{"id":5476,"date":"2022-06-02T18:43:08","date_gmt":"2022-06-02T18:43:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5476"},"modified":"2022-11-01T05:04:52","modified_gmt":"2022-11-01T05:04:52","slug":"scalar-line-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/scalar-line-integrals\/","title":{"raw":"Scalar Line Integrals","rendered":"Scalar Line Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Calculate a scalar line integral along a curve.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793928878\">A\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term241\" data-type=\"term\">line integral<\/span>\u00a0gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let\u2019s look at scalar line integrals first.<\/p>\r\n<p id=\"fs-id1167793607702\">A scalar line integral is defined just as a single-variable integral is defined, except that for a scalar line integral, the integrand is a function of more than one variable and the domain of integration is a curve in a plane or in space, as opposed to a curve on the [latex]x[\/latex]-axis.<\/p>\r\n<p id=\"fs-id1167794036521\">For a scalar line integral, we let [latex]C[\/latex] be a smooth curve in a plane or in space and let [latex]f[\/latex] be a function with a domain that includes [latex]C[\/latex]. We chop the curve into small pieces. For each piece, we choose point [latex]P[\/latex] in that piece and evaluate [latex]f[\/latex] at [latex]P[\/latex]. (We can do this because all the points in the curve are in the domain of [latex]f[\/latex].) We multiply [latex]{f}{(P)}[\/latex] by the arc length of the piece [latex]{\\Delta}{s}[\/latex], add the product [latex]{f}{(P)}{\\Delta}{s}[\/latex] over all the pieces, and then let the arc length of the pieces shrink to zero by taking a limit. The result is the scalar line integral of the function over the curve.<\/p>\r\n<p id=\"fs-id1167794330401\">For a formal description of a scalar line integral, let [latex]C[\/latex] be a smooth curve in space given by the parameterization [latex]{\\textbf{r}}(t) = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. Let [latex]f(x, y, z)[\/latex] be a function with a domain that includes curve [latex]C[\/latex]. To define the line integral of the function [latex]f[\/latex] over [latex]C[\/latex], we begin as most definitions of an integral begin: we chop the curve into small pieces. 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\u00a0<em>i<\/em>th interval [latex][{{t}_{i-1}},{t_i}][\/latex]. 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 curve [latex]C[\/latex] into [latex]n[\/latex] pieces [latex]{C}_{1},{C}_{2},...,{C}_{n}[\/latex], with lengths [latex]{\\Delta}{{s}_{1}}, {\\Delta}{{s}_{2}},..., {\\Delta}{{s}_{n}}[\/latex], respectively. Let [latex]{P}^{*}_{i}[\/latex] denote the endpoint of [latex]{\\textbf{r}}({t^{*}_{i}})[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex]. Now, we evaluate the function [latex]f[\/latex] at point [latex]{P}^{*}_{i}[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex]. Note that [latex]{P}^{*}_{i}[\/latex] is in piece [latex]C_1[\/latex], and therefore [latex]{P}^{*}_{i}[\/latex] is in the domain of [latex]f[\/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[\/latex] by the length [latex]{\\Delta}{{s}_{1}}[\/latex] of [latex]C_1[\/latex], which gives the area of the \u201csheet\u201d with base [latex]C_1[\/latex], and height [latex]{f}{({P}^{*}_{i})}[\/latex]. This is analogous to using rectangles to approximate area in a single-variable integral. Now, we form the sum [latex]{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex]. Note the similarity of this sum versus a Riemann sum; in fact, this definition is a generalization of a Riemann sum to arbitrary curves in space. Just as with Riemann sums and integrals of form [latex]{\\displaystyle\\int^{b}_{a}}{g}{(x)}{dx}[\/latex], we define an integral by letting the width of the pieces of the curve shrink to zero by taking a limit. The result is the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex].s<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"526\"]<img src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/ef803993941028598bec179194a610712f5ff7d6\" alt=\"A diagram of a curve in quadrant one. Several points and segments are labeled. Starting at the left, the first points are P_0 and P_1. The segment between them is labeled delta S_1. The next points are P_i-1, P_i, and P_i+1. The segments connecting them are delta S_i and delta S_j+1. Point P_i starred and point P_i+1 starred are located on each segment, respectively. The last two points are P_n-1 and P_n, connected by segment S_n.\" width=\"526\" height=\"310\" \/> Figure 1. Curve [latex]C[\/latex] has been divided into [latex]n[\/latex] pieces, and a point inside each piece has been chosen.