{"id":1119,"date":"2021-11-09T23:41:14","date_gmt":"2021-11-09T23:41:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1119"},"modified":"2022-11-01T05:07:09","modified_gmt":"2022-11-01T05:07:09","slug":"applying-line-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/applying-line-integrals\/","title":{"raw":"Applying Line Integrals","rendered":"Applying 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\">Use a line integral to compute the work done in moving an object along a curve in a vector field.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Describe the flux and circulation of a vector field.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Applications of Line Integrals<\/h2>\r\n<p id=\"fs-id1167793482851\">Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term247\" class=\"no-emphasis\" data-type=\"term\">mass of a wire<\/span>\u00a0using a scalar line integral and the work done by a force using a vector line integral.<\/p>\r\n<p id=\"fs-id1167793693424\">Suppose that a piece of wire is modeled by curve [latex]C[\/latex] in space. The mass per unit length (the linear density) of the wire is a continuous function [latex]{\\rho}{({x}, {y}, {z})}[\/latex]. We can calculate the total mass of the wire using the scalar line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex]. The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by [latex]{\\rho}{({x}^{*}, {y}^{*}, {z}^{*})}{\\Delta}{s}[\/latex] for some point [latex]({x}^{*}, {y}^{*}, {z}^{*})[\/latex] in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167794143327\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating the mass of a wire<\/h3>\r\nCalculate the mass of a spring in the shape of a helix parameterized by [latex]\\langle{t},2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq\\frac{\\pi}2[\/latex], with a density function given by [latex]\\rho(x,y,z)=e^x+yz[\/latex] kg\/m.\r\n\r\n[caption id=\"attachment_5175\" align=\"aligncenter\" width=\"421\"]<img class=\"size-full wp-image-5175\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174118\/6.21.jpg\" alt=\"A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi\/2, 0, 2). The arrow on the curve is pointing to the latter endpoint.\" width=\"421\" height=\"522\" \/> Figure 1. The wire from Example \"Calculating the Mass of a Wire\".[\/caption]\r\n\r\n[reveal-answer q=\"394875298\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394875298\"]\r\n<p id=\"fs-id1167793419129\">To calculate the mass of the spring, we must find the value of the scalar line integral [latex]\\displaystyle\\int_c(e^x+yz)ds[\/latex], where [latex]C[\/latex] is the given helix. To calculate this integral, we write it in terms of [latex]t[\/latex]\u00a0using The Scalar Line Integral Calculation Theorem Equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_ce^x+yzds&amp;=\\displaystyle\\int_0^{\\pi\/2}\\left((e^t+4\\cos{t}\\sin{t})\\sqrt{1+(-2\\cos{t}^2+(2\\sin{t})^2}\\right)dt \\\\\r\n&amp;=\\displaystyle\\int_0^{\\pi\/2}((e^t+4\\cos{t}\\sin{t})\\sqrt{5})dt \\\\\r\n&amp;=\\sqrt{5}[e^t+2\\sin^2t]_{t=0}^{t=\\pi\/2} \\\\\r\n&amp;=\\sqrt{5}(e^{\\pi\/2}+1)\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167794223613\">Therefore, the mass is [latex]\\sqrt{5}(e^{\\pi\/2}+1)[\/latex] kg.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the mass of a spring in the shape of a helix parameterized by [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex], with a density function given by\u00a0[latex]\\rho(x,y,z)=x+y+z[\/latex] kg\/m.\r\n\r\n[reveal-answer q=\"904921475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"904921475\"]\r\n\r\n[latex]18\\sqrt2\\pi^2[\/latex] kg.\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=8250316&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=WwobFvL33Jc&amp;video_target=tpm-plugin-k9958w5c-WwobFvL33Jc\" 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\/CP6.20_transcript.html\">transcript for \u201cCP 6.20\u201d here (opens in new window).<\/a><\/center>When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term248\" class=\"no-emphasis\" data-type=\"term\">work done by a vector field<\/span>\u00a0representing a force is a standard use of vector line integrals. Recall that if an object moves along curve [latex]C[\/latex] in force field\u00a0[latex]{\\bf{F}}[\/latex], then the work required to move the object is given by [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating work<\/h3>\r\nHow much work is required to move an object in vector force field [latex]{\\bf{F}}=\\langle{y}x,xy,xz\\rangle[\/latex] along path [latex]{\\bf{r}}(t)=\\langle{t}^2,t,t^4\\rangle[\/latex],\u00a0[latex]0\\leq{t}\\leq1[\/latex]?\u00a0 See\u00a0Figure 2.\r\n<div data-type=\"note\">\r\n\r\n[reveal-answer q=\"218730126\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218730126\"]\r\n<p id=\"fs-id1167793631774\">Let [latex]C[\/latex] denote the given path. We need to find the value of [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex]. To do this, we use the equation to compute vector line integrals:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}&amp;=\\displaystyle\\int_0^1(\\langle{t}^5,t^3,t^6\\rangle\\cdot\\langle2t,1,4t^3\\rangle)dt\\\\\r\n&amp;=\\displaystyle\\int_0^1(2t^6+t^3+4t^9) \\ dt \\\\\r\n&amp;=\\left[\\frac{2t^7}7+\\frac{t^4}4+\\frac{2t^{10}}5\\right]_{t=0}^{t=1}=\\frac{131}{140}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_5178\" align=\"aligncenter\" width=\"316\"]<img class=\"size-full wp-image-5178\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174320\/6.22.jpg\" alt=\"A three dimensional diagram of the curve and vector field for the example. The curve is an increasing concave up curve starting close to the origin and above the x axis. As the curve goes left above the (x,y) plane, the height also increases. The arrows in the vector field get longer as the z component becomes larger.