Vector Line Integrals

Learning Objectives

Calculate a vector line integral along an oriented curve in space.

The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let

${\textbf{F}}({x},{y},{z}) = {{P}({x},{y},{z}){\textbf{i}}} + {{Q}({x},{y},{z}){\textbf{j}}} + {{R}({x},{y},{z}){\textbf{k}}}$

be a continuous vector field in ${\mathbb{R}}^{3}$ that represents a force on a particle, and let $C$ be a smooth curve in ${\mathbb{R}}^{3}$ contained in the domain of ${\textbf{F}}$ . How would we compute the work done by ${\textbf{F}}$ in moving a particle along $C$ ?

To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve $C$ ; such a specified direction is called an orientation of a curve. The specified direction is the positive direction along $C$ ; the opposite direction is the negative direction along $C$ . When $C$ has been given an orientation, $C$ is called an oriented curve (Figure 1). The work done on the particle depends on the direction along the curve in which the particle is moving.

A closed curve is one for which there exists a parameterization ${\textbf{r}}{(t)}, {a} \leq {t} \leq {b}$ , such that ${\textbf{r}}{(a)} = {\textbf{r}}{(b)}$ and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain $(a, b)$ .

Two images, labeled A and B. Image A shows a curve C that is an oriented curve. It is a curve that connects two points; it is a line segment with curves. Image B, on the other hand, is a closed curve. It has no endpoints and completely encloses an area.

Figure 1. (a) An oriented curve between two points. (b) A closed oriented curve.

Let ${\textbf{r}}{(t)}$ be a parameterization of $C$ for ${a} \leq {t} \leq {b}$ such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along $C$ . Divide the parameter interval $[{a},{b}]$ into $n$ subintervals $[{{t}_{{i} - {1}}}, {t}_{i}], {0} \leq {i} \leq n$ , of equal width. Denote the endpoints of ${\textbf{r}}{(t_{0})}, {\textbf{r}}{(t_{1})},..., {\textbf{r}}{(t_{n})}$ by ${{P}_{0}},...,{{P}_{n}}$ . Points ${P}_{i}$ divide $C$ into $n$ pieces. Denote the length of the piece from ${P}_{{i} - {1}}$ to ${P}_{i}$ by ${\Delta}{{s}_{i}}$ . For each $i$ , choose a value ${t}^{*}_{i}$ in the subinterval $[{{t}_{{i} - {1}}}, {t}_{i}]$ . Then, the endpoint of ${\textbf{r}}{({t}^{*}_{i})}$ is a point in the piece of $C$ between ${P}_{{i} - {1}}$ and ${P}_{i}$ (Figure 2). If ${\Delta}{{s}_{i}}$ is small, then as the particle moves from ${P}_{{i} - {1}}$ to ${P}_{i}$ along $C$ , it moves approximately in the direction of ${\textbf{T}}{({P}_{i})}$ , the unit tangent vector at the endpoint of ${\textbf{r}}{({t}^{*}_{i})}$ . Let ${P}^{*}_{i}$ denote the endpoint of ${\textbf{r}}{({t}^{*}_{i})}$ . Then, the work done by the force vector field in moving the particle from ${P}_{{i} - {1}}$ to ${P}_{i}$ is ${\textbf{F}}{({P}^{*}_{i})} \cdot {({\Delta}{{s}_{i}}{\textbf{T}}{({P}^{*}_{i})})}$ , so the total work done along $C$ is

${\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot ({{\Delta}{s}_{i}}{\textbf{T}}{({P}^{*}_{i})}) = {\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}{{\Delta}{s}_{i}}$ .

An image of a concave down curve – initially increasing, but later decreasing. Several points are labeled along the curve, as are arrowheads along the curve pointing in the direction of increasing P value. The points are: P_0, P_1, P_i-1, P_i starred, P_i, P_n-1, and Pn. Two arrows have their endpoints at P_i. The first is an increasing tangent vector labeled T(P_i starred). The second is labeled F(P_i starred) and points up and to the left.

Figure 2. Curve $C$ is divided into n pieces, and a point inside each piece is chosen. The dot product of any tangent vector in the ith piece with the corresponding vector $F$ is approximated by ${\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}$ .

Letting the arc length of the pieces of

C

$C$ get arbitrarily small by taking a limit as

n \to \infty

${n}{\rightarrow}{\infty}$ gives us the work done by the field in moving the particle along

C

$C$ . Therefore, the work done by F in moving the particle in the positive direction along

C

$C$ is defined as

${W} = {\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds},$

which gives us the concept of a vector line integral.

definition

The of vector field F along oriented smooth curve $C$ is

${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds} = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}{{\Delta}{s}_{i}}$

if that limit exists.

