The specified direction is the positive direction<\/strong> along [latex]C[\/latex], and the opposite direction is the negative direction<\/strong>. When [latex]C[\/latex] has been given an orientation, we call it an oriented curve<\/strong> (Figure 1). The work done on the particle depends on the direction along the curve in which the particle is moving.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"506\"] 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 and the particle moves in the positive direction along [latex]C[\/latex]. We'll build up the work calculation step by step.<\/p>\r\n Divide the parameter interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals [latex][t_{i-1}, t_i][\/latex], [latex]0 \\leq i \\leq n[\/latex], of equal width. Denote the endpoints [latex]{\\textbf{r}}(t_0), {\\textbf{r}}(t_1), \\ldots, {\\textbf{r}}(t_n)[\/latex] by [latex]P_0, \\ldots, P_n[\/latex]. These points divide [latex]C[\/latex] into [latex]n[\/latex] pieces.<\/p>\r\n Let [latex]{\\Delta}s_i[\/latex] denote the length of the piece from [latex]P_{i-1}[\/latex] to [latex]P_i[\/latex]. For each [latex]i[\/latex], choose a value [latex]{t}^{*}_{i}[\/latex] in the subinterval [latex][{{t}_{{i} - {1}}}, {t}_{i}][\/latex]. 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).<\/p>\r\n 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].<\/p>\r\n 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]. Therefore, the total work done along [latex]C[\/latex] is:<\/p>\r\n [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].<\/p>\r\n\r\n Letting the arc length of the pieces of [latex]C[\/latex] get arbitrarily small by taking a limit as [latex]n \\to \\infty[\/latex] gives us the work done by the field in moving the particle along [latex]C[\/latex]. The work done by [latex]{\\textbf{F}}[\/latex] in moving the particle in the positive direction along [latex]C[\/latex] is:<\/p>\r\n [latex]W = \\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex]<\/p>\r\n This integral is called a vector line integral<\/strong>.<\/p>\r\n\r\n The vector line integral<\/strong> of vector field [latex]{\\textbf{F}}[\/latex] along oriented smooth curve [latex]C[\/latex] is:<\/p>\r\n [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]<\/p>\r\n if that limit exists.<\/p>\r\n\r\n<\/section>\r\n With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once, the value of the line integral is unchanged.<\/p>\r\n With vector line integrals, orientation is critical. If you hike up a mountain, the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, Earth's gravitational force does positive work on you. Reversing the path changes the work value from negative to positive.<\/p>\r\n\r\n If [latex]C[\/latex] is an oriented curve, then [latex]-C[\/latex] represents the same curve but with opposite orientation.<\/p>\r\n\r\n<\/section>\r\n As with scalar line integrals, it's easier to compute a vector line integral using the parameterization function [latex]{\\textbf{r}}[\/latex] and the variable [latex]t[\/latex].<\/p>\r\n To translate the integral [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex] in terms of [latex]t[\/latex], recall that the unit tangent vector [latex]{\\textbf{T}}[\/latex] along [latex]C[\/latex] is:<\/p>\r\n [latex]{\\textbf{T}} = \\frac{{\\textbf{r}}'(t)}{\\left | {\\textbf{r}}'(t) \\right |}[\/latex]<\/p>\r\n (assuming [latex]\\left | {\\textbf{r}}'(t) \\right | \\neq 0[\/latex])<\/p>\r\n Since [latex]ds = \\left | {\\textbf{r}}'(t) \\right | dt[\/latex], as we saw when discussing scalar line integrals, we have:<\/p>\r\n [latex]{\\textbf{F}} \\cdot {\\textbf{T}} ds = {\\textbf{F}}({\\textbf{r}}(t)) \\cdot \\frac{{\\textbf{r}}'(t)}{\\left | {\\textbf{r}}'(t) \\right |}\\left | {\\textbf{r}}'(t) \\right | dt = {\\textbf{F}}({\\textbf{r}}(t)) \\cdot {\\textbf{r}}'(t) dt[\/latex]<\/p>\r\n This gives us a practical formula for computing vector line integrals.