\" width=\"316\" height=\"597\" \/> Figure 2. The curve and vector field for Example \"Calculating Work\".[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2 data-type=\"title\">Flux and Circulation<\/h2>\r\n<p id=\"fs-id1167793299873\">We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green\u2019s theorem, and in higher dimensions for Stokes\u2019 theorem and the divergence theorem.<\/p>\r\n<p id=\"fs-id1167793299884\">Let [latex]C[\/latex] be a plane curve and let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field in the plane. Imagine [latex]C[\/latex] is a membrane across which fluid flows, but [latex]C[\/latex] does not impede the flow of the fluid. In other words, [latex]C[\/latex] is an idealized membrane invisible to the fluid. Suppose\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing [latex]C[\/latex]?<\/p>\r\n<p id=\"fs-id1167793299924\">Recall that the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] is [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]\u2014in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector [latex]{\\bf{N}}(t)[\/latex] and instead compute integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds[\/latex], we determine the flux across [latex]C[\/latex]. To be precise, the definition of integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds[\/latex] is the same as integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex], except the\u00a0[latex]{\\bf{T}}[\/latex]\u00a0in the Riemann sum is replaced with\u00a0[latex]{\\bf{N}}[\/latex]. Therefore, the flux across [latex]C[\/latex] is defined as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds=\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{i=1}^n{\\bf{F}}(P_i^*)\\cdot{\\bf{N}}(P_i^*)\\Delta{s_i}}[\/latex],<\/p>\r\nwhere [latex]P_i^*[\/latex] and [latex]\\Delta{s}[\/latex] are defined as they were for integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]. Therefore, a flux integral is an integral that is\u00a0<em data-effect=\"italics\">perpendicular<\/em>\u00a0to a vector line integral, because [latex]{\\bf{N}}[\/latex] and\u00a0[latex]{\\bf{T}}[\/latex]\u00a0are perpendicular vectors.\r\n<p id=\"fs-id1167793551306\">If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a velocity field of a fluid and [latex]C[\/latex] is a curve that represents a membrane, then the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]C[\/latex] is the quantity of fluid flowing across [latex]C[\/latex] per unit time, or the rate of flow.<\/p>\r\n<p id=\"fs-id1167793750503\">More formally, let [latex]C[\/latex] be a plane curve parameterized by [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex] be the vector that is normal to [latex]C[\/latex] at the endpoint of [latex]{\\bf{r}}(t)[\/latex] and points to the right as we traverse [latex]C[\/latex] in the positive direction (Figure 3). Then, [latex]{\\bf{N}}(t)=\\frac{{\\bf{n}}(t)}{||{\\bf{n}}(t)||}[\/latex] is the unit normal vector to [latex]C[\/latex] at the endpoint of [latex]{\\bf{r}}(t)[\/latex] that points to the right as we traverse [latex]C[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167793460505\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term249\" data-type=\"term\">flux<\/span>\u00a0<\/strong>of\u00a0[latex]\\bf{F}[\/latex]\u00a0across [latex]C[\/latex] is line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}(t)}{||{\\bf{n}}(t)||}[ds[\/latex].\r\n\r\n<\/div>\r\n[caption id=\"attachment_5242\" align=\"aligncenter\" width=\"522\"]<img class=\"size-full wp-image-5242\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222343\/6.23.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/f73f7f23dee876375a79ebe5a3321b2c37df15fd&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A simple diagram of an increasing concave down curve C in vector field F, with no coordinate plane. Towards the top of the curve, the normal n is drawn perpendicular to the curve C. Another arrow F is drawn sharing n\u2019s endpoint. This flux points up and to the right at about a 90-degree angle to n. The arrows in the vector field to the left of n are drawn pointing straight up. The arrows after n point in the same direction as the flux.&quot; id=&quot;37&quot;&gt;\" width=\"522\" height=\"491\" \/> Figure 3. The flux of vector field [latex]\\bf{F}[\/latex]\u00a0across curve [latex]C[\/latex] is computed by an integral similar to a vector line integral.[\/caption]We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see\u00a0the equation to compute vector line integrals).