[\/caption]You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths [latex]{\\Delta}{{s}_{1}}, {\\Delta}{{s}_{2}},..., {\\Delta}{{s}_{n}}[\/latex] aren\u2019t necessarily the same; in the definition of a single-variable integral, the curve in the [latex]x[\/latex]-axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793842996\">Let [latex]f[\/latex] be a function with a domain that includes the smooth curve [latex]C[\/latex] that is parameterized by [latex]{\\textbf{r}}{(t)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. The\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term242\" data-type=\"term\">scalar line integral<\/span><\/strong>\u00a0of[latex]f[\/latex] along [latex]C[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{f}{({x},{y},{z})}{ds} = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793603762\">if this limit exists ([latex]{t}^{*}_{i}[\/latex] and [latex]{\\Delta}{s}_{i}[\/latex] are defined as in the previous paragraphs). If [latex]C[\/latex] is a planar curve, then [latex]C[\/latex] can be represented by the parametric equations [latex]x=x(t)[\/latex], [latex]y=y(t)[\/latex], and [latex]{a} \\leq {t} \\leq {b}[\/latex]. If [latex]C[\/latex] is smooth and [latex]f(x, y)[\/latex] is a function of two variables, then the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex] is defined similarly as<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds} = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex],<\/p>\r\n<p id=\"fs-id1167794296597\">if this limit exists.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793417464\">If [latex]f[\/latex] is a continuous function on a smooth curve [latex]C[\/latex], then [latex]{\\displaystyle\\int_C}{f}{d}{s}[\/latex] always exists. Since [latex]{\\displaystyle\\int_C}{f}{d}{s}[\/latex] is defined as a limit of Riemann sums, the continuity of [latex]f[\/latex] is enough to guarantee the existence of the limit, just as the integral [latex]{\\displaystyle\\int^{b}_{a}}{g{(x)}}{dx}[\/latex] exists if [latex]g[\/latex] is continuous over [latex][{a},{b}][\/latex].<\/p>\r\n<p id=\"fs-id1167793432079\">Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that [latex]{f}{({x},{y})} \\geq {0}[\/latex] for all points [latex](x, y)[\/latex] on a smooth planar curve [latex]C[\/latex]. Imagine taking curve [latex]C[\/latex] and projecting it \u201cup\u201d to the surface defined by [latex]f(x, y)[\/latex], thereby creating a new curve [latex]{C}^{\\prime}[\/latex] that lies in the graph of [latex]f(x, y)[\/latex] (Figure 2). Now we drop a \u201csheet\u201d from [latex]{C}^{\\prime}[\/latex] down to the [latex]xy[\/latex]-plane. The area of this sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex]. If [latex]{f}{({x},{y})} \\leq {0}[\/latex] for some points in [latex]C[\/latex], then the value of [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex] is the area above the [latex]xy[\/latex]-plane less the area below the [latex]xy[\/latex]-plane. (Note the similarity with integrals of the form [latex]{\\displaystyle\\int^{b}_{a}}{g}{(x)}{dx}[\/latex].)<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"405\"]<img src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/d8f17ee89085c3bbc74b92268857dbfff02aa02d\" alt=\"A diagram in three dimensions. The original curve C in the (x,y) plane looks like a parabola opening to the left with vertex in quadrant 1. The surface defined by f(x,y) is shown always above the (x,y) plane. A curve on the surface directly above the original curve C is labeled as C\u2019. A blue sheet stretches down from C\u2019 to C.\" width=\"405\" height=\"386\" \/> Figure 2. The area of the blue sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex].[\/caption]From this geometry, we can see that line integral [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex] does not depend on the parameterization [latex]{\\textbf{r}}{(t)}[\/latex] of [latex]C[\/latex]. As long as the curve is traversed exactly once by the parameterization, the area of the sheet formed by the function and the curve is the same. This same kind of geometric argument can be extended to show that the line integral of a three-variable function over a curve in space does not depend on the parameterization of the curve.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the value of a line integral<\/h3>\r\nFind the value of integral [latex]{\\displaystyle\\int_{C}}{2}{ds}[\/latex], where [latex]C[\/latex] is the upper half of the unit circle.\r\n\r\n[reveal-answer q=\"374562045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374562045\"]\r\n\r\nThe 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].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"424\"]<img src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/b8c8f5c10b679e69719de9b1a42580cf84b973b0\" alt=\"A graph in three dimensions. There is a flat plane just above the (x,y) plane. The upper half of the unit circle in quadrants 1 and 2 of the (x,y) plane is raised up to form a semicircle sheet into the z-plane.\" width=\"424\" height=\"277\" \/> 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<p id=\"fs-id1167793911831\">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\u00a0<strong data-effect=\"bold\">r<\/strong>. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{f}{ds} &amp; = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{f}{({\\textbf{r}}{({t}^{*}_{i})})}{\\Delta}{s_{i}} \\\\\r\n&amp; = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{2}{\\Delta}{s_{i}} \\\\\r\n&amp; = {2}{\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{\\Delta}{s_{i}} \\\\\r\n&amp; = {2}{\\text{(length of C)}} \\\\\r\n&amp; = {2}{\\pi}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the value of [latex]{\\displaystyle\\int_{C}}{({x} + {y})}{ds}[\/latex], where [latex]C[\/latex] is the curve parameterized by [latex]x=t[\/latex],\u00a0[latex]y=t[\/latex],\u00a0[latex]{0} \\leq {t} \\leq {1}[\/latex].