<\/p>\r\n\r\n [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds = \\displaystyle\\int^{b}_{a}{\\textbf{F}}({\\textbf{r}}(t)) \\cdot {\\textbf{r}}'(t) dt[\/latex]<\/p>\r\n We often use the notation [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot d{\\textbf{r}}[\/latex] for the line integral [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex].<\/p>\r\n[latex]\\\\[\/latex]\r\n If [latex]{\\textbf{r}} = \\langle x(t), y(t), z(t) \\rangle[\/latex], then [latex]d{\\textbf{r}}[\/latex] denotes the vector differential [latex]\\langle x'(t), y'(t), z'(t) \\rangle dt[\/latex].<\/p>\r\n\r\n<\/section> We can use\u00a0the vector line integrals equation to convert the variable of integration from [latex]s[\/latex] to [latex]t[\/latex]. We then have<\/p>\r\n [latex]{\\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}[\/latex].<\/p>\r\n Therefore,<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\displaystyle\\int_{0}^{\\pi}}{{\\sin}^{2}{t}} + {{\\cos}^{2}{t}}{dt} \\\\\r\n& = {\\displaystyle\\int_{0}^{\\pi}}{1}{dt} = {\\pi}.\r\n\\end{aligned}[\/latex].<\/p>\r\n See\u00a0Figure 3.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"] Notice that this is the same problem as\u00a0Example \"Evaluating a Vector Line Integral\", except the orientation of the curve has been traversed. In this example, the parameterization starts at [latex]{\\textbf{r}}{(0)} = {\\left \\langle -{1},{0} \\right \\rangle}[\/latex] and ends at [latex]{\\textbf{r}}{(\\pi)} = {\\left \\langle {1},{0} \\right \\rangle}[\/latex]. By the equation to compute vector line integrals,<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\displaystyle\\int^{\\pi}_{0}}{\\left \\langle -{\\sin{t}}, -{\\cos{t}} \\right \\rangle} \\cdot {\\left \\langle {\\sin{t}}, {\\cos{t}} \\right \\rangle}{dt} \\\\\r\n& = {\\displaystyle\\int^{\\pi}_{0}}{\\left ( -{\\sin^{2}{t}} - {\\cos^{2}{t}} \\right )}{dt} \\\\\r\n& = {\\displaystyle\\int^{\\pi}_{0}}{-1}{dt} \\\\\r\n& = {-}{\\pi}\r\n\\end{aligned}.[\/latex]<\/p>\r\n Notice that this is the negative of the answer in the previous example. It makes sense that this answer is negative because the orientation of the curve goes against the \u201cflow\u201d of the vector field.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n The previous examples illustrate an important property:<\/p>\r\n [latex]\\displaystyle\\int_{-C}{\\textbf{F}} \\cdot d{\\textbf{r}} = -\\displaystyle\\int_{C}{\\textbf{F}} \\cdot d{\\textbf{r}}[\/latex]<\/p>\r\n Reversing the orientation of a curve changes the sign of a line integral. This makes physical sense when thinking about work\u2014if you reverse direction, the work done changes sign.<\/p>\r\n\r\n Another standard notation for integral [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot d{\\textbf{r}}[\/latex] is [latex]\\displaystyle\\int_{C}P dx + Q dy + R dz[\/latex]. In this notation, [latex]P[\/latex], [latex]Q[\/latex], and [latex]R[\/latex] are functions, and we think of [latex]d{\\textbf{r}}[\/latex] as vector [latex]\\langle dx, dy, dz \\rangle[\/latex].<\/p>\r\n To justify this convention, recall that [latex]d{\\textbf{r}} = {\\textbf{T}} ds = {\\textbf{r}}'(t) dt = \\left \\langle \\dfrac{dx}{dt}, \\dfrac{dy}{dt}, \\dfrac{dz}{dt} \\right \\rangle dt[\/latex]. Therefore:<\/p>\r\n [latex]{\\textbf{F}} \\cdot d{\\textbf{r}} = \\langle P, Q, R \\rangle \\cdot \\langle dx, dy, dz \\rangle = P dx + Q dy + R dz[\/latex]<\/p>\r\n If [latex]d{\\textbf{r}} = \\langle dx, dy, dz \\rangle[\/latex], then [latex]\\dfrac{d{\\textbf{r}}}{dt} = \\left \\langle \\dfrac{dx}{dt}, \\dfrac{dy}{dt}, \\dfrac{dz}{dt} \\right \\rangle[\/latex], which implies that [latex]d{\\textbf{r}} = \\left \\langle \\dfrac{dx}{dt}, \\dfrac{dy}{dt}, \\dfrac{dz}{dt} \\right \\rangle dt[\/latex]. Therefore:<\/p>\r\n [latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{\\textbf{F}} \\cdot {d}{\\textbf{r}} & = {\\displaystyle\\int_{C}}{Pdx} + {Qdy} + {Rdz} \\\\\r\n& = {\\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}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of [latex]t[\/latex]. In this case,<\/p>\r\n[latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{\\textbf{F}} \\cdot {d}{\\textbf{r}} & = {\\displaystyle\\int_{C}}{Pdx} + {Qdy} + {Rdz} \\\\\r\n& = {\\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}.