With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once by the parameterization, the value of the line integral is unchanged. With vector line integrals, the orientation of the curve does matter. If we think of the line integral as computing work, then this makes sense: if you hike up a mountain, then the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, then Earth’s gravitational force does positive work on you. In other words, reversing the path changes the work value from negative to positive in this case. Note that if $C$ is an oriented curve, then we let $-C$ represent the same curve but with opposite orientation.

As with scalar line integrals, it is easier to compute a vector line integral if we express it in terms of the parameterization function r and the variable $t$ . To translate the integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}$ in terms of t, note that unit tangent vector T along $C$ is given by ${\textbf{T}} = {\frac{{\textbf{r}}^{\prime}{(t)}}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}}$ (assuming ${\left \| {\textbf{r}}^{\prime}{(t)} \right \|} \neq {0}$ ). Since ${ds} = {\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt}$ , as we saw when discussing scalar line integrals, we have

${\textbf{F}} \cdot {\textbf{T}}{ds} = {\textbf{F}}{({\textbf{r}}{(t)})} \cdot {\frac{{\textbf{r}}^{\prime}{(t)}}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt} = {\textbf{F}}{({\textbf{r}}{(t)})} \cdot {\textbf{r}}^{\prime}{(t)}{dt}$ .

Thus, we have the following formula for computing vector line integrals:

${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds} = {\displaystyle\int^{b}_{a}}{\textbf{F}}{({\textbf{r}}{(t)})} \cdot {\textbf{r}}^{\prime}{(t)}{dt}$ .

Because of the vector line integrals equation above, we often use the notation ${\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}$ for the line integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}$ .

If ${\textbf{r}} = {\left \langle {x}{(t)}, {y}{(t)}, {z}{(t)} \right \rangle}$ , then ${d}{\textbf{r}}$ denotes vector differential ${\left \langle {{x}^{\prime}}{(t)}, {{y}^{\prime}}{(t)}, {{z}^{\prime}}{(t)} \right \rangle}{dt}$ .

Example: evaluating a vector line integral

Find the value of integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}$ , where $C$ is the semicircle parameterized by ${\textbf{r}}{(t)} = {\left \langle {\cos{t}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}$ and ${\textbf{F}} = {\left \langle {-y}, {x} \right \rangle}$ .

Show Solution

We can use the vector line integrals equation to convert the variable of integration from $s$ to $t$ . We then have

${\textbf{F}}{({\textbf{r}}{(t)})} = {\left \langle {{-}{\sin{t}}}, {\cos{t}} \right \rangle}{\text{ and }}{\textbf{r}}^{\prime}{(t)} = {\left \langle {{-}{\sin{t}}}, {\cos{t}} \right \rangle}$ .

Therefore,

$\begin{aligned} {\displaystyle\int_{C}}{\textbf{F}}{\cdot}{d}{\textbf{r}} & = {\displaystyle\int_{0}^{\pi}}{\left \langle {-}{\sin{t}}, {\cos{t}} \right \rangle}{\cdot}{\left \langle {-}{\sin{t}}, {\cos{t}} \right \rangle}{dt} \\ & = {\displaystyle\int_{0}^{\pi}}{{\sin}^{2}{t}} + {{\cos}^{2}{t}}{dt} \\ & = {\displaystyle\int_{0}^{\pi}}{1}{dt} = {\pi}. \end{aligned}$ .

See Figure 3.

A vector field in two dimensions. The closer the arrows are to the origin, the smaller they are. The further away they are, the longer they are. The arrows surround the origin in a radial pattern. A single curve is plotted and follows the radial pattern in quadrants 1 and 2 over the interval [-1,1]. It is a concave down arch that looks like a downward opening parabola.

Figure 3. This figure shows curve ${\textbf{r}}{(t)} = {\left \langle {\cos{t}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}$ in vector field ${\textbf{F}} = {\left \langle {-y}, {x} \right \rangle}.$

Example: reversing orientation

Find the value of integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}$ , where $C$ is the semicircle parameterized by ${\textbf{r}}{(t)} = {\left \langle {\cos{({t} + {\pi})}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}$ , and ${\textbf{F}} = {\left \langle -{y}, {x} \right \rangle}$ .