\r\n\r\n[reveal-answer q=\"948375296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"948375296\"]\r\n<p style=\"text-align: center;\">[latex]\\sqrt{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793418697\">Note that in a scalar line integral, the integration is done with respect to arc length [latex]s[\/latex], which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate [latex]{\\displaystyle\\int_{C}}{f}{ds}[\/latex] to an integral with a variable of integration that is [latex]t[\/latex].<\/p>\r\n<p id=\"fs-id1167793937004\">Let [latex]{\\textbf{r}}{(t)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}[\/latex] for [latex]{a} \\leq {t} \\leq {b}[\/latex] be a parameterization of [latex]C[\/latex]. Since we are assuming that [latex]C[\/latex] is smooth, [latex]{\\textbf{r}}^{\\prime}{(t)} = {\\left \\langle {x}{\\prime}{(t)}, {y}{\\prime}{(t)}, {z}{\\prime}{(t)} \\right \\rangle}[\/latex] is continuous for all [latex]t[\/latex] in [latex][{a},{b}][\/latex]. In particular, [latex]{x}{\\prime}{(t)}, {y}^{\\prime}{(t)}[\/latex], and [latex]{z}^{\\prime}{(t)}[\/latex] exist for all [latex]t[\/latex] in [latex][{a},{b}][\/latex]. According to the arc length formula, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\text{length}}{({C}_{i})} = {\\Delta}{{s}_{i}} = {\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{d}{t}[\/latex].<\/p>\r\n<p id=\"fs-id1167794210535\">If width [latex]{\\Delta}{{t}_{i}} = {{t}_{i}} - {{t}_{{i} - {1}}}[\/latex] is small, then function [latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\| {\\Delta}{{t}_{i}} \\right \\|} {\\textbf{r}}^{\\prime}{(t)} \\right \\|}[\/latex] is almost constant over the interval [latex]\\left [ {{t}_{{i} - {1}}},{{t}_{i}} \\right ][\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\right \\|} {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}},[\/latex]<\/p>\r\nand we have\r\n<p style=\"text-align: center;\">[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<p id=\"fs-id1167793881085\">See\u00a0Figure 4.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"419\"]<img src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/6ac27b2f44bbefe4d435b72eb0d064b555debbef\" alt=\"A segment of an increasing concave down curve labeled C. A small segment of the curved is boxed and labeled as delta t_i. In the zoomed-in insert, this boxed segment of the curve is almost linear.\" width=\"419\" height=\"201\" \/> 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<p id=\"fs-id1167793915895\">Note that<\/p>\r\n<p style=\"text-align: center;\">[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\nIn other words, as the widths of intervals [latex][{{t}_{{i} - {1}}},{{t}_{i}}][\/latex] shrink to zero, the sum [latex]{\\displaystyle\\sum^{n}_{{i} = {1}}}{f}{({\\textbf{r}}{({t^{*}_{i}})})}{\\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}}[\/latex] converges to the integral [latex]{\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)})}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt}[\/latex]. Therefore, we have the following theorem.\r\n<div id=\"fs-id1167794058922\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: evaluating a scalar line integral<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794058930\">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<p style=\"text-align: center;\">[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<\/div>\r\n<p id=\"fs-id1167793925270\">Although we have labeled\u00a0[latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\right \\|} {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}},[\/latex] as an equation, it is more accurately considered an approximation because we can show that the left-hand side of the equation approaches the right-hand side as [latex]{n}{\\rightarrow}{\\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. Since<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|} = {\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2},}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793380170\">we obtain the following theorem, which we use to compute scalar line integrals.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: scalar line integral calculation<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793939706\">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)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. Then<\/p>\r\n<p style=\"text-align: center;\">[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<p id=\"fs-id1167793944241\">Similarly,<\/p>\r\n<p style=\"text-align: center;\">[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<p id=\"fs-id1167794137230\">if [latex]C[\/latex] is a planar curve and [latex]f[\/latex] is a function of two variables.<\/p>\r\n\r\n<\/div>\r\nNote that a consequence of this theorem is the equation [latex]{ds} = {\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{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]{\\textbf{r}}^{\\prime}{(t)}[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a line integral<\/h3>\r\nFind 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].\r\n\r\n[reveal-answer q=\"384725475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384725475\"]\r\n<p id=\"fs-id1167793221724\">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 Theorem Equation\u00a0to 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<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} &amp; = {\\sqrt{{({-}{\\sin{(t)}})^{2}} + {\\cos}^{2}{(t)} + {1}}} \\\\\r\n&amp; = {\\sqrt{2}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793930422\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} = {\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt}[\/latex].