{"id":1347,"date":"2025-10-17T13:48:06","date_gmt":"2025-10-17T13:48:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1347"},"modified":"2025-11-05T18:04:20","modified_gmt":"2025-11-05T18:04:20","slug":"line-integrals-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/line-integrals-learn-it-2\/","title":{"raw":"Line Integrals: Learn It 2","rendered":"Line Integrals: Learn It 2"},"content":{"raw":"

Scalar Line Integrals<\/h2>\r\n

A scalar line integral is defined similarly to a single-variable integral, except the integrand is a function of more than one variable and the domain of integration is a curve rather than an interval on the [latex]x[\/latex]-axis.<\/p>\r\n\r\n

\r\n

Riemann Sums for Single-Variable Integrals<\/strong><\/p>\r\n

For [latex]\\int_a^b g(x) dx[\/latex], we:<\/p>\r\n\r\n

    \r\n \t
  1. Partition [latex][a,b][\/latex] into subintervals of width [latex]\\Delta x[\/latex]<\/li>\r\n \t
  2. Choose sample points [latex]x_i^*[\/latex] in each subinterval<\/li>\r\n \t
  3. Form the sum [latex]\\sum g(x_i^*) \\Delta x[\/latex]<\/li>\r\n \t
  4. Take the limit as [latex]\\Delta x \\to 0[\/latex]<\/li>\r\n<\/ol>\r\n

    Scalar line integrals follow the same pattern, but over curves instead of interval<\/p>\r\n\r\n<\/section>\r\n

    For a scalar line integral, we start with a smooth curve [latex]C[\/latex] and a function [latex]f[\/latex] whose domain includes [latex]C[\/latex]. Here's the process:<\/p>\r\n\r\n

      \r\n \t
    1. Chop the curve into small pieces<\/li>\r\n \t
    2. For each piece, choose a point [latex]P[\/latex] in each piece and evaluate [latex]f(P)[\/latex]<\/li>\r\n \t
    3. Multiply [latex]f(P)[\/latex] by the arc length [latex]\\Delta s[\/latex] of that piece<\/li>\r\n \t
    4. Sum all products [latex]f(P)\\Delta s[\/latex]<\/li>\r\n \t
    5. Let the arc lengths shrink to zero by taking a limit<\/li>\r\n<\/ol>\r\n

      The result is the scalar line integral of the function over the curve.<\/p>\r\n\r\n

      Formal Definition<\/h3>\r\n

      Let [latex]C[\/latex] be a smooth curve in space given by the parameterization [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex], [latex]a \\leq t \\leq b[\/latex]. Let [latex]f(x,y,z)[\/latex] be a function whose domain includes curve [latex]C[\/latex].<\/p>\r\n

      Partition the parameter interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals [latex][t_{i-1}, t_i][\/latex] of equal width for [latex]1 \\leq i \\leq n[\/latex], where [latex]t_0 = a[\/latex] and [latex]t_n = b[\/latex] (Figure 1). Let [latex]t_i^*[\/latex] be a value in the [latex]i[\/latex]th interval [latex][t_{i-1}, t_i][\/latex].<\/p>\r\n

      Denote the endpoints [latex]\\mathbf{r}(t_0), \\mathbf{r}(t_1), \\ldots, \\mathbf{r}(t_n)[\/latex] by [latex]P_0, \\ldots, P_n[\/latex]. These points divide curve [latex]C[\/latex] into [latex]n[\/latex] pieces [latex]C_1, C_2, \\ldots, C_n[\/latex] with lengths [latex]\\Delta s_1, \\Delta s_2, \\ldots, \\Delta s_n[\/latex]. Let [latex]{P}^{*}_{i}[\/latex] denote the endpoint of [latex]{\\textbf{r}}({t^{*}_{i}})[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"526\"]\"A Figure 1. Curve [latex]C[\/latex] has been divided into [latex]n[\/latex] pieces, and a point inside each piece has been chosen.[\/caption]\r\n

      Now we evaluate function [latex]f[\/latex] at point [latex]{P}^{*}_{i}[\/latex] foreach [latex]i[\/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[\/latex]\u00a0by the length [latex]\\Delta s_i[\/latex] of [latex]C_i[\/latex], which gives the area of a \"sheet\" with base [latex]C_i[\/latex] and height [latex]f(P_i^*)[\/latex]. This is analogous to using rectangles to approximate area in single-variable integrals.<\/p>\r\n