\r\n\\end{aligned}[\/latex]\r\n\r\nallows us to make this change:\r\n [latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{zdx} + {xdy} + {ydz} & = {\\displaystyle\\int^{4}_{1}}{\\left ( {t}{(2t)} + {t^{2}}(\\frac{1}{2{\\sqrt{t}}}) + {\\sqrt{t}} \\right )}{dt} \\\\\r\n& = {\\displaystyle\\int^{4}_{1}}{({2t^{2}} + {\\frac{t^{3\/2}}{2}} + {\\sqrt{t}})}{dt} \\\\\r\n& = {\\left [ {\\frac{2t^3}{3}} + {\\frac{t^{5\/2}}{5}} + {\\frac{2t^{3\/2}}{3}} \\right ]}^{t = 4}_{t = 1} \\\\\r\n& = {\\frac{793}{15}}.\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n 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 [latex]C[\/latex] is a piecewise smooth curve<\/strong>.<\/p>\r\n 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}, \\ldots, 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 this condition, we write their union as [latex]C_{1} + C_{2} + \\cdots + C_{n}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"227\"] Let [latex]{\\textbf{F}}[\/latex] and [latex]{\\textbf{G}}[\/latex] be continuous vector fields with domains that include the oriented smooth curve [latex]C[\/latex]. Then:<\/p>\r\n\r\n Notice the similarities between these properties and the properties of single-variable integrals.<\/p>\r\n Properties 1 and 2 say that line integrals are linear<\/strong>, which is true of single-variable integrals as well. Property 3 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\u2014if the particle moves backward rather than forward, 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].<\/p>\r\n Finally, if [latex][a_1, a_2][\/latex], [latex][a_2, a_3][\/latex], [latex]\\ldots[\/latex], [latex][a_{n-1}, a_n][\/latex] are intervals, then:<\/p>\r\n [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]<\/p>\r\n This is analogous to property 4.<\/p>\r\n\r\n Note that curve [latex]C[\/latex] is the union of its four sides, and each side is smooth. Therefore\u00a0C<\/em>\u00a0is piecewise smooth. Let [latex]C_1[\/latex] represent the side from [latex](0, 0)[\/latex] to [latex](2, 0)[\/latex], let [latex]C_2[\/latex] represent the side from [latex](2, 0)[\/latex] to [latex](2, 1)[\/latex], let [latex]C_3[\/latex] represent the side from [latex](2, 1)[\/latex] to [latex](0, 1)[\/latex], and let [latex]C_4[\/latex] represent the side from [latex](0, 1)[\/latex] to [latex](0, 0)[\/latex] (Figure 5). Then,<\/p>\r\n [latex]{\\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}}[\/latex].<\/p>\r\n 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 [latex]C[\/latex] counterclockwise):<\/p>\r\n [latex]\\begin{aligned}\r\n{C_1} & : {\\langle {t}, {0}\\rangle}, {0} \\leq {t} \\leq {2} \\\\\r\n{C_2} & : {\\langle {2}, {t} \\rangle}, {0} \\leq {t} \\leq {1} \\\\\r\n{C_3} & : {\\langle {2} - {t}, {1}\\rangle}, {0} \\leq {t} \\leq {2} \\\\\r\n{C_4} & : {\\langle {0},{1} - {t}\\rangle}, {0} \\leq {t} \\leq {1}.\r\n\\end{aligned}[\/latex].<\/p>\r\n Therefore,<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\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} \\\\\r\n& = {\\left [ {\\frac{t^{2}}{2}} \\right ]}^{2}_{0} = {2}.\r\n\\end{aligned}[\/latex].<\/p>\r\n Notice that the value of this integral is positive, which should not be surprising. As we move along curve [latex]C_1[\/latex] from left to right, our movement flows in the general direction of the vector field itself. At any point along [latex]C_1[\/latex], the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90\u00b0. Therefore, the tangent vector and the force vector have a positive dot product all along [latex]C_1[\/latex], and the line integral will have positive value.