<\/p>\r\n<p id=\"fs-id1167794328944\">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<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt} &amp; = {\\left [ {\\sqrt{2}}{t} + {\\frac{{\\sqrt{2}}{t}^{2}}{2}} \\right ]}^{{2}{\\pi}}_{0} \\\\\r\n&amp; = {2}{\\sqrt{2}}{\\pi} + {2}{\\sqrt{2}}{\\pi}^{2},\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793633032\">we have<\/p>\r\n<p style=\"text-align: center;\">[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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds}[\/latex], where [latex]C[\/latex] is the curve with parameterization\u00a0[latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\sin}{(3t)}, {\\cos}{(3t), {t}} \\right \\rangle}, {0} \\leq {t} \\leq {2}{\\pi}[\/latex].\r\n\r\n[reveal-answer q=\"094320778\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"094320778\"]\r\n<p style=\"text-align: center;\">[latex]2\\sqrt{10}\\pi+2\\sqrt{10}\\pi^2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: independence of parameterization<\/h3>\r\nFind 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]. 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.\r\n\r\n[reveal-answer q=\"371932814\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371932814\"]\r\n<p id=\"fs-id1167793978351\">As with the previous example, we use\u00a0The Scalar Line Integral Calculation Theorem Equation 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<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} &amp; = {\\sqrt{({-}{\\sin{t}} + {\\cos{t}} + {4})}} \\\\\r\n&amp; = {2}{\\sqrt{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793244536\">so we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} &amp; = {2}{\\sqrt{2}}{\\displaystyle\\int^{\\pi}_{0}}{({1} + {2t})}{dt} \\\\\r\n&amp; = {2}{\\sqrt{2}}{[{t} + {t}^{2}]}^{\\pi}_{0} \\\\\r\n&amp; = {2}{\\sqrt{2}}{({\\pi} + {\\pi}^{2})}.\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793959042\">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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate line integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {yz})}{ds}[\/latex], where [latex]C[\/latex] is the line with parameterization [latex]{\\textbf{r}}{(t)} = {\\left \\langle {2t}, {5t}, {-t} \\right \\rangle}, {0} \\leq {t} \\leq {10}[\/latex]. Reparameterize [latex]C[\/latex] with parameterization [latex]{\\textbf{s}}{(t)} = {\\left \\langle {4t}, {10t}, {-2t} \\right \\rangle}, {0} \\leq {t} \\leq {5}[\/latex], recalculate line integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {yz})}{ds}[\/latex], and notice that the change of parameterization had no effect on the value of the integral.\r\n\r\n[reveal-answer q=\"872471336\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"872471336\"]\r\n\r\nBoth line integrals equal\u00a0[latex]-\\frac{1,000\\sqrt{30}}3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250313&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3EIXmDq7zfY&amp;video_target=tpm-plugin-ojc8ic91-3EIXmDq7zfY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.15_transcript.html\">transcript for \u201cCP 6.15\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167793401189\">Now that we can evaluate line integrals, we can use them to calculate arc length. If [latex]f(x, y, z)=1[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\displaystyle\\int_{C}}{f}{({x}, {y}, {z})}{ds} &amp; = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}\\displaystyle{\\sum^{n}_{i = 1}}{f}{({t}^{*}_{i})}{\\Delta}{{s}_{i}} \\\\\r\n&amp; = \\displaystyle{\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{\\Delta}{{s}_{i}} \\\\\r\n&amp; = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\text{length}}{(C)} \\\\\r\n&amp; = {\\text{length}}{(C)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793420990\">Therefore, [latex]{\\displaystyle\\int_{C}}{1}{ds}[\/latex] is the arc length of [latex]C[\/latex].<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating arc length<\/h3>\r\nA 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.\r\n\r\n[reveal-answer q=\"092499841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092499841\"]\r\n<p id=\"fs-id1167793383514\">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<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\text{The length of the wire}} &amp; = {\\displaystyle\\int_{C}}{1}{ds} \\\\\r\n&amp; = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt} \\\\\r\n&amp; = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{({-}{\\sin{t}})^{2} + {\\cos}^{2}{t} + {tdt}}} \\\\\r\n&amp; = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{{1} + {tdt}}} \\\\\r\n&amp; = {\\left [ {\\frac{{2}{({1} + {t})^{3\/2}}}{3}} \\right ]}^{{4}{\\pi}}_{0} \\\\\r\n&amp; = {\\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the length of a wire with parameterization [latex]{\\textbf{r}}{(t)} = {\\left \\langle {3t} + {1}, {4} - {2t},{5} + {2t} \\right \\rangle}, {0} \\leq {t} \\leq {4}[\/latex].\r\n\r\n[reveal-answer q=\"442837426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"442837426\"]\r\n<p style=\"text-align: center;\">[latex]4\\sqrt{17}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Calculate a scalar line integral along a curve.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793928878\">A\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term241\" data-type=\"term\">line integral<\/span>\u00a0gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let\u2019s look at scalar line integrals first.