\norm{\rrp(t)}}\norm{\rrp(t)}\dd{t} \\ \amp = \FF(\rr(t))\cdot\rrp(t)\dd{t} \\ \amp = \FF\cdot\dd{\rr} \text{.} \end{align*}
Therefore
\begin{equation} \int_C\FF\cdot\TT\dd{s} = \int_a^b\FF\cdot\dd{\rr}\text{.}\tag{2} \end{equation}

Example 4.1.6. Computing a Circulation.

Find the circulation of \(\FF = (x^2+y^2)^{-1}\mqty[x \amp y]\) along the curve \(C:\rr(t) = \mqty[t \amp 4t], 1\leq t\leq 10\text{.}\)
Solution.
First, note that \(C\) is a smooth curve on which \(\FF\) is continuous and the line integral of \(\FF\) along \(C\) is therefore defined. By (2), it follows that
\begin{equation*} \int_C\FF\cdot\TT\dd{s} = \int_1^{10}\FF\cdot\dd{\rr} \end{equation*}
where
\begin{align*} \FF \amp = \FF(\rr(t)) = \frac{1}{17t^2}\mqty[t \amp 4t] \\ \dd{\rr} \amp = \rrp(t)\dd{t} = \mqty[1 \amp 4]\dd{t} \text{.} \end{align*}
Therefore
\begin{align*} \int_1^{10}\FF\cdot\dd{\rr} \amp = \int_1^{10}\frac{1}{17t^2}(t + 16t)\dd{t} \\ \amp = \int_1^{10}\frac{1}{t}\dd{t} \\ \amp = \ln(10) \text{.} \end{align*}
Note that this is greater than \(0\text{,}\) which indicates that the vector field \(\FF\) and the curve \(C\) tend to be aligned. This can be verified using the following plot produced by Sage:
The vector field \(\FF\) over \(C\text{.}\)
Figure 4.1.7. The vector field \(\FF = (x^2+y^2)^{-1}\mqty[x\amp y]\) over \(C\text{.}\)
Curves that “follow” vector fields precisely are known as flow lines or streamlines. The curve \(C\) in Example 4.1.6 is an example of a streamline of the vector field \(\FF\) in the same example, as can be seen from Figure 4.1.7. It can be shown that all other streamlines of \(\FF\) are curves of the form \(y = cx\) for some constant \(c\text{.}\)
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