      Form the sum [latex]\\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]. As with Riemann sums, we define the integral by letting the width of the pieces shrink to zero. The result is the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex].<\/p>\r\n\r\n

      Unlike single-variable integrals where we partition into equal lengths, the arc lengths [latex]\\Delta s_1, \\Delta s_2, \\ldots, \\Delta s_n[\/latex] aren't necessarily equal. However, this difference doesn't affect the limit\u2014as arc lengths shrink to zero, their values become close enough that small differences become irrelevant.<\/section>
      \r\n

      scalar line integral<\/h3>\r\n

      Let [latex]f[\/latex] be a function with a domain that includes the smooth curve [latex]C[\/latex] parameterized by [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex], [latex]a \\leq t \\leq b[\/latex]. The scalar line integral<\/strong> of [latex]f[\/latex] along [latex]C[\/latex] is:<\/p>\r\n

      [latex]\\int_C f(x,y,z) ds = \\lim_{n \\to \\infty} \\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]<\/p>\r\n

      if this limit exists.<\/p>\r\n

      For a planar curve with parameterization [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex], [latex]a \\leq t \\leq b[\/latex]:<\/p>\r\n

      [latex]\\int_C f(x,y) ds = \\lim_{n \\to \\infty} \\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]<\/p>\r\n\r\n<\/section>\r\n

      If [latex]f[\/latex] is continuous on smooth curve [latex]C[\/latex], then [latex]\\int_C f \\, ds[\/latex] always exists. The continuity of [latex]f[\/latex] guarantees the existence of the limit, just as [latex]\\int_a^b g(x) dx[\/latex] exists when [latex]g[\/latex] is continuous over [latex][a,b][\/latex].<\/p>\r\n\r\n

      Geometric Interpretation<\/h3>\r\n

      Before learning how to compute line integrals, let's examine the geometry they capture.<\/p>\r\n

      Suppose [latex]f(x,y) \\geq 0[\/latex] for all points [latex](x, y)[\/latex]\u00a0on a smooth planar curve [latex]C[\/latex]. Imagine projecting curve [latex]C[\/latex] \"up\" to the surface defined by [latex]f(x,y)[\/latex], creating a new curve [latex]C'[\/latex] that lies on the graph of [latex]f(x,y)[\/latex] (Figure 2). Now drop a \"sheet\" from [latex]C'[\/latex] down to the [latex]xy[\/latex]-plane. The area of this sheet is [latex]\\int_C f(x,y) ds[\/latex].<\/p>\r\n

      If [latex]f(x,y) \\leq 0[\/latex] for some points in [latex]C[\/latex], then [latex]\\int_C f(x,y) ds[\/latex] equals the area above the [latex]xy[\/latex]-plane minus the area below the [latex]xy[\/latex]-plane. This is analogous to single-variable integrals [latex]\\int_a^b g(x) dx[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"405\"]\"A Figure 2. The area of the blue sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex].[\/caption]

      From this geometry, we see that line integral [latex]\\int_C f(x,y) ds[\/latex] does not depend on the parameterization [latex]\\mathbf{r}(t)[\/latex] of [latex]C[\/latex]. As long as the curve is traversed exactly once, the area of the sheet formed by the function and curve is the same. This geometric argument extends to three-dimensional curves as well.<\/section>
      Find the value of integral [latex]{\\displaystyle\\int_{C}}{2}{ds}[\/latex], where [latex]C[\/latex] is the upper half of the unit circle.[reveal-answer q=\"374562045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374562045\"]The integrand is [latex]f(x, y)=2[\/latex].\u00a0Figure 3\u00a0shows the graph of [latex]f(x, y)=2[\/latex], curve [latex]C[\/latex], and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width [latex]\\pi[\/latex] and length 2. Therefore, [latex]{\\displaystyle\\int_{C}}{2}{ds} = {2}{\\pi}[\/latex].[caption id=\"\" align=\"aligncenter\" width=\"424\"]\"A Figure 3. The sheet that is formed by the upper half of the unit circle in a plane and the graph of [latex]{f}{({x},{y})} = {2}[\/latex].[\/caption]\r\n