<\/p>\n<p id=\"fs-id1167793607702\">A scalar line integral is defined just as a single-variable integral is defined, except that for a scalar line integral, the integrand is a function of more than one variable and the domain of integration is a curve in a plane or in space, as opposed to a curve on the [latex]x[\/latex]-axis.<\/p>\n<p id=\"fs-id1167794036521\">For a scalar line integral, we let [latex]C[\/latex] be a smooth curve in a plane or in space and let [latex]f[\/latex] be a function with a domain that includes [latex]C[\/latex]. We chop the curve into small pieces. For each piece, we choose point [latex]P[\/latex] in that piece and evaluate [latex]f[\/latex] at [latex]P[\/latex]. (We can do this because all the points in the curve are in the domain of [latex]f[\/latex].) We multiply [latex]{f}{(P)}[\/latex] by the arc length of the piece [latex]{\\Delta}{s}[\/latex], add the product [latex]{f}{(P)}{\\Delta}{s}[\/latex] over all the pieces, and then let the arc length of the pieces shrink to zero by taking a limit. The result is the scalar line integral of the function over the curve.<\/p>\n<p id=\"fs-id1167794330401\">For a formal description of a scalar line integral, let [latex]C[\/latex] be a smooth curve in space given by the parameterization [latex]{\\textbf{r}}(t) = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. Let [latex]f(x, y, z)[\/latex] be a function with a domain that includes curve [latex]C[\/latex]. To define the line integral of the function [latex]f[\/latex] over [latex]C[\/latex], we begin as most definitions of an integral begin: we chop the curve into small pieces. 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\u00a0<em>i<\/em>th interval [latex][{{t}_{i-1}},{t_i}][\/latex]. 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 curve [latex]C[\/latex] into [latex]n[\/latex] pieces [latex]{C}_{1},{C}_{2},...,{C}_{n}[\/latex], with lengths [latex]{\\Delta}{{s}_{1}}, {\\Delta}{{s}_{2}},..., {\\Delta}{{s}_{n}}[\/latex], respectively. Let [latex]{P}^{*}_{i}[\/latex] denote the endpoint of [latex]{\\textbf{r}}({t^{*}_{i}})[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex]. Now, we evaluate the function [latex]f[\/latex] at point [latex]{P}^{*}_{i}[\/latex] for [latex]{1} \\leq {i} \\leq {n}[\/latex]. Note that [latex]{P}^{*}_{i}[\/latex] is in piece [latex]C_1[\/latex], and therefore [latex]{P}^{*}_{i}[\/latex] is in the domain of [latex]f[\/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[\/latex] by the length [latex]{\\Delta}{{s}_{1}}[\/latex] of [latex]C_1[\/latex], which gives the area of the \u201csheet\u201d with base [latex]C_1[\/latex], and height [latex]{f}{({P}^{*}_{i})}[\/latex]. This is analogous to using rectangles to approximate area in a single-variable integral. Now, we form the sum [latex]{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex]. Note the similarity of this sum versus a Riemann sum; in fact, this definition is a generalization of a Riemann sum to arbitrary curves in space. Just as with Riemann sums and integrals of form [latex]{\\displaystyle\\int^{b}_{a}}{g}{(x)}{dx}[\/latex], we define an integral by letting the width of the pieces of the curve shrink to zero by taking a limit. The result is the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex].s<\/p>\n<div style=\"width: 536px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/ef803993941028598bec179194a610712f5ff7d6\" alt=\"A diagram of a curve in quadrant one. Several points and segments are labeled. Starting at the left, the first points are P_0 and P_1. The segment between them is labeled delta S_1. The next points are P_i-1, P_i, and P_i+1. The segments connecting them are delta S_i and delta S_j+1. Point P_i starred and point P_i+1 starred are located on each segment, respectively. The last two points are P_n-1 and P_n, connected by segment S_n.\" width=\"526\" height=\"310\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Curve [latex]C[\/latex] has been divided into [latex]n[\/latex] pieces, and a point inside each piece has been chosen.<\/p>\n<\/div>\n<p>You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths [latex]{\\Delta}{{s}_{1}}, {\\Delta}{{s}_{2}},..., {\\Delta}{{s}_{n}}[\/latex] aren\u2019t necessarily the same; in the definition of a single-variable integral, the curve in the [latex]x[\/latex]-axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167793842996\">Let [latex]f[\/latex] be a function with a domain that includes the smooth curve [latex]C[\/latex] that is parameterized by [latex]{\\textbf{r}}{(t)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. The\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term242\" data-type=\"term\">scalar line integral<\/span><\/strong>\u00a0of[latex]f[\/latex] along [latex]C[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{f}{({x},{y},{z})}{ds} = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex]<\/p>\n<p id=\"fs-id1167793603762\">if this limit exists ([latex]{t}^{*}_{i}[\/latex] and [latex]{\\Delta}{s}_{i}[\/latex] are defined as in the previous paragraphs). If [latex]C[\/latex] is a planar curve, then [latex]C[\/latex] can be represented by the parametric equations [latex]x=x(t)[\/latex], [latex]y=y(t)[\/latex], and [latex]{a} \\leq {t} \\leq {b}[\/latex]. If [latex]C[\/latex] is smooth and [latex]f(x, y)[\/latex] is a function of two variables, then the scalar line integral of [latex]f[\/latex] along [latex]C[\/latex] is defined similarly as<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds} = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\\Delta}{s_{i}}[\/latex],<\/p>\n<p id=\"fs-id1167794296597\">if this limit exists.