📂Calculus

Scalar Field Line Integral

Line Integral over a Plane Curve¹

Buildup

Given a function as in $y = f(x)$, its definite integral is defined by the idea of adding up all the function values $f(x)$ along the $x$ axis. Thus, the integral value is obtained along a straight line on the $x$ axis.

Now, consider a two-variable function $z=f(x,y)$. Unlike in the case of single-variable functions, since the variable moves over the $xy-$ plane, the integration interval does not necessarily have to be a straight line. One can think of integrating $z=f(x,y)$ along any freely shaped line. Let’s say there’s a smooth curve $C$ represented by the parametric equation $x=x(t), y=y(t), a\le t \le b$ as shown in the picture below.

Let $\Delta s_{i}$ be the length of the arc divided by points $P_{i}$, and let an arbitrary point within it be $(x_{i}^{\ast}, y_{i}^{\ast})$. Then, the area of the $f$ graph along the curve $C$ can be approximated as follows.

$$\sum \limits_{i=1}^{n} f(x_{i}^{\ast}, y_{i}^{\ast})\Delta s_{i}$$

As $n$ increases, it will get closer to the actual area.

Definition

Let’s say curve $C$ is a smooth curve expressed by the parametric equation $x=x(t), y=y(t), a\le t \le b$. Let’s say $f$ is a function defined on $C$. If the following limit exists, it is defined as the line integral of $f$ along $C$ and is denoted as follows.

$$\int_{C} f(x,y) ds = \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} f(x_{i}^{\ast}, y_{i}^{\ast})\Delta s_{i}$$

Explanation

Length of a Curve

$$L = \int_{C} ds = \int_{a}^{b} \sqrt{\left( \dfrac{d x}{d t} \right)^{2} + \left( \dfrac{d y}{d t} \right)^{2}} dt$$

This can be thought of as multiplying the function value $f(x,y)$ by a weight to calculate the length of the curve. Therefore, the following equation is obtained.

$$\int_{C} f(x,y) ds = \int_{a}^{b} f\left( x(t), y(t) \right)\sqrt{\left(\dfrac{d x}{d t}\right)^{2} + \left(\dfrac{d y}{d t}\right)^{2}}dt$$

Also, if we set $\mathbf{r}(t) = \left( x(t), y(t) \right)$, since $\mathbf{r}^{\prime} = \left( x^{\prime}(t), y^{\prime}(t) \right)$, the following holds.

$$\begin{align*} \int_{C} f(x,y) ds =&\ \int_{a}^{b} f\left( x(t), y(t) \right)\sqrt{\left(\dfrac{d x}{d t}\right)^{2} + \left(\dfrac{d y}{d t}\right)^{2}}dt \\ =&\ \int_{a}^{b} f\left( \mathbf{r}(t) \right) \left| \mathbf{r}^{\prime}(t) \right| dt \end{align*}$$

It is possible to think of integration in terms of lengths $dx, dy$ at each coordinate axis, rather than the length of the curve $ds$, as follows.

$$\begin{align*} \int_{C} f(x,y) dx =&\ \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} f(x_{i}^{\ast}, y_{i}^{\ast})\Delta x_{i} = \int_{a}^{b} f\left( x(t), y(t) \right) x^{\prime}(t) dt \\ \int_{C} f(x,y) dy =&\ \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} f(x_{i}^{\ast}, y_{i}^{\ast})\Delta y_{i} = \int_{a}^{b} f\left( x(t), y(t) \right) y^{\prime}(t) dt \end{align*}$$

Such forms can be encountered in the line integral of a vector field.

Line Integral over a Space Curve

The line integral of a three-dimensional function is naturally defined in the same way as the two-dimensional line integral.

Definition

Let’s say curve $C$ is a smooth curve expressed by the parametric equation $x=x(t), y=y(t), z=z(t), a\le t \le b$. Let’s say $f$ is a function defined on $C$. If the following limit exists, it is defined as the line integral of $f$ along $C$ and is denoted as follows.

$$\int_{C} f(x,y,z) ds = \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} f(x_{i}^{\ast}, y_{i}^{\ast}, z_{i}^{\ast})\Delta s_{i}$$

Explanation

The formulas for two dimensions are also valid for three dimensions.

$$\int_{C} f(x,y,z) ds = \int_{a}^{b} f\left( x(t), y(t), z(t) \right)\sqrt{\left(\dfrac{d x}{d t}\right)^{2} + \left(\dfrac{d y}{d t}\right)^{2} + \left(\dfrac{d z}{d t}\right)^{2}}dt$$