      To see that [latex]{\\displaystyle\\int_{C}}{2}{ds} = {2}{\\pi}[\/latex] using the definition of line integral, we let [latex]{\\textbf{r}}{(t)}[\/latex] be a parameterization of [latex]C[\/latex]. Then, [latex]{f}{({\\textbf{r}}{(t_{i})})} = {2}[\/latex] for any number [latex]{t_{i}}[\/latex] in the domain of\u00a0r<\/strong>. Therefore,<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{f}{ds} & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{f}{({\\textbf{r}}{({t}^{*}_{i})})}{\\Delta}{s_{i}} \\\\\r\n& = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{2}{\\Delta}{s_{i}} \\\\\r\n& = {2}{\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{\\Delta}{s_{i}} \\\\\r\n& = {2}{\\text{(length of C)}} \\\\\r\n& = {2}{\\pi}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Computing Scalar Line Integrals<\/h3>\r\n

      Scalar line integrals integrate with respect to arc length [latex]s[\/latex], which can make calculations difficult. We can simplify by translating [latex]\\int_C f \\, ds[\/latex] to an integral with variable [latex]t[\/latex].<\/p>\r\n

      Let [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex] for [latex]a \\leq t \\leq b[\/latex] be a parameterization of smooth curve [latex]C[\/latex]. Since [latex]C[\/latex] is smooth, [latex]\\mathbf{r}'(t) = \\langle x'(t), y'(t), z'(t) \\rangle[\/latex] is continuous for all [latex]t[\/latex] in [latex][a,b][\/latex].<\/p>\r\n

      According to the arc length formula:<\/p>\r\n

      [latex]\\text{length}(C_i) = \\Delta s_i = \\int_{t_{i-1}}^{t_i} ||\\mathbf{r}'(t)|| dt[\/latex]<\/p>\r\n

      If width [latex]\\Delta t_i = t_i - t_{i-1}[\/latex] is small, then [latex]||\\mathbf{r}'(t)||[\/latex] is nearly constant over interval [latex][t_{i-1}, t_i][\/latex]. Therefore:<\/p>\r\n

      [latex]\\int_{t_{i-1}}^{t_i} ||\\mathbf{r}'(t)|| dt \\approx ||\\mathbf{r}'(t_i^*)|| \\Delta t_i[\/latex]<\/p>\r\n

      This gives us:<\/p>\r\n

      [latex]{\\displaystyle\\sum^{n}_{{i} = {1}}}{f}{({\\textbf{r}}{({t^{*}_{i}})})}{\\Delta}{{s}_{i}} = {\\displaystyle\\sum^{n}_{{i} = {1}}}{f}{({\\textbf{r}}{({t^{*}_{i}})})}{\\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"419\"]\"A Figure 4. If we zoom in on the curve enough by making [latex]{\\Delta}{{t}_{i}}[\/latex]\u00a0very small, then the corresponding piece of the curve is approximately linear.[\/caption]\r\n

      Taking the limit as [latex]n \\to \\infty[\/latex]:<\/p>\r\n

      [latex]{\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{{i} = {1}}}{f}{({\\textbf{r}}{({t^{*}_{i}})})}{\\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}} = {\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)})}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt}[\/latex].<\/p>\r\n

      This gives us our main computational theorem.<\/p>\r\n\r\n

      \r\n

      evaluating a scalar line integral<\/h3>\r\n

      Let [latex]f[\/latex] be a continuous function with a domain that includes the smooth curve [latex]C[\/latex] with parameterization [latex]{\\textbf{r}}{(t)}, {a} \\leq {t} \\leq {b}[\/latex]. Then:<\/p>\r\n

      [latex]{\\displaystyle\\int_C}{f}{d}{s} = {\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)})}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt}[\/latex].<\/p>\r\n\r\n<\/section>

      In our derivation, we wrote [latex]\\int_{t_{i-1}}^{t_i} ||\\mathbf{r}'(t)|| dt \\approx ||\\mathbf{r}'(t_i^*)|| \\Delta t_i[\/latex]. This is an approximation, not an equation. The left-hand side approaches the right-hand side as [latex]n \\to \\infty[\/latex]. In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum.<\/section>\r\n

      Since [latex]||\\mathbf{r}'(t)|| = \\sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}[\/latex], we obtain a practical formula for computation.<\/p>\r\n\r\n

      \r\n

      scalar line integral calculation<\/h3>\r\n

      Let [latex]f[\/latex] be continuous with domain including smooth curve [latex]C[\/latex] with parameterization [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex], [latex]a \\leq t \\leq b[\/latex]. Then:<\/p>\r\n