Show Solution

Notice that this is the same problem as Example “Evaluating a Vector Line Integral”, except the orientation of the curve has been traversed. In this example, the parameterization starts at ${\textbf{r}}{(0)} = {\left \langle -{1},{0} \right \rangle}$ and ends at ${\textbf{r}}{(\pi)} = {\left \langle {1},{0} \right \rangle}$ . By the equation to compute vector line integrals,

$\begin{aligned} {\displaystyle\int_{C}}{\textbf{F}} \cdot {dr} & = {\displaystyle\int^{\pi}_{0}}{\left \langle -{\sin{t}}, {\cos{({t} + {\pi})}} \right \rangle} \cdot {\left \langle -{\sin{({t} + {\pi})}}, {\cos{t}} \right \rangle}{dt} \\ & = {\displaystyle\int^{\pi}_{0}}{\left \langle -{\sin{t}}, -{\cos{t}} \right \rangle} \cdot {\left \langle {\sin{t}}, {\cos{t}} \right \rangle}{dt} \\ & = {\displaystyle\int^{\pi}_{0}}{\left ( -{\sin^{2}{t}} - {\cos^{2}{t}} \right )}{dt} \\ & = {\displaystyle\int^{\pi}_{0}}{-1}{dt} \\ & = {-}{\pi} \end{aligned}.$

Notice that this is the negative of the answer in Example “Evaluating a Vector Line Integral”. It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

Let $C$ be an oriented curve and let − $C$ denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

${\displaystyle\int_{-{C}}}{\textbf{F}} \cdot {dr} = {-}{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}$ .

That is, reversing the orientation of a curve changes the sign of a line integral.

try it

Let ${\textbf{F}} = {x}{\textbf{i}} + {y}{\textbf{j}}$ be a vector field and let $C$ be the curve with parameterization ${\left \langle {t},{{t}^{2}} \right \rangle}$ for ${0} \leq {t} \leq {2}$ . Which is greater: ${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}$ or ${\displaystyle\int_{-{C}}}{\textbf{F}} \cdot {\textbf{T}}{ds}$ ?

Show Solution

$\displaystyle\int_C{\bf{\text{F}}}\cdot{\bf{\text{T}}}ds$

Another standard notation for integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}$ is ${\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz}$ . In this notation, $P, Q,$ and $R$ are functions, and we think of ${d}{\textbf{r}}$ as vector ${\left \langle {dx},{dy},{dz} \right \rangle}$ . To justify this convention, recall that ${d}{\textbf{r}} = {\textbf{T}}{ds} = {\textbf{r}}{\prime}{(t)}{dt} = {\left \langle {\frac{dx}{dt}}, {\frac{dy}{dt}}, {\frac{dz}{dt}} \right \rangle}{dt}$ . Therefore,

${\textbf{F}} \cdot {d}{\textbf{r}} = {\left \langle {P}, {Q}, {R} \right \rangle} \cdot {\left \langle {dx},{dy},{dz} \right \rangle} = {Pdx} + {Qdy} + {Rdz}$ .

If ${d}{\textbf{r}} = {\left \langle {dx}, {dy}, {dz} \right \rangle}$ , then ${\frac{{d}{\textbf{r}}}{dt}} = {\left \langle {\frac{dx}{dt}}, {\frac{dy}{dt}}, {\frac{dz}{dt}} \right \rangle}$ , which implies that ${d}{\textbf{r}} = {\left \langle {\frac{dx}{dt}}, {\frac{dy}{dt}}, {\frac{dz}{dt}} \right \rangle}{dt}$ . Therefore

$\begin{aligned} {\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz} \\ & = {\displaystyle\int}{\left ( {P}{({\textbf{r}}{(t)})}{\frac{dx}{dt}} + {Q}{({\textbf{r}}{(t)})}{\frac{dy}{dt}} + {R}{({\textbf{r}}{(t)})}{\frac{dz}{dt}} \right )}{dt}. \end{aligned}$

Example: finding the value of an integral of the form ${\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz}$

Find the value of integral ${\displaystyle\int_{C}}{zdx} + {xdy} + {ydz}$ , where $C$ is the curve parameterized by ${\textbf{r}}{(t)} = {\left \langle {t^{2}}, {\sqrt{t}},{t} \right \rangle}, {1} \leq {t} \leq {4}$ .

