{"id":5485,"date":"2022-06-02T18:47:09","date_gmt":"2022-06-02T18:47:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5485"},"modified":"2022-11-01T05:09:10","modified_gmt":"2022-11-01T05:09:10","slug":"fundamental-theorem-for-line-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/fundamental-theorem-for-line-integrals\/","title":{"raw":"Fundamental Theorem for Line Integrals","rendered":"Fundamental Theorem for 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\">Describe simple and closed curves; define connected and simply connected regions.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167794160226\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Curves and Regions<\/h2>\r\n<p id=\"fs-id1167793876161\">Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.<\/p>\r\n<p id=\"fs-id1167793516148\">We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure 1).<\/p>\r\n\r\n<div id=\"fs-id1167793928878\" 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\nCurve [latex]C[\/latex] is a\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term251\" data-type=\"term\">closed curve<\/span><\/strong>\u00a0if there is a parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] of [latex]C[\/latex] such that the parameterization traverses the curve exactly once and [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex]. Curve [latex]C[\/latex] is a\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term252\" data-type=\"term\">simple curve<\/span><\/strong>\u00a0if [latex]C[\/latex] does not cross itself. That is, [latex]C[\/latex] is simple if there exists a parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] of [latex]C[\/latex] such that\u00a0[latex]{\\bf{r}}[\/latex]\u00a0is one-to-one over [latex](a, b)[\/latex]. It is possible for [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex], meaning that the simple curve is also closed.\r\n\r\n<\/div>\r\n\r\n[caption id=\"attachment_5246\" align=\"aligncenter\" width=\"929\"]<img class=\"size-full wp-image-5246\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222810\/6.25.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/6b7ab47eeaf8153b1a5df319d257a4e13e034e1d&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;An image showing eight curves and their types. The first curve is neither simple nor closed; it has two endpoints and crosses itself twice. The second curve is simple but not closed; it does not cross itself and has two endpoints. The third curve is closed but is not simple; it crosses itself a few times. The fourth is a simple closed curve; it does not cross itself and has no endpoints. The fifth is a simple, not closed curve; it does not cross itself, but it has endpoints. The sixth is a simple, closed curve; it does not cross itself and has no endpoints. The seventh is closed but not a simple curve; it crosses itself but has no endpoints. The last is not simple and not closed; it crosses itself and has endpoints.&quot; id=&quot;3&quot;&gt;\" width=\"929\" height=\"379\" \/> Figure 1. Types of curves that are simple or not simple and closed or not closed.[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: determining whether a curve is simple and closed<\/h3>\r\nIs the curve with parameterization [latex]{\\bf{r}}(t)=\\left\\langle\\cos{t},\\frac{\\sin{(2t)}}2\\right\\rangle, \\ 0\\leq{t}\\leq2\\pi[\/latex] a simple closed curve?\r\n\r\n[reveal-answer q=\"834725833\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834725833\"]\r\n\r\nNote that [latex]{\\bf{r}}(0)=\\langle0,1\\rangle[\/latex]; therefore, the curve is closed. The curve is not simple, however. To see this, note that [latex]{\\bf{r}}\\left(\\frac{\\pi}2\\right)=\\langle0,0\\rangle={\\bf{r}}\\left(\\frac{3\\pi}2\\right)[\/latex], and therefore the curve crosses itself at the origin (Figure 2).\r\n\r\n[caption id=\"attachment_5709\" align=\"aligncenter\" width=\"642\"]<img class=\"wp-image-5709 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2022\/06\/03210940\/6.26.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220509.174553\/resources\/e3c0f528ccac9907f5cd1fba1bcfa807ed43ebe7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram in the (x,y) coordinate plane that shows a closed but not simple curve. It looks like a horizontal figure eight with the crossing point at the origin.&quot; id=&quot;5&quot;&gt;\" width=\"642\" height=\"371\" \/> Figure 2. A curve that is closed but not simple.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIs the curve given by parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},3\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex] a simple closed curve?\r\n\r\n[reveal-answer q=\"814582267\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"814582267\"]\r\n\r\nYes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793361387\">Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region\u2019s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.<\/p>\r\n\r\n<div id=\"fs-id1167793847738\" 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\nA region [latex]D[\/latex] is a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term253\" data-type=\"term\">connected region<\/span>\u00a0if, for any two points [latex]P_1[\/latex] and [latex]P_2[\/latex], there is a path from [latex]P_1[\/latex] to [latex]P_2[\/latex] with a trace contained entirely inside [latex]D[\/latex]. A region [latex]D[\/latex] is a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term254\" data-type=\"term\">simply connected region<\/span>\u00a0if\u00a0<em data-effect=\"italics\">D<\/em>\u00a0is connected for any simple closed curve [latex]C[\/latex] that lies inside [latex]D[\/latex], and curve [latex]C[\/latex] can be shrunk continuously to a point while staying entirely inside [latex]D[\/latex]. In two dimensions, a region is simply connected if it is connected and has no holes.