      [latex]{\\displaystyle\\int_{C}}{f}{({x},{y},{z})}{ds} = {\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)}}){\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}}{dt}[\/latex].<\/p>\r\n

      For a planar curve:<\/p>\r\n

      [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds} = {\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)}}){\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2}}}{dt}[\/latex]<\/p>\r\n\r\n<\/section>A key consequence of this theorem is the equation [latex]ds = ||\\mathbf{r}'(t)|| dt[\/latex]. In other words, the change in arc length can be viewed as a change in the [latex]t[\/latex] domain, scaled by the magnitude of vector [latex]\\mathbf{r}'(t)[\/latex].\r\n\r\n

      Find the value of integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds}[\/latex], where [latex]C[\/latex] is part of the helix parameterized by [latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\cos {t}}, {\\sin {t}}, {t} \\right \\rangle}, {0} \\leq {t} \\leq {2}{\\pi}[\/latex].[reveal-answer q=\"384725475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384725475\"]\r\n

      To compute a scalar line integral, we start by converting the variable of integration from arc length [latex]s[\/latex] to [latex]t[\/latex]. Then, we can use the scalar line integral calculation to compute the integral with respect to [latex]t[\/latex]. Note that [latex]{f}{({\\textbf{r}}{(t)})} = {{\\cos}^{2}{t}} + {{\\sin}^{2}{t}} + {t} = {1} + {t}[\/latex] and<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} & = {\\sqrt{{({-}{\\sin{(t)}})^{2}} + {\\cos}^{2}{(t)} + {1}}} \\\\\r\n& = {\\sqrt{2}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n

      Therefore,<\/p>\r\n

      [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} = {\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt}[\/latex].<\/p>\r\n

      Notice that\u00a0The Scalar Line Integral Calculation Theorem Equation\u00a0translated the original difficult line integral into a manageable single-variable integral. Since<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt} & = {\\left [ {\\sqrt{2}}{t} + {\\frac{{\\sqrt{2}}{t}^{2}}{2}} \\right ]}^{{2}{\\pi}}_{0} \\\\\r\n& = {2}{\\sqrt{2}}{\\pi} + {2}{\\sqrt{2}}{\\pi}^{2},\r\n\\end{aligned}[\/latex]<\/p>\r\n

      we have<\/p>\r\n

      [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} = {2}{\\sqrt{2}}{\\pi} + {2}{\\sqrt{2}}{\\pi}^{2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

      Find the value of integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds}[\/latex], where [latex]C[\/latex] is part of the helix parameterized by [latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\cos}{(2t)}, {\\sin}{(2t), {2}{t}} \\right \\rangle}, {0} \\leq {t} \\leq {\\pi}[\/latex]. [latex]\\\\[\/latex]Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.[reveal-answer q=\"371932814\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371932814\"]\r\n

      As with the previous example, we use the scalar line integral calculation to compute the integral with respect to [latex]t[\/latex]. Note that [latex]{f}{({\\textbf{r}}{(t)})} = {\\cos}^{2}{(2t)} + {\\sin}^{2}{(2t)} + {2t} = {2t} + {1}[\/latex] and<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} & = {\\sqrt{({-}{\\sin{t}} + {\\cos{t}} + {4})}} \\\\\r\n& = {2}{\\sqrt{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n

      so we have<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} & = {2}{\\sqrt{2}}{\\displaystyle\\int^{\\pi}_{0}}{({1} + {2t})}{dt} \\\\\r\n& = {2}{\\sqrt{2}}{[{t} + {t}^{2}]}^{\\pi}_{0} \\\\\r\n& = {2}{\\sqrt{2}}{({\\pi} + {\\pi}^{2})}.\r\n\\end{aligned}[\/latex].<\/p>\r\n

      Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Arc Length as a Line Integral<\/h3>\r\n

      Line integrals provide a way to calculate arc length. If [latex]f(x,y,z) = 1[\/latex], then:<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{f}{({x}, {y}, {z})}{ds} & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}\\displaystyle{\\sum^{n}_{i = 1}}{f}{({t}^{*}_{i})}{\\Delta}{{s}_{i}} \\\\\r\n& = \\displaystyle{\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{\\Delta}{{s}_{i}} \\\\\r\n& = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\text{length}}{(C)} \\\\\r\n& = {\\text{length}}{(C)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n

      Therefore, [latex]{\\displaystyle\\int_{C}}{1}{ds}[\/latex] is the arc length of [latex]C[\/latex].<\/p>\r\n\r\n