\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: calculating flux across a curve<\/h3>\r\n<p id=\"fs-id1167793662617\">Let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field and let [latex]C[\/latex] be a smooth curve with parameterization [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex]. The flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]C[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Proof<\/h2>\r\n<p id=\"fs-id1167793482875\">The proof of The Calculating Flux Across A Curve Theorem Equation\u00a0is similar to the proof of The Scalar Line Integration Calculation Theorem Equation. Before deriving the formula, note that [latex]||{\\bf{n}}(t)||=||\\langle y'(t),-x'(t)\\rangle||=\\sqrt{(y^\\prime(t))^2+(x^\\prime(t))^2}=||{\\bf{r}}^\\prime(t)||[\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&amp;=\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}ds \\\\\r\n&amp;=\\displaystyle\\int_a^b{\\bf{F}}\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}||{\\bf{r}}^\\prime(t)||dt \\\\\r\n&amp;=\\displaystyle\\int_a^b{\\bf{F}}[({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt\r\n\\end{aligned}[[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: flux across a curve<\/h3>\r\nCalculate the flux of [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex] across a unit circle oriented counterclockwise (Figure 4).\r\n\r\n[caption id=\"attachment_5244\" align=\"aligncenter\" width=\"496\"]<img class=\"size-full wp-image-5244\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222518\/6.24.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/1284547517bb19bc89545262451458a9b835e511&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A unit circle in a vector field in two dimensions. The arrows point away from the origin in a radial pattern. Shorter vectors are near the origin, and longer ones are further away. A unit circle is drawn around the origin to fit the pattern, and arrowheads are drawn on the circle in a counterclockwise manner.&quot; id=&quot;40&quot;&gt;\" width=\"496\" height=\"272\" \/> Figure 4. A unit circle in vector field [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex].[\/caption][reveal-answer q=\"589274091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589274091\"]\r\n<p id=\"fs-id1167793671536\">To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization [latex]{\\bf{r}}(t)\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. The normal vector to a unit circle is [latex]\\langle\\cos{t},\\sin{t}\\rangle[\/latex]. Therefore, the flux is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&amp;=\\displaystyle\\int_0^{2\\pi}\\langle2\\cos{t},2\\sin{t}\\rangle\\cdot\\langle\\cos{t},\\sin{t}\\rangle \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}(2\\cos^2{t},2\\sin^2{t}) \\ dt = 2\\displaystyle\\int_0^{2\\pi}(\\cos^2{t},\\sin^2{t}) \\ dt \\\\\r\n&amp;=2\\displaystyle\\int_0^{2\\pi} \\ dt = 4\\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\nCalculate the flux of [latex]{\\bf{F}}=\\langle{x}+y,2y\\rangle[\/latex] across the line segment from [latex](0, 0)[\/latex] to [latex](2, 3)[\/latex] where the curve is oriented from left to right.\r\n\r\n[reveal-answer q=\"203497590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203497590\"]\r\n\r\n[latex]3\/2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793594106\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle[\/latex] be a two-dimensional vector field. Recall that integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{T}}ds[\/latex] is sometimes written as [latex]\\displaystyle\\int_CPdx+Qdy[\/latex]. Analogously, flux [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds[\/latex] is sometimes written in the notation [latex]\\displaystyle\\int_C-Qdx+Pdy[\/latex], because the unit normal vector\u00a0[latex]\\bf{N}[\/latex]\u00a0is perpendicular to the unit tangent\u00a0[latex]\\bf{T}[\/latex]. Rotating the vector [latex]d{\\bf{r}}=\\langle{d}x,dy\\rangle[\/latex] by 90\u00b0 results in vector [latex]\\langle{d}y,-dx\\rangle[\/latex]. Therefore, the line integral in\u00a0Example \"Using Properties to Compute a Vector Line Integral\"\u00a0can be written as [latex]\\displaystyle\\int_C-2ydx+2xdy[\/latex].<\/p>\r\n<p id=\"fs-id1167793617594\">Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field\u00a0[latex]\\bf{F}[\/latex]\u00a0along an oriented closed curve is called the\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term250\" data-type=\"term\">circulation<\/span><\/strong>\u00a0of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex]. Circulation line integrals have their own notation: [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds[\/latex]. The circle on the integral symbol denotes that [latex]C[\/latex] is \u201ccircular\u201d in that it has no endpoints.\u00a0Example \"Evaluating a Vector Line Integral\"\u00a0shows a calculation of circulation.<\/p>\r\n<p id=\"fs-id1167793706001\">To see where the term\u00a0<em data-effect=\"italics\">circulation<\/em>\u00a0comes from and what it measures, let\u00a0[latex]\\bf{v}[\/latex]\u00a0represent the velocity field of a fluid and let [latex]C[\/latex] be an oriented closed curve. At a particular point [latex]P[\/latex], the closer the direction of [latex]{\\bf{v}}(P)[\/latex] is to the direction of [latex]{\\bf{T}}(P)[\/latex], the larger the value of the dot product [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex]. The maximum value of [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex] occurs when the two vectors are pointing in the exact same direction; the minimum value of [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex] occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation [latex]\\displaystyle\\oint_C{\\bf{v}}\\cdot{\\bf{T}}ds[\/latex] measures the tendency of the fluid to move in the direction of [latex]C[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating circulation<\/h3>\r\nLet [latex]{\\bf{F}}=\\langle-y,x\\rangle[\/latex] be the vector field from\u00a0Example \"Independence of Parameterization\"\u00a0and let [latex]C[\/latex] represent the unit circle oriented counterclockwise. Calculate the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex].