Line Integrals of Vector Fields

Recall from the Line Integrals page that if $z = f(x, y)$ is a two variable real-valued function and $C$ is a smooth curve parameterized as $\vec{r}(t) = (x(t), y(t))$ for $a ≤ t ≤ b$, then the line integral of $f$ along $C$ is given by the following formula:

(1)

\begin{align} \quad \int_C f(x, y) \: ds = \int_a^b f(x(t), y(t)) \sqrt{ \left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2 } \: dt \end{align}

Similarly if $w = f(x, y, z)$ is a three variable real-valued function and $C$ is a smooth curve parameterized as $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$, then the line integral of $f$ along $C$ is given by the following formula:

(2)

\begin{align} \quad \int_C f(x, y, z) \: ds = \int_a^b f(x(t), y(t), z(t)) \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2 + \left ( \frac{dz}{dt} \right )^2} \: dt \end{align}

Now it is sometimes useful to evaluate line integrals over vector fields $\mathbf{F}$. Such line integrals appear frequently in physics. Let's first define a line integral over a vector field on $\mathbb{R}^3$. The results below have analogous counterparts in $\mathbb{R}^2$. Let $\mathbf{F} (x, y, z) = P(x, y, z)\vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ be continuous on $\mathbb{R}^3$. We will shorten this as simple $\mathbf{F} = P\vec{i} + Q \vec{j} + R \vec{k}$. Let $C$ be a smooth curve that is given by the parametric equations $x = x(t)$, $y = y(t)$, and $z = z(t)$ for $a ≤ t ≤ b$ and define $\vec{r}(t) = (x(t), y(t), z(t))$.

Now take the parameter $t$'s interval $[a, b]$ and divide it into $n$ subintervals of equal width. The endpoints of these subintervals correspond to points on the curve $P_0$, $P_1$, …, $P_n$. These points divide the curve $C$ into $n$ sub arcs which are arcs between the points $P_{i-1}$ and $P_i$ on the curve. Let $\Delta s_i$ be the lengths of each of these arc. Now choose any point $P_i^*(x_i^*, y_i^*, z_i^*)$ between $P_{i-1}$ and $P_i$ that corresponds to a $t_i^* \in [t_{i-1}, t_i]$.

Now for small $\Delta s_i$, then the progression as we move from $P_{i-1}$ to $P_i$ is approximately in the direction of the unit tangent vector at $P_i^*$, that is, approximately the direction of $\mathbf{\hat{T}} (t_i^*)$. We define a Riemann sum of the dot products between the field vector $\mathbf{F} (x_i^*, y_i^*, z_i^*)$ at $(x_i^*, y_i^*, z_i^*)$ and these tangent vectors with lengths $\Delta s_i$, $\mathbf{\hat{T}}(t_i^*)$ at $(x_i^*, y_i^*, z_i^*)$:

(3)

\begin{align} \quad \sum_{i=1}^{n} F(x_i^*, y_i^*, z_i^*) \cdot [\Delta s_i \mathbf{\hat{T}} (x_i^*, y_i^*, z_i^*)] = \sum_{i=1}^{n} [F(x_i^*, y_i^*, z_i^*) \cdot \mathbf{\hat{T}} (x_i^*, y_i^*, z_i^*)] \Delta s_i \end{align}

If we let $n \to \infty$, then we obtain a line integral which we define below.

Definition: Let $\mathbf{F}(x, y, z) = P\vec{i} +Q\vec{j} + R \vec{k}$ be a continuous field on $\mathbb{R}^3$ and let $C$ be a smooth curve parameterized by the equations $x = x(t)$, $y = y(t)$, and $z = z(t)$ where $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$. Then the Line Integral of $\mathbf{F}$ Along $C$ is defined as $\int_C \mathbf{F} (x, y, z) \cdot \mathbf{\hat{T}} (x, y, z) \: ds = \lim_{n \to \infty} \sum_{i=1}^{n} [\mathbf{F}(x_i^*, y_i^*, z_i^*) \cdot \mathbf{\hat{T}}(x_i^*, y_i^*, z_i^*)] \Delta s_i$ provided that this limit exists.

Another name for the integral above is the Line Integral of The Tangential Component of $\mathbf{F}$. A short hand form to rewrite the integral above is $\int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \int_C \mathbf{F} (x, y, z) \cdot \mathbf{\hat{T}} (x, y, z) \: ds$. The following notation $\int_C \mathbf{F} \cdot d\vec{r} = \int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds$ is also commonly used.

Now note that the curve $C$ is given by $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$. From the Unit Tangent Vectors to a Space Curve page, we saw that thus $\mathbf{\hat{T}}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. Further, we note that $\frac{ds}{dt} = \| \vec{r'}(t) \|$ and so $ds = \| \vec{r'}(t) \| \:dt$ and so:

(4)

\begin{align} \quad \int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \int_a^b \mathbf{F}(r(t)) \cdot \mathbf{\hat{T}(t)} \: ds = \int_C \mathbf{F} \cdot \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} \| \vec{r'}(t) \| \: dt = \int_a^b \mathbf{F}(r(t)) \cdot \vec{r'}(t) \: dt \end{align}

Now the following theorem will draw a connection between the line integral of a vector field and line integrals of scalar fields.

Theorem 1: If $\mathbf{F}(x, y, z) = P(x, y, z)\vec{i} + Q(x, y, z)\vec{j}$ is a continuous vector field on $\mathbb{R}^3$ and if $C$ is a smooth curve parameterized by the equations $x = x(t)$, $y = y(t)$, and $z = z(t)$ where $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$, then $\int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \int_C P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \: dz$.

Of course we can write this in short form as $\int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \int_C P \: dx + Q \: dy + R \: dz$ or $\int_C \mathbf{F} \cdot d\vec{r} = \int_C P \: dx + Q \: dy + R \: dz$.

• Proof: Showing Theorem 1 is true can be done by definition.

(5)

\begin{align} \quad \int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \int_a^b \mathbf{F}(\vec{r}(t)) \cdot \vec{r'}(t) \: dt = \int_a^b (P\vec{i}, Q\vec{j}, R\vec{k}) \cdot (x'(t)\vec{i}, y'(t)\vec{j}, z'(t)\vec{k}) \: dt \\ = \int_a^b P(x(t), y(t), z(t)) x'(t) + Q(x(t), y(t), z(t)) y'(t) + R(x(t), y(t), z(t)) z'(t) \: dt \\ = \int_a^b P(x(t), y(t), z(t)) x'(t) \: dt + \int_a^b Q(x(t), y(t), z(t)) y'(t) \: dt + \int_a^b R(x(t), y(t), z(t)) z'(t) \: dt \\ = \int_C P(x, y, z) \: dx + \int_C Q(x, y, z) \: dy + \int_C R(x, y, z) \: dz = \int_C P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \: dz \quad \blacksquare \end{align}

Remark 1: We should note that if the direction along $C$ is reversed, denote it as $-C$, then the line integral above changes sign, that is $\int_{-C} \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = - \int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds$.

Remark 2: If $C$ is a closed curve, then often times a little circle on the integral symbol is used notationally to represent the integral above, that is $\int_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds = \oint_C \mathbf{F} \cdot \mathbf{\hat{T}} \: ds$, or equivalently, $\int_C \mathbf{F} \cdot dr = \oint_C \mathbf{F} \cdot dr$, which denotes the Circulation of the vector field $\mathbb{F}$ around $C$.