      A wire has a shape that can be modeled with the parameterization [latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\cos{t}}, {\\sin{t}}, {\\frac{2}{3}}{t}^{3\/2} \\right \\rangle}, {0} \\leq {t} \\leq {4}{\\pi}[\/latex]. Find the length of the wire.[reveal-answer q=\"092499841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092499841\"]\r\n

      The length of the wire is given by [latex]{\\displaystyle\\int_{C}}{1}{ds}[\/latex], where [latex]C[\/latex] is the curve with parameterization\u00a0[latex]{\\textbf{r}}[\/latex]. Therefore,<\/p>\r\n

      [latex]\\begin{aligned}\r\n{\\text{The length of the wire}} & = {\\displaystyle\\int_{C}}{1}{ds} \\\\\r\n& = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt} \\\\\r\n& = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{({-}{\\sin{t}})^{2} + {\\cos}^{2}{t} + {tdt}}} \\\\\r\n& = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{{1} + {tdt}}} \\\\\r\n& = {\\left [ {\\frac{{2}{({1} + {t})^{3\/2}}}{3}} \\right ]}^{{4}{\\pi}}_{0} \\\\\r\n& = {\\frac{2}{3}}{\\left (({1} + {4{\\pi}})^{3\/2} - {1} \\right )}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

      Scalar Line Integrals<\/h2>\n

      A scalar line integral is defined similarly to a single-variable integral, except the integrand is a function of more than one variable and the domain of integration is a curve rather than an interval on the [latex]x[\/latex]-axis.<\/p>\n

      \n

      Riemann Sums for Single-Variable Integrals<\/strong><\/p>\n

      For [latex]\\int_a^b g(x) dx[\/latex], we:<\/p>\n

        \n
      1. Partition [latex][a,b][\/latex] into subintervals of width [latex]\\Delta x[\/latex]<\/li>\n
      2. Choose sample points [latex]x_i^*[\/latex] in each subinterval<\/li>\n
      3. Form the sum [latex]\\sum g(x_i^*) \\Delta x[\/latex]<\/li>\n
      4. Take the limit as [latex]\\Delta x \\to 0[\/latex]<\/li>\n<\/ol>\n

        Scalar line integrals follow the same pattern, but over curves instead of interval<\/p>\n<\/section>\n

        For a scalar line integral, we start with a smooth curve [latex]C[\/latex] and a function [latex]f[\/latex] whose domain includes [latex]C[\/latex]. Here’s the process:<\/p>\n

          \n
        1. Chop the curve into small pieces<\/li>\n
        2. For each piece, choose a point [latex]P[\/latex] in each piece and evaluate [latex]f(P)[\/latex]<\/li>\n
        3. Multiply [latex]f(P)[\/latex] by the arc length [latex]\\Delta s[\/latex] of that piece<\/li>\n
        4. Sum all products [latex]f(P)\\Delta s[\/latex]<\/li>\n
        5. Let the arc lengths shrink to zero by taking a limit<\/li>\n<\/ol>\n

          The result is the scalar line integral of the function over the curve.<\/p>\n

          Formal Definition<\/h3>\n

          Let [latex]C[\/latex] be a smooth curve in space given by the parameterization [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex], [latex]a \\leq t \\leq b[\/latex]. Let [latex]f(x,y,z)[\/latex] be a function whose domain includes curve [latex]C[\/latex].<\/p>\n

          Partition the parameter interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals [latex][t_{i-1}, t_i][\/latex] of equal width for [latex]1 \\leq i \\leq n[\/latex], where [latex]t_0 = a[\/latex] and [latex]t_n = b[\/latex] (Figure 1). Let [latex]t_i^*[\/latex] be a value in the [latex]i[\/latex]th interval [latex][t_{i-1}, t_i][\/latex].<\/p>\n

          Denote the endpoints [latex]\\mathbf{r}(t_0), \\mathbf{r}(t_1), \\ldots, \\mathbf{r}(t_n)[\/latex] by [latex]P_0, \\ldots, P_n[\/latex]. These points divide curve [latex]C[\/latex] into [latex]n[\/latex] pieces [latex]C_1, C_2, \\ldots, C_n[\/latex] with lengths [latex]\\Delta s_1, \\Delta s_2, \\ldots, \\Delta s_n[\/latex]. Let [latex]{P}^{*}_{i}[\/latex] denote the endpoint of [latex]{\\textbf{r}}({t^{*}_{i}})[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex].<\/p>\n