\r\n\r\n[reveal-answer q=\"039258487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"039258487\"]\r\n<p id=\"fs-id1167793355114\">We use the standard parameterization of the unit circle: [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex] and [latex]{\\bf{r}}'(t)=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds&amp;=\\displaystyle\\int_0^{2\\pi}\\langle-\\sin{t},\\cos{t}\\rangle\\cdot\\langle-\\sin{t},\\cos{t}\\rangle \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}(\\sin^2{t}+\\cos^2{t}) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi} \\ dt = 2\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793662249\">Notice that the circulation is positive. The reason for this is that the orientation of [latex]C[\/latex] \u201cflows\u201d with the direction of\u00a0[latex]\\bf{F}[\/latex]. At any point along the circle, the tangent vector and the vector from\u00a0[latex]\\bf{F}[\/latex]\u00a0form an angle of less than 90\u00b0, and therefore the corresponding dot product is positive.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793924502\">In\u00a0Example \"Calculating Circulation\", what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by [latex]-C[\/latex]. Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_{-C}{\\bf{F}}\\cdot{\\bf{T}} \\ ds}=-\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}=-2\\pi[\/latex].<\/p>\r\n<p id=\"fs-id1167793641713\">Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of\u00a0[latex]\\bf{F}[\/latex].<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the circulation of [latex]{\\bf{F}}(x,y)=\\left\\langle-\\frac{y}{x^2+y^2},\\frac{y}{x^2+y^2}\\right\\rangle[\/latex] along a unit circle oriented counterclockwise.\r\n\r\n[reveal-answer q=\"834752048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834752048\"]\r\n\r\n[latex]2\\pi[\/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=8250315&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oWRQU5Oqtl0&amp;video_target=tpm-plugin-lldll002-oWRQU5Oqtl0\" 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\/CP6.22_transcript.html\">transcript for \u201cCP 6.22\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating work<\/h3>\r\nCalculate the work done on a particle that traverses circle [latex]C[\/latex] of radius 2 centered at the origin, oriented counterclockwise, by field [latex]{\\bf{F}}(x,y)=\\langle-2,y\\rangle[\/latex]. Assume the particle starts its movement at [latex](1, 0)[\/latex].\r\n\r\n[reveal-answer q=\"802581366\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"802581366\"]\r\n<p id=\"fs-id1167794090726\">The work done by\u00a0[latex]\\bf{F}[\/latex]\u00a0on the particle is the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex]: [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]. We use the parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex] for [latex]C[\/latex]. Then, [latex]{\\bf{r}}^\\prime(t)=\\langle-2\\cos{t},2\\sin{t}\\rangle[\/latex] and [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-2,2\\sin{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds&amp;=\\displaystyle\\int_0^{2\\pi}\\langle-2,2\\sin{t}\\rangle\\cdot\\langle-2,2\\cos{t}\\rangle \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}(4\\sin{t}+4\\sin{t}\\cos{t}) \\ dt \\\\\r\n&amp;=\\left[-4\\cos{t}+4\\sin^2{t}\\right]_0^{2\\pi} \\\\\r\n&amp;=(-4\\cos(2\\pi)+2\\sin^2(2\\pi))-(-4\\cos(0)+4\\sin^2(0)) \\\\\r\n&amp;=-4+4=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793398124\">The force field does zero work on the particle.<\/p>\r\n<p id=\"fs-id1167793398127\">Notice that the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is zero. Furthermore, notice that since\u00a0[latex]\\bf{F}[\/latex]\u00a0is the gradient of [latex]f(x,y)=-2x+\\frac{y^2}2[\/latex],\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the work done by field [latex]{\\bf{F}}(x,y)=\\langle2x,3y\\rangle[\/latex] on a particle that traverses the unit circle. Assume the particle begins its movement at [latex](-1, 0)[\/latex].\r\n\r\n[reveal-answer q=\"734256411\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734256411\"]\r\n\r\n[latex]0[\/latex].\r\n\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\">Use a line integral to compute the work done in moving an object along a curve in a vector field.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Describe the flux and circulation of a vector field.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Applications of Line Integrals<\/h2>\n<p id=\"fs-id1167793482851\">Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term247\" class=\"no-emphasis\" data-type=\"term\">mass of a wire<\/span>\u00a0using a scalar line integral and the work done by a force using a vector line integral.