<\/p>\r\n The calculations for the three other line integrals are done similarly:<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\displaystyle\\int^{1}_{0}}{({t} - {2})}{dt} \\\\\r\n& = {\\left [ {\\frac{t^{2}}{2}} - {2{t}} \\right ]}^{1}_{0} = -{\\frac{3}{2}},\r\n\\end{aligned}[\/latex]<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\displaystyle\\int^{2}_{0}}{tdt} = {2},\r\n\\end{aligned}[\/latex]<\/p>\r\n and<\/p>\r\n [latex]\\begin{aligned}\r\n{\\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} \\\\\r\n& = {\\displaystyle\\int^{1}_{0}}{({t} - {1})}{dt} \\\\\r\n& = {\\left [ {\\frac{t^{2}}{2}} - {t} \\right ]}^{1}_{0} = -{\\frac{1}{2}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n Thus, we have [latex]{\\displaystyle\\int_{C}}{\\textbf{F}} \\cdot {d}{\\textbf{r}} = {2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/div>","rendered":" The second type of line integrals are vector line integrals. Vector line integrals allow us to integrate along a curve through a vector field. Consider a continuous vector field in [latex]{\\mathbb{R}}^{3}[\/latex]:<\/p>\n [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]<\/p>\n Suppose [latex]{\\textbf{F}}[\/latex] represents a force on a particle, and [latex]C[\/latex] is a smooth curve in [latex]{\\mathbb{R}}^{3}[\/latex] contained in the domain of [latex]{\\textbf{F}}[\/latex]. How do we compute the work done by [latex]{\\textbf{F}}[\/latex] in moving a particle along [latex]C[\/latex]?<\/p>\n To answer this question, first note that a particle can travel in two directions along a curve: forward or backward. The work done by the vector field depends on which direction the particle moves. We must specify a direction along curve [latex]C[\/latex], called an orientation<\/strong>.<\/p>\n The specified direction is the positive direction<\/strong> along [latex]C[\/latex], and the opposite direction is the negative direction<\/strong>. When [latex]C[\/latex] has been given an orientation, we call it an oriented curve<\/strong> (Figure 1). The work done on the particle depends on the direction along the curve in which the particle is moving.<\/p>\n 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 and the particle moves in the positive direction along [latex]C[\/latex]. We’ll build up the work calculation step by step.<\/p>\n Divide the parameter interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals [latex][t_{i-1}, t_i][\/latex], [latex]0 \\leq i \\leq n[\/latex], of equal width. Denote the endpoints [latex]{\\textbf{r}}(t_0), {\\textbf{r}}(t_1), \\ldots, {\\textbf{r}}(t_n)[\/latex] by [latex]P_0, \\ldots, P_n[\/latex]. These points divide [latex]C[\/latex] into [latex]n[\/latex] pieces.<\/p>\n Let [latex]{\\Delta}s_i[\/latex] denote the length of the piece from [latex]P_{i-1}[\/latex] to [latex]P_i[\/latex]. For each [latex]i[\/latex], choose a value [latex]{t}^{*}_{i}[\/latex] in the subinterval [latex][{{t}_{{i} - {1}}}, {t}_{i}][\/latex]. 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).<\/p>\n 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].<\/p>\n 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]. Therefore, the total work done along [latex]C[\/latex] is:<\/p>\n [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].<\/p>\n Letting the arc length of the pieces of [latex]C[\/latex] get arbitrarily small by taking a limit as [latex]n \\to \\infty[\/latex] gives us the work done by the field in moving the particle along [latex]C[\/latex]. The work done by [latex]{\\textbf{F}}[\/latex] in moving the particle in the positive direction along [latex]C[\/latex] is:<\/p>\n [latex]W = \\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex]<\/p>\n This integral is called a vector line integral<\/strong>.