<\/p>\n<\/div>\n<p id=\"fs-id1167793417464\">If [latex]f[\/latex] is a continuous function on a smooth curve [latex]C[\/latex], then [latex]{\\displaystyle\\int_C}{f}{d}{s}[\/latex] always exists. Since [latex]{\\displaystyle\\int_C}{f}{d}{s}[\/latex] is defined as a limit of Riemann sums, the continuity of [latex]f[\/latex] is enough to guarantee the existence of the limit, just as the integral [latex]{\\displaystyle\\int^{b}_{a}}{g{(x)}}{dx}[\/latex] exists if [latex]g[\/latex] is continuous over [latex][{a},{b}][\/latex].<\/p>\n<p id=\"fs-id1167793432079\">Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that [latex]{f}{({x},{y})} \\geq {0}[\/latex] for all points [latex](x, y)[\/latex] on a smooth planar curve [latex]C[\/latex]. Imagine taking curve [latex]C[\/latex] and projecting it \u201cup\u201d to the surface defined by [latex]f(x, y)[\/latex], thereby creating a new curve [latex]{C}^{\\prime}[\/latex] that lies in the graph of [latex]f(x, y)[\/latex] (Figure 2). Now we drop a \u201csheet\u201d from [latex]{C}^{\\prime}[\/latex] down to the [latex]xy[\/latex]-plane. The area of this sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex]. If [latex]{f}{({x},{y})} \\leq {0}[\/latex] for some points in [latex]C[\/latex], then the value of [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex] is the area above the [latex]xy[\/latex]-plane less the area below the [latex]xy[\/latex]-plane. (Note the similarity with integrals of the form [latex]{\\displaystyle\\int^{b}_{a}}{g}{(x)}{dx}[\/latex].)<\/p>\n<div style=\"width: 415px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/d8f17ee89085c3bbc74b92268857dbfff02aa02d\" alt=\"A diagram in three dimensions. The original curve C in the (x,y) plane looks like a parabola opening to the left with vertex in quadrant 1. The surface defined by f(x,y) is shown always above the (x,y) plane. A curve on the surface directly above the original curve C is labeled as C\u2019. A blue sheet stretches down from C\u2019 to C.\" width=\"405\" height=\"386\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The area of the blue sheet is [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex].<\/p>\n<\/div>\n<p>From this geometry, we can see that line integral [latex]{\\displaystyle\\int_{C}}{f}{({x},{y})}{ds}[\/latex] does not depend on the parameterization [latex]{\\textbf{r}}{(t)}[\/latex] of [latex]C[\/latex]. As long as the curve is traversed exactly once by the parameterization, the area of the sheet formed by the function and the curve is the same. This same kind of geometric argument can be extended to show that the line integral of a three-variable function over a curve in space does not depend on the parameterization of the curve.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the value of a line integral<\/h3>\n<p>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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374562045\">Show Solution<\/span><\/p>\n<div id=\"q374562045\" class=\"hidden-answer\" style=\"display: none\">\n<p>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].<\/p>\n<div style=\"width: 434px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/b8c8f5c10b679e69719de9b1a42580cf84b973b0\" alt=\"A graph in three dimensions. There is a flat plane just above the (x,y) plane. The upper half of the unit circle in quadrants 1 and 2 of the (x,y) plane is raised up to form a semicircle sheet into the z-plane.\" width=\"424\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">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].<\/p>\n<\/div>\n<p id=\"fs-id1167793911831\">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\u00a0<strong data-effect=\"bold\">r<\/strong>. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\displaystyle\\int_{C}}{f}{ds} & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{f}{({\\textbf{r}}{({t}^{*}_{i})})}{\\Delta}{s_{i}} \\\\  & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{2}{\\Delta}{s_{i}} \\\\  & = {2}{\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{{\\displaystyle\\sum}^{n}_{i = 1}}{\\Delta}{s_{i}} \\\\  & = {2}{\\text{(length of C)}} \\\\  & = {2}{\\pi}.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the value of [latex]{\\displaystyle\\int_{C}}{({x} + {y})}{ds}[\/latex], where [latex]C[\/latex] is the curve parameterized by [latex]x=t[\/latex],\u00a0[latex]y=t[\/latex],\u00a0[latex]{0} \\leq {t} \\leq {1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q948375296\">Show Solution<\/span><\/p>\n<div id=\"q948375296\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793418697\">Note that in a scalar line integral, the integration is done with respect to arc length [latex]s[\/latex], which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate [latex]{\\displaystyle\\int_{C}}{f}{ds}[\/latex] to an integral with a variable of integration that is [latex]t[\/latex].