Show Solution

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of $t$ . In this case,

allows us to make this change:

$\begin{aligned} {\displaystyle\int_{C}}{zdx} + {xdy} + {ydz} & = {\displaystyle\int^{4}_{1}}{\left ( {t}{(2t)} + {t^{2}}(\frac{1}{2{\sqrt{t}}}) + {\sqrt{t}} \right )}{dt} \\ & = {\displaystyle\int^{4}_{1}}{({2t^{2}} + {\frac{t^{3/2}}{2}} + {\sqrt{t}})}{dt} \\ & = {\left [ {\frac{2t^3}{3}} + {\frac{t^{5/2}}{5}} + {\frac{2t^{3/2}}{3}} \right ]}^{t = 4}_{t = 1} \\ & = {\frac{793}{15}}. \end{aligned}$ .

try it

Find the value of ${\displaystyle\int_{C}}{4x}{dx} + {zdy} + {4y^{2}}{dz}$ , where $C$ is the curve parameterized by ${\textbf{r}}{(t)} = {\left \langle {4} \ {\cos{(2t)}}, {2} \ {\sin{(2t)}}, {3} \right \rangle}, {0} \leq {t} \leq {\frac{\pi}{4}}$ .

Show Solution

$-26$ .

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.18” here (opens in new window).

We have learned how to integrate smooth oriented curves. Now, suppose that $C$ is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve. To be precise, curve $C$ is piecewise smooth if $C$ can be written as a union of $n$ smooth curves ${{C}_{1}}, {{C}_{2}},...,{{C}_{n}}$ such that the endpoint of $C_i$ is the starting point of ${C}_{{i} + {1}}$ (Figure 4). When curves $C_i$ satisfy the condition that the endpoint of $C_i$ is the starting point of ${C}_{{i} + {1}}$ , we write their union as ${{C}_{1}} + {{C}_{2}} + ... + {{C}_{n}}$ .

Three curves: C_1, C_2, and C_3. One of the endpoints of C_2 is also an endpoint of C_1, and the other endpoint of C_2 is also an end point of C_3. C_1’s and C_3’s other endpoints are not connect to any other curve. C_1 and C_3 appear to be nearly straight lines while C_2 is an increasing concave down curve. There are three arrowheads on each curve segment all pointing in the same direction: C_1 to C_2, C_2 to C_3, and C_3 to its other endpoint.

Figure 4. The union of ${{C}_{1}}, {{C}_{2}}, {{C}_{3}}$ is a piecewise smooth curve.

The next theorem summarizes several key properties of vector line integrals.

theorem: properties of vector line integrals

Let F and G be continuous vector fields with domains that include the oriented smooth curve C. Then

${\displaystyle\int_{C}}{({\textbf{F}} + {\textbf{G}})} \cdot {d}{\textbf{r}} = {\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} + {\displaystyle\int_{C}}{\textbf{G}} \cdot {dr}$
${\displaystyle\int_{C}}{k}{\textbf{F}} \cdot {d}{\textbf{r}} = {k}{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}$ , where $k$ is a constant
${\displaystyle\int_{-C}}{\textbf{F}} \cdot {d}{\textbf{r}} = -{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}$
Suppose instead that $C$ is a piecewise smooth curve in the domains of F and G, where ${C} = {C_1} + {C_2} + ... + {C_n}$ and ${C_1}, {C_2}, ..., {C_n}$ are smooth curves such that the endpoint of $C_i$ is the starting point of ${C}_{{i} + {1}}$ . Then ${\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{s}} = {\displaystyle\int_{C_1}}{\textbf{F}} \cdot {d}{\textbf{s}} + {\displaystyle\int_{C_2}}{\textbf{F}} \cdot {d}{\textbf{s}} + ... + {\displaystyle\int_{C_n}}{\textbf{F}} \cdot {d}{\textbf{s}}$ .

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along $C$ , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation ${\displaystyle\int^{b}_{a}}{f}{(x)}{dx} = {-}{\displaystyle\int^{a}_{b}}{f}{(x)}{dx}$ . Finally, if ${[{a_1}, {a_2}]}, {[{a_2}, {a_3}]}, ..., {[{a_{n-1}}, {a_n}]}$ are intervals, then

${\displaystyle\int^{a_n}_{a_1}}{f}{(x)}{dx} = {\displaystyle\int^{a_2}_{a_1}}{f}{(x)}{dx} + {\displaystyle\int^{a_3}_{a_1}}{f}{(x)}{dx} + ... + {\displaystyle\int^{a_n}_{a_{n-1}}}{f}{(x)}{dx},$

which is analogous to property iv.

Example: using properties to compute a vector line integral

Find the value of integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}$ , where $C$ is the rectangle (oriented counterclockwise) in a plane with vertices $(0, 0)$ , $(2, 0)$ , $(2, 1)$ ,and $(0, 1)$ and where ${\textbf{F}} = {\left \langle {x} - {2{y}}, {y} - {x} \right \rangle}$ (Figure 5).

