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
[latex]{\textbf{F}}({x},{y},{z}) = {{P}({x},{y},{z}){\textbf{i}}} + {{Q}({x},{y},{z}){\textbf{j}}} + {{R}({x},{y},{z}){\textbf{k}}}[/latex]
be a continuous vector field in [latex]{\mathbb{R}}^{3}[/latex] that represents a force on a particle, and let [latex]C[/latex] be a smooth curve in [latex]{\mathbb{R}}^{3}[/latex] contained in the domain of [latex]{\textbf{F}}[/latex]. How would we compute the work done by [latex]{\textbf{F}}[/latex] in moving a particle along [latex]C[/latex]?
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 [latex]C[/latex]; such a specified direction is called an orientation of a curve. The specified direction is the positive direction along [latex]C[/latex]; the opposite direction is the negative direction along [latex]C[/latex]. When [latex]C[/latex] has been given an orientation, [latex]C[/latex] 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 [latex]{\textbf{r}}{(t)}, {a} \leq {t} \leq {b}[/latex], such that [latex]{\textbf{r}}{(a)} = {\textbf{r}}{(b)}[/latex] and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain [latex](a, b)[/latex].
Let [latex]{\textbf{r}}{(t)}[/latex] be a parameterization of [latex]C[/latex] for [latex]{a} \leq {t} \leq {b}[/latex] such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along [latex]C[/latex]. Divide the parameter interval [latex][{a},{b}][/latex] into [latex]n[/latex] subintervals [latex][{{t}_{{i} - {1}}}, {t}_{i}], {0} \leq {i} \leq n[/latex], of equal width. Denote the endpoints of [latex]{\textbf{r}}{(t_{0})}, {\textbf{r}}{(t_{1})},…, {\textbf{r}}{(t_{n})}[/latex] by [latex]{{P}_{0}},…,{{P}_{n}}[/latex]. Points [latex]{P}_{i}[/latex] divide[latex]C[/latex] into [latex]n[/latex] pieces. Denote the length of the piece from [latex]{P}_{{i} - {1}}[/latex] to [latex]{P}_{i}[/latex] by [latex]{\Delta}{{s}_{i}}[/latex]. For each [latex]i[/latex], choose a value [latex]{t}^{*}_{i}[/latex] in the subinterval [latex][{{t}_{{i} - {1}}}, {t}_{i}][/latex]. Then, the endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex] is a point in the piece of [latex]C[/latex] between [latex]{P}_{{i} - {1}}[/latex] and [latex]{P}_{i}[/latex] (Figure 2). If [latex]{\Delta}{{s}_{i}}[/latex] is small, then as the particle moves from [latex]{P}_{{i} - {1}}[/latex] to [latex]{P}_{i}[/latex] along [latex]C[/latex], it moves approximately in the direction of [latex]{\textbf{T}}{({P}_{i})}[/latex], the unit tangent vector at the endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex]. Let [latex]{P}^{*}_{i}[/latex] denote the endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex]. Then, the work done by the force vector field in moving the particle from [latex]{P}_{{i} - {1}}[/latex] to [latex]{P}_{i}[/latex] is [latex]{\textbf{F}}{({P}^{*}_{i})} \cdot {({\Delta}{{s}_{i}}{\textbf{T}}{({P}^{*}_{i})})}[/latex], so the total work done along [latex]C[/latex] is
[latex]{\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}}[/latex].
[latex]{W} = {\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds},[/latex]
which gives us the concept of a vector line integral.
definition
The vector line integral of vector field F along oriented smooth curve [latex]C[/latex] is
[latex]{\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}}[/latex]
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 [latex]C[/latex] is an oriented curve, then we let [latex]-C[/latex] 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 [latex]t[/latex]. To translate the integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex] in terms of t, note that unit tangent vector T along [latex]C[/latex] is given by [latex]{\textbf{T}} = {\frac{{\textbf{r}}^{\prime}{(t)}}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}}[/latex] (assuming [latex]{\left \| {\textbf{r}}^{\prime}{(t)} \right \|} \neq {0}[/latex]). Since [latex]{ds} = {\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt}[/latex], as we saw when discussing scalar line integrals, we have
[latex]{\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}[/latex].
Thus, we have the following formula for computing vector line integrals:
[latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds} = {\displaystyle\int^{b}_{a}}{\textbf{F}}{({\textbf{r}}{(t)})} \cdot {\textbf{r}}^{\prime}{(t)}{dt}[/latex].
Because of the vector line integrals equation above, we often use the notation [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex] for the line integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex].
If [latex]{\textbf{r}} = {\left \langle {x}{(t)}, {y}{(t)}, {z}{(t)} \right \rangle}[/latex], then [latex]{d}{\textbf{r}}[/latex] denotes vector differential [latex]{\left \langle {{x}^{\prime}}{(t)}, {{y}^{\prime}}{(t)}, {{z}^{\prime}}{(t)} \right \rangle}{dt}[/latex].
Example: evaluating a vector line integral
Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex], where [latex]C[/latex] is the semicircle parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos{t}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}[/latex] and [latex]{\textbf{F}} = {\left \langle {-y}, {x} \right \rangle}[/latex].
Example: reversing orientation
Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex], where [latex]C[/latex] is the semicircle parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos{({t} + {\pi})}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}[/latex], and [latex]{\textbf{F}} = {\left \langle -{y}, {x} \right \rangle}[/latex].