<\/p>\n<p id=\"fs-id1167793693424\">Suppose that a piece of wire is modeled by curve [latex]C[\/latex] in space. The mass per unit length (the linear density) of the wire is a continuous function [latex]{\\rho}{({x}, {y}, {z})}[\/latex]. We can calculate the total mass of the wire using the scalar line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex]. The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by [latex]{\\rho}{({x}^{*}, {y}^{*}, {z}^{*})}{\\Delta}{s}[\/latex] for some point [latex]({x}^{*}, {y}^{*}, {z}^{*})[\/latex] in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex].<\/p>\n<div id=\"fs-id1167794143327\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: calculating the mass of a wire<\/h3>\n<p>Calculate the mass of a spring in the shape of a helix parameterized by [latex]\\langle{t},2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq\\frac{\\pi}2[\/latex], with a density function given by [latex]\\rho(x,y,z)=e^x+yz[\/latex] kg\/m.<\/p>\n<div id=\"attachment_5175\" style=\"width: 431px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5175\" class=\"size-full wp-image-5175\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174118\/6.21.jpg\" alt=\"A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi\/2, 0, 2). The arrow on the curve is pointing to the latter endpoint.\" width=\"421\" height=\"522\" \/><\/p>\n<p id=\"caption-attachment-5175\" class=\"wp-caption-text\">Figure 1. The wire from Example &#8220;Calculating the Mass of a Wire&#8221;.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q394875298\">Show Solution<\/span><\/p>\n<div id=\"q394875298\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793419129\">To calculate the mass of the spring, we must find the value of the scalar line integral [latex]\\displaystyle\\int_c(e^x+yz)ds[\/latex], where [latex]C[\/latex] is the given helix. To calculate this integral, we write it in terms of [latex]t[\/latex]\u00a0using The Scalar Line Integral Calculation Theorem Equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_ce^x+yzds&=\\displaystyle\\int_0^{\\pi\/2}\\left((e^t+4\\cos{t}\\sin{t})\\sqrt{1+(-2\\cos{t}^2+(2\\sin{t})^2}\\right)dt \\\\  &=\\displaystyle\\int_0^{\\pi\/2}((e^t+4\\cos{t}\\sin{t})\\sqrt{5})dt \\\\  &=\\sqrt{5}[e^t+2\\sin^2t]_{t=0}^{t=\\pi\/2} \\\\  &=\\sqrt{5}(e^{\\pi\/2}+1)  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167794223613\">Therefore, the mass is [latex]\\sqrt{5}(e^{\\pi\/2}+1)[\/latex] kg.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the mass of a spring in the shape of a helix parameterized by [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex], with a density function given by\u00a0[latex]\\rho(x,y,z)=x+y+z[\/latex] kg\/m.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q904921475\">Show Solution<\/span><\/p>\n<div id=\"q904921475\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]18\\sqrt2\\pi^2[\/latex] kg.<\/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=8250316&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=WwobFvL33Jc&amp;video_target=tpm-plugin-k9958w5c-WwobFvL33Jc\" 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\/CP6.20_transcript.html\">transcript for \u201cCP 6.20\u201d here (opens in new window).<\/a><\/div>\n<p>When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term248\" class=\"no-emphasis\" data-type=\"term\">work done by a vector field<\/span>\u00a0representing a force is a standard use of vector line integrals. Recall that if an object moves along curve [latex]C[\/latex] in force field\u00a0[latex]{\\bf{F}}[\/latex], then the work required to move the object is given by [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: calculating work<\/h3>\n<p>How much work is required to move an object in vector force field [latex]{\\bf{F}}=\\langle{y}x,xy,xz\\rangle[\/latex] along path [latex]{\\bf{r}}(t)=\\langle{t}^2,t,t^4\\rangle[\/latex],\u00a0[latex]0\\leq{t}\\leq1[\/latex]?\u00a0 See\u00a0Figure 2.<\/p>\n<div data-type=\"note\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q218730126\">Show Solution<\/span><\/p>\n<div id=\"q218730126\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793631774\">Let [latex]C[\/latex] denote the given path. We need to find the value of [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex]. To do this, we use the equation to compute vector line integrals:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}&=\\displaystyle\\int_0^1(\\langle{t}^5,t^3,t^6\\rangle\\cdot\\langle2t,1,4t^3\\rangle)dt\\\\  &=\\displaystyle\\int_0^1(2t^6+t^3+4t^9) \\ dt \\\\  &=\\left[\\frac{2t^7}7+\\frac{t^4}4+\\frac{2t^{10}}5\\right]_{t=0}^{t=1}=\\frac{131}{140}.  \\end{aligned}[\/latex]<\/p>\n<div id=\"attachment_5178\" style=\"width: 326px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5178\" class=\"size-full wp-image-5178\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174320\/6.22.jpg\" alt=\"A three dimensional diagram of the curve and vector field for the example. The curve is an increasing concave up curve starting close to the origin and above the x axis. As the curve goes left above the (x,y) plane, the height also increases. The arrows in the vector field get longer as the z component becomes larger.\" width=\"316\" height=\"597\" \/><\/p>\n<p id=\"caption-attachment-5178\" class=\"wp-caption-text\">Figure 2. The curve and vector field for Example &#8220;Calculating Work&#8221;.