Line Integrals

# Line Integrals of Function of Two Variables

We will now introduce a new type of integral known as a line integral. Let $z = f(x, y)$ be a two variable real-valued function, and consider a curve $C$ given parametrically by the equations $x = x(t)$ and $y = y(t)$ for $a ≤ t ≤ b$ that is contained in the domain of $f$. We can merge these two parametric equations nicely into a vector equation $\vec{r}(t) = (x(t), y(t)) = x(t) \vec{i} + y(t) \vec{j}$. Suppose further that $C$ is a smooth curve, that is, $\vec{r'}(t) \neq 0$ and $\vec{r'}(t)$ is continuous for $a ≤ t ≤ b$. Now we will take the interval $[a, b]$ for the parameter $t$ and divide it up into $n$ equal-width subintervals $[t_{i-1}, t_i]$. Then let $\vec{r}(t_i) = (x(t_i), y(t_i)) = (x_i, y_i)$ be the point $P_i(x_i, y_i)$ on $C$. The curve of $C$ is thus divided into $n$ sub arcs that are partitioned by the points $P_i$ on $C$. The lengths of each of these arcs will be $\Delta s_1, \Delta s_2, ...,\Delta s_n$. Now choose a sample number $t_i^* \in [t_{i-1}, t_i]$. This value $t_i^*$ corresponds to the point $P_i^*(x_i^*, y_i^*)$ on $C$ and between $P_{i-1}$ and $P_i$. Now we evaluate $f$ at the point $(x_i^*, y_i^*)$ and multiply by the length $\Delta s_i$ then we get the area of of rectangle that is standing on the $xy$ plane with height $f(x_i^*, y_i^*)$ and length $\Delta s_i$. Thus the area of each of these rectangles is $f(x_i^*, y_i^*)$. We now sum up all of these rectangles:

(1)
\begin{align} \quad \sum_{i=1}^{n} f(x_i^*, y_i^*) \Delta s_i \end{align} If $f(x, y) ≥ 0$ for points $(x, y)$ on $C$, then as $n \to \infty$, this sum represents the area of the upright "fence" standing on the $xy$-plane. We formally define this limit to be the line integral of $f$ along the curve $C$ which we define below.

 Definition: Let $z = f(x, y)$ be a two variable real-valued function, and let $C$ be a smooth curve given by the parametric equations $x = x(t)$ and $y = y(t)$ for $a ≤ t ≤ b$. Then the Line Integral of $f$ Along $C$ is defined as $\int_C f(x, y) \: ds = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*, y_i^*) \Delta s_i$ provided that this limit exists.

Now recall that $\frac{ds}{dt}$ is precisely the speed function of $\vec{r}(t)$, that is:

(2)
\begin{align} \quad \frac{ds}{dt} = \| \vec{r'}(t) \| = \sqrt{ \left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2} \end{align}

Therefore $ds = \sqrt{ \left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2} \: dt$, and so we can evaluate a line integral with the following formula:

(3)
\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 = \int_a^b f(\vec{r}(t)) \| \vec{r'}(t) \| \: dt \end{align}

Furthermore, we have generalized the definite integral of a single variable function to a two variable function. Note that if we want to compute the line integral of $f$ from the point $(a, 0)$ to $(b, 0)$ then we have that this curve can be parameterized as $\vec{r}(t) = (x(t), y(t)) = (x, 0)$ and so $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = 0$ so:

(4)
\begin{align} \quad \int_C f(x, y) \: ds = \int_a^b f(x, 0) \sqrt{1} \: dx = \int_a^b f(x, 0) \: dx \end{align}

It is important to note that the value of the line integral of $f$ along $C$ does NOT depend of the parameterization of $C$. We prove this statement in the following proposition.

 Proposition 1: If $z = f(x, y)$ is a two variable real-valued function and $C$ is a smooth plane curve contained in the domain of $f$ and parameterized as $\vec{r}(t) = (x(t), y(t))$ for $a ≤ t ≤ b$ and $\vec{r^*}(u)$ for $\alpha ≤ u ≤ \beta$ then $\int_a^b f(\vec{r}(t)) \| \vec{r'}(t) \| \: dt = \int_{\alpha}^{\beta} f(\vec{r^*}(u)) \| \vec{r^*{'}}(u) \| \: du$.
• Proof: Suppose that $\vec{r}(t) = (x(t), y(t))$ for $a ≤ t ≤ b$ and $\vec{r^*}(u)$ for $\alpha ≤ u ≤ \beta$ both represent the plane curve $C$. Then any point $\vec{r}(t)$ on the curve $C$ can be represented in terms of the parameterization $\vec{r^*}(u)$ where $u$ depends on $u$ depends on $t$, that is $u = u(t)$.
• Now there are two cases to consider. If $\vec{r^*}(u)$ traces $C$ in the same direction as $\vec{r}(t)$ then we have that $u(a) = \alpha$, $u(b) = \beta$, and $\frac{du}{dt} ≥ 0$. Thus:
(5)
\begin{align} \quad \int_a^b f(\vec{r}(t)) \biggr \| \frac{d \vec{r}}{dt} \biggr \| \: dt = \int_a^b f(\vec{r^*}(u(t)) \biggr \| \frac{d \vec{r^*}}{du} \frac{du}{dt} \biggr \| \: dt= \int_a^b f(\vec{r^*}(u(t)) \biggr \| \frac{d \vec{r^*}}{du} \biggr \| \frac{du}{dt} \: dt = \int_{\alpha}^{\beta} f(\vec{r^*}(u)) \biggr \| \frac{d \vec{r^*}}{du} \biggr \| \: du \end{align}
• Similarly, if $\vec{r^*}(u)$ traces $C$ in the opposite direction a $\vec{r}(t)$ then we have that $u(a) = \beta$ and $u(b) = \alpha$ and $\frac{du}{dt} ≤ 0$, so:
(6)
\begin{align} \quad \: \int_a^b f(\vec{r}(t)) \biggr \| \frac{d \vec{r}}{dt} \biggr \| \: dt = \int_a^b f(\vec{r^*}(u(t)) \biggr \| \frac{d \vec{r^*}}{du} \frac{du}{dt} \biggr \| \: dt= \int_a^b f(\vec{r^*}(u(t)) \biggr \| \frac{d \vec{r^*}}{du} \biggr \| \left ( -\frac{du}{dt} \right ) \: dt = -\int_{\beta}^{\alpha} f(\vec{r^*}(u)) \biggr \| \frac{d \vec{r^*}}{du} \biggr \| \: du = \int_{\alpha}^{\beta} f(\vec{r^*}(u)) \biggr \| \frac{d \vec{r^*}}{du} \biggr \| \: du \end{align}
• Thus our proof is complete. $\blacksquare$

We should also be weary of whether or not this curve is traversed once for $a ≤ t ≤ b$.

# Line Integrals of Functions of Three Variables

 Definition: Let $w = f(x, y, z)$ be a three variable real-valued function, and let $C$ be a smooth space curve given by the parametric equations $x = x(t)$, $y = y(t)$, $z = z(t)$ for $a ≤ t ≤ b$. Then the Line Integral of $f$ Along $C$ is defined as $\int_C f(x, y) \: ds = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*, y_i^*, z_i^*) \Delta s_i$ provided that this limit exists.

Like with line integrals of functions of two variables, the following formula allows us to readily evaluate such line integrals:

(7)
\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 = \int_a^b f(\vec{r}(t)) \| \vec{r'}(t) \| \: dt \end{align}