<\/p>\n<p id=\"fs-id1167793937004\">Let [latex]{\\textbf{r}}{(t)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}[\/latex] for [latex]{a} \\leq {t} \\leq {b}[\/latex] be a parameterization of [latex]C[\/latex]. Since we are assuming that [latex]C[\/latex] is smooth, [latex]{\\textbf{r}}^{\\prime}{(t)} = {\\left \\langle {x}{\\prime}{(t)}, {y}{\\prime}{(t)}, {z}{\\prime}{(t)} \\right \\rangle}[\/latex] is continuous for all [latex]t[\/latex] in [latex][{a},{b}][\/latex]. In particular, [latex]{x}{\\prime}{(t)}, {y}^{\\prime}{(t)}[\/latex], and [latex]{z}^{\\prime}{(t)}[\/latex] exist for all [latex]t[\/latex] in [latex][{a},{b}][\/latex]. According to the arc length formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{length}}{({C}_{i})} = {\\Delta}{{s}_{i}} = {\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{d}{t}[\/latex].<\/p>\n<p id=\"fs-id1167794210535\">If width [latex]{\\Delta}{{t}_{i}} = {{t}_{i}} - {{t}_{{i} - {1}}}[\/latex] is small, then function [latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\| {\\Delta}{{t}_{i}} \\right \\|} {\\textbf{r}}^{\\prime}{(t)} \\right \\|}[\/latex] is almost constant over the interval [latex]\\left [ {{t}_{{i} - {1}}},{{t}_{i}} \\right ][\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\right \\|} {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}},[\/latex]<\/p>\n<p>and we have<\/p>\n<p style=\"text-align: center;\">[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>\n<p id=\"fs-id1167793881085\">See\u00a0Figure 4.<\/p>\n<div style=\"width: 429px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/apps\/archive\/20220118.185250\/resources\/6ac27b2f44bbefe4d435b72eb0d064b555debbef\" alt=\"A segment of an increasing concave down curve labeled C. A small segment of the curved is boxed and labeled as delta t_i. In the zoomed-in insert, this boxed segment of the curve is almost linear.\" width=\"419\" height=\"201\" \/><\/p>\n<p class=\"wp-caption-text\">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.<\/p>\n<\/div>\n<p id=\"fs-id1167793915895\">Note that<\/p>\n<p style=\"text-align: center;\">[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>\n<p>In other words, as the widths of intervals [latex][{{t}_{{i} - {1}}},{{t}_{i}}][\/latex] shrink to zero, the sum [latex]{\\displaystyle\\sum^{n}_{{i} = {1}}}{f}{({\\textbf{r}}{({t^{*}_{i}})})}{\\left \\| {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}}[\/latex] converges to the integral [latex]{\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)})}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt}[\/latex]. Therefore, we have the following theorem.<\/p>\n<div id=\"fs-id1167794058922\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: evaluating a scalar line integral<\/h3>\n<hr \/>\n<p id=\"fs-id1167794058930\">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>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_C}{f}{d}{s} = {\\displaystyle\\int^{b}_{a}}{f}{({\\textbf{r}}{(t)})}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793925270\">Although we have labeled\u00a0[latex]{\\displaystyle\\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} {\\left \\| {d}{t}{\\approx} \\right \\|} {\\textbf{r}}^{\\prime}{({t^{*}_{i}})} \\right \\|}{\\Delta}{{t}_{i}},[\/latex] as an equation, it is more accurately considered an approximation because we can show that the left-hand side of the equation approaches the right-hand side as [latex]{n}{\\rightarrow}{\\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. Since<\/p>\n<p style=\"text-align: center;\">[latex]{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|} = {\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2},}}[\/latex]<\/p>\n<p id=\"fs-id1167793380170\">we obtain the following theorem, which we use to compute scalar line integrals.<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: scalar line integral calculation<\/h3>\n<hr \/>\n<p id=\"fs-id1167793939706\">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)} = {\\left \\langle {x}{(t)}, {y}{(t)}, {z}{(t)} \\right \\rangle}, {a} \\leq {t} \\leq {b}[\/latex]. Then<\/p>\n<p style=\"text-align: center;\">[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>\n<p id=\"fs-id1167793944241\">Similarly,<\/p>\n<p style=\"text-align: center;\">[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>\n<p id=\"fs-id1167794137230\">if [latex]C[\/latex] is a planar curve and [latex]f[\/latex] is a function of two variables.<\/p>\n<\/div>\n<p>Note that a consequence of this theorem is the equation [latex]{ds} = {\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{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]{\\textbf{r}}^{\\prime}{(t)}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a line integral<\/h3>\n<p>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].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384725475\">Show Solution<\/span><\/p>\n<div id=\"q384725475\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793221724\">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 Theorem Equation\u00a0to 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>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} & = {\\sqrt{{({-}{\\sin{(t)}})^{2}} + {\\cos}^{2}{(t)} + {1}}} \\\\  & = {\\sqrt{2}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793930422\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} = {\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt}[\/latex].