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Flux and Circulation<\/h2>\n<p id=\"fs-id1167793299873\">We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green\u2019s theorem, and in higher dimensions for Stokes\u2019 theorem and the divergence theorem.<\/p>\n<p id=\"fs-id1167793299884\">Let [latex]C[\/latex] be a plane curve and let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field in the plane. Imagine [latex]C[\/latex] is a membrane across which fluid flows, but [latex]C[\/latex] does not impede the flow of the fluid. In other words, [latex]C[\/latex] is an idealized membrane invisible to the fluid. Suppose\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing [latex]C[\/latex]?<\/p>\n<p id=\"fs-id1167793299924\">Recall that the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] is [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]\u2014in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector [latex]{\\bf{N}}(t)[\/latex] and instead compute integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds[\/latex], we determine the flux across [latex]C[\/latex]. To be precise, the definition of integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds[\/latex] is the same as integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex], except the\u00a0[latex]{\\bf{T}}[\/latex]\u00a0in the Riemann sum is replaced with\u00a0[latex]{\\bf{N}}[\/latex]. Therefore, the flux across [latex]C[\/latex] is defined as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds=\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{i=1}^n{\\bf{F}}(P_i^*)\\cdot{\\bf{N}}(P_i^*)\\Delta{s_i}}[\/latex],<\/p>\n<p>where [latex]P_i^*[\/latex] and [latex]\\Delta{s}[\/latex] are defined as they were for integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]. Therefore, a flux integral is an integral that is\u00a0<em data-effect=\"italics\">perpendicular<\/em>\u00a0to a vector line integral, because [latex]{\\bf{N}}[\/latex] and\u00a0[latex]{\\bf{T}}[\/latex]\u00a0are perpendicular vectors.<\/p>\n<p id=\"fs-id1167793551306\">If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a velocity field of a fluid and [latex]C[\/latex] is a curve that represents a membrane, then the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]C[\/latex] is the quantity of fluid flowing across [latex]C[\/latex] per unit time, or the rate of flow.<\/p>\n<p id=\"fs-id1167793750503\">More formally, let [latex]C[\/latex] be a plane curve parameterized by [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex] be the vector that is normal to [latex]C[\/latex] at the endpoint of [latex]{\\bf{r}}(t)[\/latex] and points to the right as we traverse [latex]C[\/latex] in the positive direction (Figure 3). Then, [latex]{\\bf{N}}(t)=\\frac{{\\bf{n}}(t)}{||{\\bf{n}}(t)||}[\/latex] is the unit normal vector to [latex]C[\/latex] at the endpoint of [latex]{\\bf{r}}(t)[\/latex] that points to the right as we traverse [latex]C[\/latex].<\/p>\n<div id=\"fs-id1167793460505\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>The\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term249\" data-type=\"term\">flux<\/span>\u00a0<\/strong>of\u00a0[latex]\\bf{F}[\/latex]\u00a0across [latex]C[\/latex] is line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}(t)}{||{\\bf{n}}(t)||}[ds[\/latex].<\/p>\n<\/div>\n<div id=\"attachment_5242\" style=\"width: 532px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5242\" class=\"size-full wp-image-5242\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222343\/6.23.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/f73f7f23dee876375a79ebe5a3321b2c37df15fd&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A simple diagram of an increasing concave down curve C in vector field F, with no coordinate plane. Towards the top of the curve, the normal n is drawn perpendicular to the curve C. Another arrow F is drawn sharing n\u2019s endpoint. This flux points up and to the right at about a 90-degree angle to n. The arrows in the vector field to the left of n are drawn pointing straight up. The arrows after n point in the same direction as the flux.&quot; id=&quot;37&quot;&gt;\" width=\"522\" height=\"491\" \/><\/p>\n<p id=\"caption-attachment-5242\" class=\"wp-caption-text\">Figure 3. The flux of vector field [latex]\\bf{F}[\/latex]\u00a0across curve [latex]C[\/latex] is computed by an integral similar to a vector line integral.<\/p>\n<\/div>\n<p>We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see\u00a0the equation to compute vector line integrals).<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: calculating flux across a curve<\/h3>\n<p id=\"fs-id1167793662617\">Let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field and let [latex]C[\/latex] be a smooth curve with parameterization [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex]. The flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]C[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt[\/latex].<\/p>\n<\/div>\n<h2 data-type=\"title\">Proof<\/h2>\n<p id=\"fs-id1167793482875\">The proof of The Calculating Flux Across A Curve Theorem Equation\u00a0is similar to the proof of The Scalar Line Integration Calculation Theorem Equation. Before deriving the formula, note that [latex]||{\\bf{n}}(t)||=||\\langle y'(t),-x'(t)\\rangle||=\\sqrt{(y^\\prime(t))^2+(x^\\prime(t))^2}=||{\\bf{r}}^\\prime(t)||[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}ds \\\\  &=\\displaystyle\\int_a^b{\\bf{F}}\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}||{\\bf{r}}^\\prime(t)||dt \\\\  &=\\displaystyle\\int_a^b{\\bf{F}}[({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt  \\end{aligned}[[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: flux across a curve<\/h3>\n<p>Calculate the flux of [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex] across a unit circle oriented counterclockwise (Figure 4).