A vector field in two dimensions. The arrows following roughly a 90-degree angle to the origin in quadrants 1 and 3 point to the origin. As the arrows deviate from this angle, they point away from the angle ad become smaller. Above, they point up and to the left, and below, they point down and to the right. A rectangle is drawn in quadrant 1 from 0 to 2 on the x axis and from 0 to 1 on the y axis. C_1 is the base, C_2 is the right leg, C_3 is the top, and C_4 is the left leg.

Figure 5. Rectangle and vector field.

Show Solution

Note that curve $C$ is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let $C_1$ represent the side from $(0, 0)$ to $(2, 0)$ , let $C_2$ represent the side from $(2, 0)$ to $(2, 1)$ , let $C_3$ represent the side from $(2, 1)$ to $(0, 1)$ , and let $C_4$ represent the side from $(0, 1)$ to $(0, 0)$ (Figure 5). Then,

${\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} = {\displaystyle\int_{C_1}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_2}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_3}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_4}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}}$ .

We want to compute each of the four integrals on the right-hand side using The Scalar Line Integral Calculation Theorem Equation. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse $C$ counterclockwise):

$\begin{aligned} {C_1} & : {\langle {t}, {0}\rangle}, {0} \leq {t} \leq {2} \\ {C_2} & : {\langle {2}, {t} \rangle}, {0} \leq {t} \leq {1} \\ {C_3} & : {\langle {2} - {t}, {1}\rangle}, {0} \leq {t} \leq {2} \\ {C_4} & : {\langle {0},{1} - {t}\rangle}, {0} \leq {t} \leq {1}. \end{aligned}$ .

Therefore,

$\begin{aligned} {\displaystyle\int_{{C}_{1}}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} & = {\displaystyle\int^{2}_{0}}{\textbf{F}}{({\textbf{r}}({t}))} \cdot {\textbf{r}}{\prime}{(t)}{dt} \\ & = {\displaystyle\int^{2}_{0}}{\left \langle {t} - {2}{(0)}, {0} - {t} \right \rangle} \cdot {\left \langle {1}, {0} \right \rangle}{dt} = {\displaystyle\int^{1}_{0}}{t}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} \right ]}^{2}_{0} = {2}. \end{aligned}$ .

Notice that the value of this integral is positive, which should not be surprising. As we move along curve $C_1$ from left to right, our movement flows in the general direction of the vector field itself. At any point along $C_1$ , the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along $C_1$ , and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

$\begin{aligned} {\displaystyle\int_{{C}_{2}}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int^{1}_{0}}{\left \langle {2} - {2{t}}, {t} - {2} \right \rangle} \cdot {\left \langle {0}, {1} \right \rangle}{dt} \\ & = {\displaystyle\int^{1}_{0}}{({t} - {2})}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} - {2{t}} \right ]}^{1}_{0} = -{\frac{3}{2}}, \end{aligned}$

$\begin{aligned} {\displaystyle\int_{{C}_{3}}}{\textbf{F}} \cdot {\textbf{T}}{ds} & = {\displaystyle\int^{2}_{0}}{\left \langle ({2} - {t}) - {2}, {1} - ({2} - {t}) \right \rangle} \cdot {\left \langle {-1}, {0} \right \rangle}{dt} \\ & = {\displaystyle\int^{2}_{0}}{tdt} = {2}, \end{aligned}$

and

$\begin{aligned} {\displaystyle\int_{{C}_{4}}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int^{1}_{0}}{\left \langle {-2}(1 - t) , {1 - t}\right \rangle} \cdot {\left \langle {0}, {-1} \right \rangle}{dt} \\ & = {\displaystyle\int^{1}_{0}}{({t} - {1})}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} - {t} \right ]}^{1}_{0} = -{\frac{1}{2}}. \end{aligned}$

Thus, we have ${\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} = {2}$ .

try it

Calculate line integral ${\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}$ , where F is vector field $\left \langle {y^{2}}, {2xy} + {1} \right \rangle$ and $C$ is a triangle with vertices $(0, 0)$ , $(4, 0)$ , and $(0, 5)$ , oriented counterclockwise.

Show Solution

$0$ .

Module 6: Vector Calculus

Learning Objectives

definition

Example: evaluating a vector line integral

Example: reversing orientation

try it

Example: finding the value of an integral of the form ${\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz}$

try it

theorem: properties of vector line integrals

Example: using properties to compute a vector line integral

try it

Candela Citations

Module 6: Vector Calculus