Line Integrals of Vector Fields Examples 1

# Line Integrals of Vector Fields Examples 1

Recall from the Line Integrals of Vector Fields page that if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a continuous vector field on $\mathbb{R}^3$ and if $C$ is a smooth curve given parametrically by $x = x(t)$, $y = y(t)$, and $z = z(t)$ for $a ≤ t , b$, then we take the line integral of the field $\mathbf{F}$ along $C$ and it is defined to be:

(1)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_C P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \: dz \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_a^b \left [ 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) \right ] \: dt \end{align}

Of course, line integrals of fields in lower and higher dimensions can be evaluated similarly.

We will now look at some examples of evaluating the line integral of a field $\mathbf{F}$ along a curve $C$.

## Example 1

Let $\mathbf{F}(x, y) = (\sin x, \cos y, xz)$ and let $\vec{r}(t) = (t^3, -t^2, t)$ trace the curve $C$ for $0 ≤ t ≤ 1$. Evaluate $\int_C \mathbf{F} \cdot d \vec{r}$.

Using the formula directly and we have that:

(2)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_C \sin x \: dx + \cos y \: dy + xz \: dz \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_0^1 \sin (t^3) (3t^2) \: dt + \int_0^1 \cos (-t^2) (-2t) \: dt + \int_0^1 (t^3)(t) (1) \: dt \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = 3\int_0^1 t^2 \sin (t^3) \: dt - 2\int_0^1 t\cos (-t^2) \: dt + \int_0^1 t^4 \: dt \end{align}

We will now use two different substitutions to evaluate the first two integrals. Let $u = t^3$. Then $du = 3t^2 \: dt$. Also let $v = -t^2$. Then $dv = -2t \: dt$ and so for some $\alpha_1, \alpha_2, \beta_1, \beta_2 \in \mathbb{R}$ we have that:

(3)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_{\alpha_1}^{\beta_1} \sin u \: du + \int_{\alpha_2}^{\beta_2} \cos v \: dv + \left [ \frac{t^5}{5} \right ]_{t=0}^{t=1} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \left [ -\cos t^3 \right ]_{t=0}^{t=1} + \left [ \sin (-t^2) \right ]_{t=0}^{t=1} + \frac{1}{5} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = 1 - \cos (1) + \sin(-1) + \frac{1}{5} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \frac{6}{5} + \sin(-1) - \cos(1) \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \frac{6}{5} - \sin (1) - \cos (1) \end{align}

The last equality comes from the fact that $\sin x$ is an odd function and so $\sin (-x) = - \sin x$.

## Example 2

Let $\mathbf{F}(x, y) = e^{x-1} \vec{i} + xy \vec{j}$ be a vector field on $\mathbb{R}^2$ and let $\vec{r}(t) = t^2 \vec{i} + t^3 \vec{j}$ parameterize the curve $C$ for $0 ≤ t ≤ 1$. Evaluate $\int_C \mathbf{F} \cdot d \vec{r}$.

Using the formula directly and we have that:

(4)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_C e^{x-1} \: dx + xy \: dy \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_0^1 e^{t^2 - 1} (2t) \: dt + \int_0^1 (t^2)(t^3) (3t^2) \: dt \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_0^1 2te^{t^2 - 1} \: dt + \int_0^1 3t^7 \: dt \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_0^1 2te^{t^2 - 1} \: dt + 3 \left \frac{t^8}{8} \right ]_{t=0}^{t=1} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_0^1 2te^{t^2 - 1} \: dt + \frac{3}{8} \end{align}

Now let $u = t^2 - 1$. Then $du = 2t \: dt$ and so for some $\alpha, \beta \in \mathbb{R}$ we have that:

(5)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_{\alpha}^{\beta} e^{u} \: du + \frac{3}{8} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \left [ e^{t^2 - 1} \right ]_{t=0}^{t=1} + \frac{3}{8} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = 1 - \frac{1}{e} + \frac{3}{8} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = \frac{11}{8} - \frac{1}{e} \end{align}