<\/p>\n<div id=\"attachment_5244\" style=\"width: 506px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5244\" class=\"size-full wp-image-5244\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222518\/6.24.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/1284547517bb19bc89545262451458a9b835e511&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A unit circle in a vector field in two dimensions. The arrows point away from the origin in a radial pattern. Shorter vectors are near the origin, and longer ones are further away. A unit circle is drawn around the origin to fit the pattern, and arrowheads are drawn on the circle in a counterclockwise manner.&quot; id=&quot;40&quot;&gt;\" width=\"496\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-5244\" class=\"wp-caption-text\">Figure 4. A unit circle in vector field [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q589274091\">Show Solution<\/span><\/p>\n<div id=\"q589274091\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793671536\">To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization [latex]{\\bf{r}}(t)\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. The normal vector to a unit circle is [latex]\\langle\\cos{t},\\sin{t}\\rangle[\/latex]. Therefore, the flux is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\int_0^{2\\pi}\\langle2\\cos{t},2\\sin{t}\\rangle\\cdot\\langle\\cos{t},\\sin{t}\\rangle \\ dt \\\\  &=\\displaystyle\\int_0^{2\\pi}(2\\cos^2{t},2\\sin^2{t}) \\ dt = 2\\displaystyle\\int_0^{2\\pi}(\\cos^2{t},\\sin^2{t}) \\ dt \\\\  &=2\\displaystyle\\int_0^{2\\pi} \\ dt = 4\\pi  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the flux of [latex]{\\bf{F}}=\\langle{x}+y,2y\\rangle[\/latex] across the line segment from [latex](0, 0)[\/latex] to [latex](2, 3)[\/latex] where the curve is oriented from left to right.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203497590\">Show Solution<\/span><\/p>\n<div id=\"q203497590\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]3\/2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793594106\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle[\/latex] be a two-dimensional vector field. Recall that integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{T}}ds[\/latex] is sometimes written as [latex]\\displaystyle\\int_CPdx+Qdy[\/latex]. Analogously, flux [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds[\/latex] is sometimes written in the notation [latex]\\displaystyle\\int_C-Qdx+Pdy[\/latex], because the unit normal vector\u00a0[latex]\\bf{N}[\/latex]\u00a0is perpendicular to the unit tangent\u00a0[latex]\\bf{T}[\/latex]. Rotating the vector [latex]d{\\bf{r}}=\\langle{d}x,dy\\rangle[\/latex] by 90\u00b0 results in vector [latex]\\langle{d}y,-dx\\rangle[\/latex]. Therefore, the line integral in\u00a0Example &#8220;Using Properties to Compute a Vector Line Integral&#8221;\u00a0can be written as [latex]\\displaystyle\\int_C-2ydx+2xdy[\/latex].<\/p>\n<p id=\"fs-id1167793617594\">Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field\u00a0[latex]\\bf{F}[\/latex]\u00a0along an oriented closed curve is called the\u00a0<strong><span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term250\" data-type=\"term\">circulation<\/span><\/strong>\u00a0of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex]. Circulation line integrals have their own notation: [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds[\/latex]. The circle on the integral symbol denotes that [latex]C[\/latex] is \u201ccircular\u201d in that it has no endpoints.\u00a0Example &#8220;Evaluating a Vector Line Integral&#8221;\u00a0shows a calculation of circulation.<\/p>\n<p id=\"fs-id1167793706001\">To see where the term\u00a0<em data-effect=\"italics\">circulation<\/em>\u00a0comes from and what it measures, let\u00a0[latex]\\bf{v}[\/latex]\u00a0represent the velocity field of a fluid and let [latex]C[\/latex] be an oriented closed curve. At a particular point [latex]P[\/latex], the closer the direction of [latex]{\\bf{v}}(P)[\/latex] is to the direction of [latex]{\\bf{T}}(P)[\/latex], the larger the value of the dot product [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex]. The maximum value of [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex] occurs when the two vectors are pointing in the exact same direction; the minimum value of [latex]{\\bf{v}}(P)\\cdot{\\bf{T}}(P)[\/latex] occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation [latex]\\displaystyle\\oint_C{\\bf{v}}\\cdot{\\bf{T}}ds[\/latex] measures the tendency of the fluid to move in the direction of [latex]C[\/latex].