<\/p>\n<p id=\"fs-id1167794328944\">Notice that\u00a0The Scalar Line Integral Calculation Theorem Equation\u00a0translated the original difficult line integral into a manageable single-variable integral. Since<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\displaystyle\\int^{{2}{\\pi}}_{0}}{({1} + {t})}{\\sqrt{2}}{dt} & = {\\left [ {\\sqrt{2}}{t} + {\\frac{{\\sqrt{2}}{t}^{2}}{2}} \\right ]}^{{2}{\\pi}}_{0} \\\\  & = {2}{\\sqrt{2}}{\\pi} + {2}{\\sqrt{2}}{\\pi}^{2},  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793633032\">we have<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} = {2}{\\sqrt{2}}{\\pi} + {2}{\\sqrt{2}}{\\pi}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate [latex]{\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds}[\/latex], where [latex]C[\/latex] is the curve with parameterization\u00a0[latex]{\\textbf{r}}{(t)} = {\\left \\langle {\\sin}{(3t)}, {\\cos}{(3t), {t}} \\right \\rangle}, {0} \\leq {t} \\leq {2}{\\pi}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q094320778\">Show Solution<\/span><\/p>\n<div id=\"q094320778\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]2\\sqrt{10}\\pi+2\\sqrt{10}\\pi^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: independence of parameterization<\/h3>\n<p>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]. 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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371932814\">Show Solution<\/span><\/p>\n<div id=\"q371932814\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793978351\">As with the previous example, we use\u00a0The Scalar Line Integral Calculation Theorem Equation 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>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\sqrt{{({x}^{\\prime}{(t)})}^{2} + {({y}^{\\prime}{(t)})}^{2} + {({z}^{\\prime}{(t)})}^{2}}} & = {\\sqrt{({-}{\\sin{t}} + {\\cos{t}} + {4})}} \\\\  & = {2}{\\sqrt{2}}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793244536\">so we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\displaystyle\\int_{C}}{({x^2} + {y^2} + {z})}{ds} & = {2}{\\sqrt{2}}{\\displaystyle\\int^{\\pi}_{0}}{({1} + {2t})}{dt} \\\\  & = {2}{\\sqrt{2}}{[{t} + {t}^{2}]}^{\\pi}_{0} \\\\  & = {2}{\\sqrt{2}}{({\\pi} + {\\pi}^{2})}.  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793959042\">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>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate line integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {yz})}{ds}[\/latex], where [latex]C[\/latex] is the line with parameterization [latex]{\\textbf{r}}{(t)} = {\\left \\langle {2t}, {5t}, {-t} \\right \\rangle}, {0} \\leq {t} \\leq {10}[\/latex]. Reparameterize [latex]C[\/latex] with parameterization [latex]{\\textbf{s}}{(t)} = {\\left \\langle {4t}, {10t}, {-2t} \\right \\rangle}, {0} \\leq {t} \\leq {5}[\/latex], recalculate line integral [latex]{\\displaystyle\\int_{C}}{({x^2} + {yz})}{ds}[\/latex], and notice that the change of parameterization had no effect on the value of the integral.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q872471336\">Show Solution<\/span><\/p>\n<div id=\"q872471336\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both line integrals equal\u00a0[latex]-\\frac{1,000\\sqrt{30}}3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250313&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3EIXmDq7zfY&amp;video_target=tpm-plugin-ojc8ic91-3EIXmDq7zfY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.15_transcript.html\">transcript for \u201cCP 6.15\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167793401189\">Now that we can evaluate line integrals, we can use them to calculate arc length. If [latex]f(x, y, z)=1[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\displaystyle\\int_{C}}{f}{({x}, {y}, {z})}{ds} & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}\\displaystyle{\\sum^{n}_{i = 1}}{f}{({t}^{*}_{i})}{\\Delta}{{s}_{i}} \\\\  & = \\displaystyle{\\lim_{{n}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{n}_{i = 1}}{\\Delta}{{s}_{i}} \\\\  & = {\\displaystyle\\lim_{{n}{\\rightarrow}{\\infty}}}{\\text{length}}{(C)} \\\\  & = {\\text{length}}{(C)}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793420990\">Therefore, [latex]{\\displaystyle\\int_{C}}{1}{ds}[\/latex] is the arc length of [latex]C[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: calculating arc length<\/h3>\n<p>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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q092499841\">Show Solution<\/span><\/p>\n<div id=\"q092499841\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793383514\">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>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\text{The length of the wire}} & = {\\displaystyle\\int_{C}}{1}{ds} \\\\  & = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\left \\| {\\textbf{r}}^{\\prime}{(t)} \\right \\|}{dt} \\\\  & = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{({-}{\\sin{t}})^{2} + {\\cos}^{2}{t} + {tdt}}} \\\\  & = {\\displaystyle\\int^{{4}{\\pi}}_{0}}{\\sqrt{{1} + {tdt}}} \\\\  & = {\\left [ {\\frac{{2}{({1} + {t})^{3\/2}}}{3}} \\right ]}^{{4}{\\pi}}_{0} \\\\  & = {\\frac{2}{3}}{\\left (({1} + {4{\\pi}})^{3\/2} - {1} \\right )}.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the length of a wire with parameterization [latex]{\\textbf{r}}{(t)} = {\\left \\langle {3t} + {1}, {4} - {2t},{5} + {2t} \\right \\rangle}, {0} \\leq {t} \\leq {4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q442837426\">Show Solution<\/span><\/p>\n<div id=\"q442837426\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]4\\sqrt{17}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5476\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 6.15. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 6.15\",\"author\":\"Ryan 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