          \"A
          Figure 1. Curve [latex]C[\/latex] has been divided into [latex]n[\/latex] pieces, and a point inside each piece has been chosen.<\/figcaption><\/figure>\n

          Now we evaluate function [latex]f[\/latex] at point [latex]{P}^{*}_{i}[\/latex] foreach [latex]i[\/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[\/latex]\u00a0by the length [latex]\\Delta s_i[\/latex] of [latex]C_i[\/latex], which gives the area of a “sheet” with base [latex]C_i[\/latex] and height [latex]f(P_i^*)[\/latex]. This is analogous to using rectangles to approximate area in single-variable integrals.<\/p>\n

          Form the sum [latex]\\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]. As with Riemann sums, we define the integral by letting the width of the pieces shrink to zero. The result is the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex].<\/p>\n

          Unlike single-variable integrals where we partition into equal lengths, the arc lengths [latex]\\Delta s_1, \\Delta s_2, \\ldots, \\Delta s_n[\/latex] aren’t necessarily equal. However, this difference doesn’t affect the limit\u2014as arc lengths shrink to zero, their values become close enough that small differences become irrelevant.<\/section>\n
          \n

          scalar line integral<\/h3>\n

          Let [latex]f[\/latex] be a function with a domain that includes the smooth curve [latex]C[\/latex] parameterized by [latex]\\mathbf{r}(t) = \\langle x(t), y(t), z(t) \\rangle[\/latex], [latex]a \\leq t \\leq b[\/latex]. The scalar line integral<\/strong> of [latex]f[\/latex] along [latex]C[\/latex] is:<\/p>\n

          [latex]\\int_C f(x,y,z) ds = \\lim_{n \\to \\infty} \\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]<\/p>\n

          if this limit exists.<\/p>\n

          For a planar curve with parameterization [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex], [latex]a \\leq t \\leq b[\/latex]:<\/p>\n

          [latex]\\int_C f(x,y) ds = \\lim_{n \\to \\infty} \\sum_{i=1}^n f(P_i^*) \\Delta s_i[\/latex]<\/p>\n<\/section>\n

          If [latex]f[\/latex] is continuous on smooth curve [latex]C[\/latex], then [latex]\\int_C f \\, ds[\/latex] always exists. The continuity of [latex]f[\/latex] guarantees the existence of the limit, just as [latex]\\int_a^b g(x) dx[\/latex] exists when [latex]g[\/latex] is continuous over [latex][a,b][\/latex].<\/p>\n

          Geometric Interpretation<\/h3>\n

          Before learning how to compute line integrals, let’s examine the geometry they capture.<\/p>\n

          Suppose [latex]f(x,y) \\geq 0[\/latex] for all points [latex](x, y)[\/latex]\u00a0on a smooth planar curve [latex]C[\/latex]. Imagine projecting curve [latex]C[\/latex] “up” to the surface defined by [latex]f(x,y)[\/latex], creating a new curve [latex]C'[\/latex] that lies on the graph of [latex]f(x,y)[\/latex] (Figure 2). Now drop a “sheet” from [latex]C'[\/latex] down to the [latex]xy[\/latex]-plane. The area of this sheet is [latex]\\int_C f(x,y) ds[\/latex].<\/p>\n

          If [latex]f(x,y) \\leq 0[\/latex] for some points in [latex]C[\/latex], then [latex]\\int_C f(x,y) ds[\/latex] equals the area above the [latex]xy[\/latex]-plane minus the area below the [latex]xy[\/latex]-plane. This is analogous to single-variable integrals [latex]\\int_a^b g(x) dx[\/latex].<\/p>\n

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          Figure 2. The area of the blue sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex].<\/figcaption><\/figure>\n
          From this geometry, we see that line integral [latex]\\int_C f(x,y) ds[\/latex] does not depend on the parameterization [latex]\\mathbf{r}(t)[\/latex] of [latex]C[\/latex]. As long as the curve is traversed exactly once, the area of the sheet formed by the function and curve is the same. This geometric argument extends to three-dimensional curves as well.<\/section>\n
          Find the value of integral [latex]{\\displaystyle\\int_{C}}{2}{ds}[\/latex], where [latex]C[\/latex] is the upper half of the unit circle.<\/p>\n