<\/p>\n The vector line integral<\/strong> of vector field [latex]{\\textbf{F}}[\/latex] along oriented smooth curve [latex]C[\/latex] is:<\/p>\n [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]<\/p>\n if that limit exists.<\/p>\n<\/section>\n With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once, the value of the line integral is unchanged.<\/p>\n With vector line integrals, orientation is critical. If you hike up a mountain, the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, Earth’s gravitational force does positive work on you. Reversing the path changes the work value from negative to positive.<\/p>\n If [latex]C[\/latex] is an oriented curve, then [latex]-C[\/latex] represents the same curve but with opposite orientation.<\/p>\n<\/section>\n As with scalar line integrals, it’s easier to compute a vector line integral using the parameterization function [latex]{\\textbf{r}}[\/latex] and the variable [latex]t[\/latex].<\/p>\n To translate the integral [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex] in terms of [latex]t[\/latex], recall that the unit tangent vector [latex]{\\textbf{T}}[\/latex] along [latex]C[\/latex] is:<\/p>\n [latex]{\\textbf{T}} = \\frac{{\\textbf{r}}'(t)}{\\left | {\\textbf{r}}'(t) \\right |}[\/latex]<\/p>\n (assuming [latex]\\left | {\\textbf{r}}'(t) \\right | \\neq 0[\/latex])<\/p>\n Since [latex]ds = \\left | {\\textbf{r}}'(t) \\right | dt[\/latex], as we saw when discussing scalar line integrals, we have:<\/p>\n [latex]{\\textbf{F}} \\cdot {\\textbf{T}} ds = {\\textbf{F}}({\\textbf{r}}(t)) \\cdot \\frac{{\\textbf{r}}'(t)}{\\left | {\\textbf{r}}'(t) \\right |}\\left | {\\textbf{r}}'(t) \\right | dt = {\\textbf{F}}({\\textbf{r}}(t)) \\cdot {\\textbf{r}}'(t) dt[\/latex]<\/p>\n This gives us a practical formula for computing vector line integrals.<\/p>\n [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds = \\displaystyle\\int^{b}_{a}{\\textbf{F}}({\\textbf{r}}(t)) \\cdot {\\textbf{r}}'(t) dt[\/latex]<\/p>\n We often use the notation [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot d{\\textbf{r}}[\/latex] for the line integral [latex]\\displaystyle\\int_{C}{\\textbf{F}} \\cdot {\\textbf{T}} ds[\/latex].<\/p>\n [latex]\\\\[\/latex]<\/p>\n If [latex]{\\textbf{r}} = \\langle x(t), y(t), z(t) \\rangle[\/latex], then [latex]d{\\textbf{r}}[\/latex] denotes the vector differential [latex]\\langle x'(t), y'(t), z'(t) \\rangle dt[\/latex].<\/p>\n<\/section>\n Figure 1. (a) An oriented curve between two points. (b) A closed oriented curve.[\/caption]\r\n
Defining the Vector Line Integral<\/h3>\r\n
Figure 2. Curve [latex]C[\/latex] is divided into n pieces, and a point inside each piece is chosen. The dot product of any tangent vector in the i<\/em>th piece with the corresponding vector [latex]F[\/latex] is approximated by [latex]{\\textbf{F}}{({P}^{*}_{i})} \\cdot {\\textbf{T}}{({P}^{*}_{i})}[\/latex].[\/caption]<\/div>\r\n
vector line integral<\/h3>\r\n
Computing Vector Line Integrals with Parameterization<\/h3>\r\n
computing vector line integrals<\/h3>\r\n
Figure 3. This figure shows curve [latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\cos{t}}, {\\sin{t}} \\right \\rangle}, {0} \\leq {t} \\leq {\\pi}[\/latex] in vector field [latex]{\\textbf{F}} = {\\left \\langle {-y}, {x} \\right \\rangle}.[\/latex][\/caption][\/hidden-answer]<\/section>
Alternative Notation<\/h3>\r\n
Piecewise Smooth Curves<\/h3>\r\n
Figure 4. The union of [latex]{{C}_{1}}, {{C}_{2}}, {{C}_{3}}[\/latex]\u00a0is a piecewise smooth curve.[\/caption]The next theorem summarizes several key properties of vector line integrals.
theorem: properties of vector line integrals<\/h3>\r\n
\r\n \t
Understanding the Properties<\/h4>\r\n
Figure 5. Rectangle and vector field.[\/caption]\r\n\r\n
Vector Line Integrals<\/h2>\n
Defining the Vector Line Integral<\/h3>\n
vector line integral<\/h3>\n
Computing Vector Line Integrals with Parameterization<\/h3>\n
computing vector line integrals<\/h3>\n