Let [latex]C[/latex] be an oriented curve and let −[latex]C[/latex] denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:
[latex]{\displaystyle\int_{-{C}}}{\textbf{F}} \cdot {dr} = {-}{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex].
That is, reversing the orientation of a curve changes the sign of a line integral.
try it
Let [latex]{\textbf{F}} = {x}{\textbf{i}} + {y}{\textbf{j}}[/latex] be a vector field and let [latex]C[/latex] be the curve with parameterization [latex]{\left \langle {t},{{t}^{2}} \right \rangle}[/latex] for [latex]{0} \leq {t} \leq {2}[/latex]. Which is greater: [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex] or [latex]{\displaystyle\int_{-{C}}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex]?
Another standard notation for integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex] is [latex]{\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz}[/latex]. In this notation, [latex]P, Q,[/latex] and [latex]R[/latex] are functions, and we think of [latex]{d}{\textbf{r}}[/latex] as vector [latex]{\left \langle {dx},{dy},{dz} \right \rangle}[/latex]. To justify this convention, recall that [latex]{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}[/latex]. Therefore,
[latex]{\textbf{F}} \cdot {d}{\textbf{r}} = {\left \langle {P}, {Q}, {R} \right \rangle} \cdot {\left \langle {dx},{dy},{dz} \right \rangle} = {Pdx} + {Qdy} + {Rdz}[/latex].
If [latex]{d}{\textbf{r}} = {\left \langle {dx}, {dy}, {dz} \right \rangle}[/latex], then [latex]{\frac{{d}{\textbf{r}}}{dt}} = {\left \langle {\frac{dx}{dt}}, {\frac{dy}{dt}}, {\frac{dz}{dt}} \right \rangle}[/latex], which implies that [latex]{d}{\textbf{r}} = {\left \langle {\frac{dx}{dt}}, {\frac{dy}{dt}}, {\frac{dz}{dt}} \right \rangle}{dt}[/latex]. Therefore
[latex]\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}[/latex]
Example: finding the value of an integral of the form [latex]{\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz}[/latex]
Find the value of integral [latex]{\displaystyle\int_{C}}{zdx} + {xdy} + {ydz}[/latex], where [latex]C[/latex] is the curve parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {t^{2}}, {\sqrt{t}},{t} \right \rangle}, {1} \leq {t} \leq {4}[/latex].
try it
Find the value of [latex]{\displaystyle\int_{C}}{4x}{dx} + {zdy} + {4y^{2}}{dz}[/latex], where [latex]C[/latex] is the curve parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {4} \ {\cos{(2t)}}, {2} \ {\sin{(2t)}}, {3} \right \rangle}, {0} \leq {t} \leq {\frac{\pi}{4}}[/latex].
Watch the following video to see the worked solution to the above Try It
We have learned how to integrate smooth oriented curves. Now, suppose that [latex]C[/latex] 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 [latex]C[/latex] is piecewise smooth if [latex]C[/latex] can be written as a union of [latex]n[/latex] smooth curves [latex]{{C}_{1}}, {{C}_{2}},…,{{C}_{n}}[/latex] such that the endpoint of [latex]C_i[/latex] is the starting point of [latex]{C}_{{i} + {1}}[/latex] (Figure 4). When curves [latex]C_i[/latex] satisfy the condition that the endpoint of [latex]C_i[/latex] is the starting point of [latex]{C}_{{i} + {1}}[/latex], we write their union as [latex]{{C}_{1}} + {{C}_{2}} + … + {{C}_{n}}[/latex].
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
- [latex]{\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}[/latex]
- [latex]{\displaystyle\int_{C}}{k}{\textbf{F}} \cdot {d}{\textbf{r}} = {k}{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}[/latex], where [latex]k[/latex] is a constant
- [latex]{\displaystyle\int_{-C}}{\textbf{F}} \cdot {d}{\textbf{r}} = -{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}[/latex]
- Suppose instead that [latex]C[/latex] is a piecewise smooth curve in the domains of F and G, where [latex]{C} = {C_1} + {C_2} + … + {C_n}[/latex] and [latex]{C_1}, {C_2}, …, {C_n}[/latex]are smooth curves such that the endpoint of [latex]C_i[/latex] is the starting point of [latex]{C}_{{i} + {1}}[/latex]. Then [latex]{\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}}[/latex].
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 [latex]C[/latex], 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 [latex]{\displaystyle\int^{b}_{a}}{f}{(x)}{dx} = {-}{\displaystyle\int^{a}_{b}}{f}{(x)}{dx}[/latex]. Finally, if [latex]{[{a_1}, {a_2}]}, {[{a_2}, {a_3}]}, …, {[{a_{n-1}}, {a_n}]}[/latex] are intervals, then
[latex]{\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},[/latex]
which is analogous to property iv.
Example: using properties to compute a vector line integral
Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex], where [latex]C[/latex] is the rectangle (oriented counterclockwise) in a plane with vertices [latex](0, 0)[/latex], [latex](2, 0)[/latex], [latex](2, 1)[/latex],and [latex](0, 1)[/latex] and where [latex]{\textbf{F}} = {\left \langle {x} - {2{y}}, {y} - {x} \right \rangle}[/latex] (Figure 5).
try it
Calculate line integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}[/latex], where F is vector field [latex]\left \langle {y^{2}}, {2xy} + {1} \right \rangle[/latex] and [latex]C[/latex] is a triangle with vertices [latex](0, 0)[/latex], [latex](4, 0)[/latex], and [latex](0, 5)[/latex], oriented counterclockwise.