Vector line integrals

5 Vector line integrals

It is also possible to form the less commonly used integrals $\int_{C} f (x, y, z) \underset{̲}{d r}$ and $\int_{C} \underset{̲}{F} (x, y, z) \times \underset{̲}{d r}$ . Each of these integrals evaluates to a vector.

Remembering that $\underset{̲}{d r} = d x \underset{̲}{i} + d y \underset{̲}{j} + d z \underset{̲}{k}$ , an integral of the form $\int_{C} f (x, y, z) \underset{̲}{d r}$ becomes

$\int_{C} f (x, y, z) d x \underset{̲}{i} + \int_{C} f (x, y, z) d y \underset{̲}{j} + \int_{C} f (x, y, z) d z \underset{̲}{k}$ . The first term can be evaluated by expressing $y$ and $z$ in terms of $x$ . Similarly the second and third terms can be evaluated by expressing all terms as functions of $y$ and $z$ respectively. Alternatively, all variables can be expressed in terms of a parameter $t$ . If an integral is two-dimensional, the term in $z$ will be absent.

Example 12

Evaluate the integral $\int_{C} x y^{2} \underset{̲}{d r}$ where $C$ represents the contour $y = x^{2}$ from $(0, 0)$ to $(1, 1)$ .

Solution

This is a two-dimensional integral so the term in $z$ will be absent.

$\begin{array}{rcl} I & = & \int_{C} x y^{2} \underset{̲}{d r} \\ = & \int_{C} x y^{2} (d x \underset{̲}{i} + d y \underset{̲}{j}) \\ = & \int_{C} x y^{2} d x \underset{̲}{i} + \int_{C} x y^{2} d y \underset{̲}{j} \\ = & \int_{x = 0}^{1} x {(x^{2})}^{2} d x \underset{̲}{i} + \int_{y = 0}^{1} y^{1 ∕ 2} y^{2} d y \underset{̲}{j} \\ = & \int_{0}^{1} x^{5} d x \underset{̲}{i} + \int_{0}^{1} y^{5 ∕ 2} d y \underset{̲}{j} \\ = & {[\frac{1}{6} x^{6}]}_{0}^{1} \underset{̲}{i} + {[\frac{2}{7} x^{7 ∕ 2}]}_{0}^{1} \underset{̲}{j} \\ = & \frac{1}{6} \underset{̲}{i} + \frac{2}{7} \underset{̲}{j} \end{array}$

Example 13

Find $I = \int_{C} x \underset{̲}{d r}$ for the contour $C$ given parametrically by $x = cos t$ , $y = sin t$ , $z = t - π$ starting at $t = 0$ and going to $t = 2 π$ , i.e. the contour starts at $(1, 0, - π)$ and finishes at $(1, 0, π)$ .

Solution

The integral becomes $\int_{C} x (d x \underset{̲}{i} + d y \underset{̲}{j} + d z \underset{̲}{k}) .$

Now, $x = cos t$ , $y = sin t$ , $z = t - π$ so $d x = - sin t d t$ , $d y = cos t d t$ and $d z = d t$ . So

$\begin{array}{rcl} I & = & \int_{0}^{2 π} cos t (- sin t d t \underset{̲}{i} + cos t d t \underset{̲}{j} + d t \underset{̲}{k}) \\ = & - \int_{0}^{2 π} cos t sin t d t \underset{̲}{i} + \int_{0}^{2 π} {cos}^{2} t d t \underset{̲}{j} + \int_{0}^{2 π} cos t d t \underset{̲}{k} \\ = & - \frac{1}{2} \int_{0}^{2 π} sin 2 t d t \underset{̲}{i} + \frac{1}{2} \int_{0}^{2 π} (1 + cos 2 t) d t \underset{̲}{j} + {[sin t]}_{0}^{2 π} \underset{̲}{k} \\ = & \frac{1}{4} {[cos 2 t]}_{0}^{2 π} \underset{̲}{i} + \frac{1}{2} {[t + \frac{1}{2} sin 2 t]}_{0}^{2 π} \underset{̲}{j} + 0 \underset{̲}{k} \\ = & 0 \underset{̲}{i} + π \underset{̲}{j} = π \underset{̲}{j} \end{array}$

Integrals of the form $\int_{C} \underset{̲}{F} \times \underset{̲}{d r}$ can be evaluated as follows. The vector field $\underset{̲}{F} = F_{1} \underset{̲}{i} + F_{2} \underset{̲}{j} + F_{3} \underset{̲}{k}$ and $\underset{̲}{d r} = d x \underset{̲}{i} + d y \underset{̲}{j} + d z \underset{̲}{k}$ so

$\begin{array}{rcl} \underset{̲}{F} \times \underset{̲}{d r} = |\begin{matrix} \underset{̲}{i} & \underset{̲}{j} & \underset{̲}{k} \\ F_{1} & F_{2} & F_{3} \\ d x & d y & d z \end{matrix}| & = & (F_{2} d z - F_{3} d y) \underset{̲}{i} + (F_{3} d x - F_{1} d z) \underset{̲}{j} + (F_{1} d y - F_{2} d x) \underset{̲}{k} \\ = & (F_{3} \underset{̲}{j} - F_{2} \underset{̲}{k}) d x + (F_{1} \underset{̲}{k} - F_{3} \underset{̲}{i}) d y + (F_{2} \underset{̲}{i} - F_{1} \underset{̲}{j}) d z \end{array}$

There are a maximum of six terms involved in one such integral; the exact details may dictate which form to use.

Example 14

Evaluate the integral $\int_{C} (x^{2} \underset{̲}{i} + 3 x y \underset{̲}{j}) \times \underset{̲}{d r}$ where $C$ represents the curve $y = 2 x^{2}$ from $(0, 0)$ to $(1, 2)$ .

Solution

Note that the $z$ component of $\underset{̲}{F}$ and $\underset{̲}{d r}$ are both zero.

So $\underset{̲}{F} \times \underset{̲}{d r} = |\begin{matrix} \underset{̲}{i} & \underset{̲}{j} & \underset{̲}{k} \\ x^{2} & 3 x y & 0 \\ d x & d y & 0 \end{matrix}| = (x^{2} d y - 3 x y d x) \underset{̲}{k}$

and $\int_{C} (x^{2} \underset{̲}{i} + 3 x y \underset{̲}{j}) \times \underset{̲}{d r} = \int_{C} (x^{2} d y - 3 x y d x) \underset{̲}{k}$

Now, on $C$ , $y = 2 x^{2}$ so $d y = 4 x d x$ and

$\begin{array}{rcl} \int_{C} (x^{2} \underset{̲}{i} + 3 x y \underset{̲}{j}) \times \underset{̲}{d r} & = & \int_{C} {x^{2} d y - 3 x y d x} \underset{̲}{k} \\ = & \int_{x = 0}^{1} \{x^{2} \times 4 x d x - 3 x \times 2 x^{2} d x\} \underset{̲}{k} \\ = & \int_{0}^{1} - 2 x^{3} d x \underset{̲}{k} \\ = & - {[\frac{1}{2} x^{4}]}_{0}^{1} \underset{̲}{k} \\ = & - \frac{1}{2} \underset{̲}{k} \end{array}$