<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: calculating circulation<\/h3>\n<p>Let [latex]{\\bf{F}}=\\langle-y,x\\rangle[\/latex] be the vector field from\u00a0Example &#8220;Independence of Parameterization&#8221;\u00a0and let [latex]C[\/latex] represent the unit circle oriented counterclockwise. Calculate the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q039258487\">Show Solution<\/span><\/p>\n<div id=\"q039258487\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793355114\">We use the standard parameterization of the unit circle: [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex] and [latex]{\\bf{r}}'(t)=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds&=\\displaystyle\\int_0^{2\\pi}\\langle-\\sin{t},\\cos{t}\\rangle\\cdot\\langle-\\sin{t},\\cos{t}\\rangle \\ dt \\\\  &=\\displaystyle\\int_0^{2\\pi}(\\sin^2{t}+\\cos^2{t}) \\ dt \\\\  &=\\displaystyle\\int_0^{2\\pi} \\ dt = 2\\pi  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793662249\">Notice that the circulation is positive. The reason for this is that the orientation of [latex]C[\/latex] \u201cflows\u201d with the direction of\u00a0[latex]\\bf{F}[\/latex]. At any point along the circle, the tangent vector and the vector from\u00a0[latex]\\bf{F}[\/latex]\u00a0form an angle of less than 90\u00b0, and therefore the corresponding dot product is positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793924502\">In\u00a0Example &#8220;Calculating Circulation&#8221;, what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by [latex]-C[\/latex]. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_{-C}{\\bf{F}}\\cdot{\\bf{T}} \\ ds}=-\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}=-2\\pi[\/latex].<\/p>\n<p id=\"fs-id1167793641713\">Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of\u00a0[latex]\\bf{F}[\/latex].<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the circulation of [latex]{\\bf{F}}(x,y)=\\left\\langle-\\frac{y}{x^2+y^2},\\frac{y}{x^2+y^2}\\right\\rangle[\/latex] along a unit circle oriented counterclockwise.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834752048\">Show Solution<\/span><\/p>\n<div id=\"q834752048\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\pi[\/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=8250315&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oWRQU5Oqtl0&amp;video_target=tpm-plugin-lldll002-oWRQU5Oqtl0\" 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\/CP6.22_transcript.html\">transcript for \u201cCP 6.22\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox exercises\">\n<h3>Example: calculating work<\/h3>\n<p>Calculate the work done on a particle that traverses circle [latex]C[\/latex] of radius 2 centered at the origin, oriented counterclockwise, by field [latex]{\\bf{F}}(x,y)=\\langle-2,y\\rangle[\/latex]. Assume the particle starts its movement at [latex](1, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q802581366\">Show Solution<\/span><\/p>\n<div id=\"q802581366\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794090726\">The work done by\u00a0[latex]\\bf{F}[\/latex]\u00a0on the particle is the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex]: [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]. We use the parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex] for [latex]C[\/latex]. Then, [latex]{\\bf{r}}^\\prime(t)=\\langle-2\\cos{t},2\\sin{t}\\rangle[\/latex] and [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-2,2\\sin{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds&=\\displaystyle\\int_0^{2\\pi}\\langle-2,2\\sin{t}\\rangle\\cdot\\langle-2,2\\cos{t}\\rangle \\ dt \\\\  &=\\displaystyle\\int_0^{2\\pi}(4\\sin{t}+4\\sin{t}\\cos{t}) \\ dt \\\\  &=\\left[-4\\cos{t}+4\\sin^2{t}\\right]_0^{2\\pi} \\\\  &=(-4\\cos(2\\pi)+2\\sin^2(2\\pi))-(-4\\cos(0)+4\\sin^2(0)) \\\\  &=-4+4=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793398124\">The force field does zero work on the particle.<\/p>\n<p id=\"fs-id1167793398127\">Notice that the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is zero. Furthermore, notice that since\u00a0[latex]\\bf{F}[\/latex]\u00a0is the gradient of [latex]f(x,y)=-2x+\\frac{y^2}2[\/latex],\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the work done by field [latex]{\\bf{F}}(x,y)=\\langle2x,3y\\rangle[\/latex] on a particle that traverses the unit circle. Assume the particle begins its movement at [latex](-1, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q734256411\">Show Solution<\/span><\/p>\n<div id=\"q734256411\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]0[\/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-1119\">\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 6.20. <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><li>CP 6.22. <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":428269,"menu_order":9,"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 6.20\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 6.22\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1119","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1119","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":123,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1119\/revisions"}],"predecessor-version":[{"id":6055,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1119\/revisions\/6055"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/24"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1119\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1119"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1119"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1119"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}