\r\n\r\n<\/div>\r\nAll simply connected regions are connected, but not all connected regions are simply connected (Figure 3).\r\n\r\n[caption id=\"attachment_5248\" align=\"aligncenter\" width=\"548\"]<img class=\"size-full wp-image-5248\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222905\/6.27.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/8d0eb3f345f768a2c295aa0f1991aaf19d3bcca2&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram showing simply connected, connected, and not connected regions. The simply connected regions have no holes. The connected regions may have holes, but a path can still be found between any two points in the region. The not connected region has some points that cannot be connected by a path in the region. Here, this is illustrated by showing two circular shapes that are defined as part of region D1 but are separated by white space.&quot; id=&quot;8&quot;&gt;\" width=\"548\" height=\"657\" \/> Figure 3. Not all connected regions are simply connected. (a) Simply connected regions have no holes. (b) Connected regions that are not simply connected may have holes but you can still find a path in the region between any two points. (c) A region that is not connected has some points that cannot be connected by a path in the region.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIs the region in the below image connected? Is the region simply connected?\r\n\r\n<img class=\"aligncenter size-full wp-image-5249\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223004\/6.25a.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/4fa7f7d2717cf20dc660c0bca880c7c7de267432&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A shaded circle with an open space in the shape of a circle inside it but very close to the boundary.&quot; id=&quot;9&quot;&gt;\" width=\"254\" height=\"254\" \/>\r\n\r\n[reveal-answer q=\"977374235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977374235\"]\r\n\r\nThe region in the figure is connected. The region in the figure is not simply connected.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]253265[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Fundamental Theorem for Line Integrals<\/h2>\r\n<p id=\"fs-id1167793786300\">Now that we understand some basic curves and regions, let\u2019s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function [latex]f[\/latex] has an antiderivative\u00a0<em data-effect=\"italics\">F<\/em>, then the integral of [latex]f[\/latex] from\u00a0<em data-effect=\"italics\">a<\/em>\u00a0to\u00a0<em data-effect=\"italics\">b<\/em>\u00a0depends only on the values of\u00a0<em data-effect=\"italics\">F<\/em>\u00a0at\u00a0<em data-effect=\"italics\">a<\/em>\u00a0and at\u00a0<em data-effect=\"italics\">b<\/em>\u2014that is,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_a^bf(x) \\ dx =F(b)-F(a)}[\/latex].<\/p>\r\n<p id=\"fs-id1167794292575\">If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a line integral and the Antiderivatives of the endpoints<\/h3>\r\nLet [latex]{\\bf{F}}(x,y)=\\langle2x,4y\\rangle[\/latex]. Calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex], where [latex]C[\/latex] is the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] (Figure 4).\r\n\r\n[reveal-answer q=\"243750312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"243750312\"]\r\n<p id=\"fs-id1167794118593\">We use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]\u00a0to calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex]. Curve [latex]C[\/latex] can be parameterized by [latex]{\\bf{r}}(t)=\\langle2t,2t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle4t,8t\\rangle[\/latex] and [latex]{\\bf{r}}^\\prime(t)=\\langle2,2\\rangle[\/latex], which implies that<\/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\\langle4t,8t\\rangle\\cdot\\langle2,2\\rangle \\ dt \\\\<\/p>\r\n&amp;=\\displaystyle\\int_0^1(8t+16t)dt=\\displaystyle\\int_0^124t \\ dt \\\\\r\n\r\n&amp;=[12t^2]_0^1=12.\r\n\r\n\\end{aligned}[\/latex]\r\n\r\n[caption id=\"attachment_5250\" align=\"aligncenter\" width=\"455\"]<img class=\"size-full wp-image-5250\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223147\/6.28.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/b801957dd45b11865bb430466bc4718add1ba07a&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows are longer the further away from the origin they are. They stretch out from the origin, forming a rectangular pattern. A line segment is drawn from P_0 at (0,0) to P_1 at (2,2).&quot; id=&quot;12&quot;&gt;\" width=\"455\" height=\"385\" \/> Figure 4. The value of line integral [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex] depends only on the value of the potential function of [latex]\\bf{F}[\/latex] at the endpoints of the curve.[\/caption]\r\n<p id=\"fs-id1167794128287\">Notice that [latex]F=\\nabla{f}[\/latex], where [latex]f(x, y)=x^{2}+2y^{2}[\/latex]. If we think of the gradient as a derivative, then [latex]f[\/latex] is an \u201cantiderivative\u201d of\u00a0[latex]\\bf{F}[\/latex]. In the case of single-variable integrals, the integral of derivative [latex]g'(x)[\/latex] is [latex]g(b)-g(a)[\/latex], where\u00a0<em data-effect=\"italics\">a<\/em>\u00a0is the start point of the interval of integration and\u00a0<em data-effect=\"italics\">b<\/em>\u00a0is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral\u00a0[latex]\\bf{F}[\/latex]\u00a0to be [latex]f(P_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of the curve of integration and [latex]P_0[\/latex] is the start point. Notice that this is the case for this example:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=12}[\/latex]<\/p>\r\n<p id=\"fs-id1167794335010\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f(2,2)-f(0,0)=4+8-0=12}[\/latex].<\/p>\r\n<p id=\"fs-id1167793832103\">In other words, the integral of a \u201cderivative\u201d can be calculated by evaluating an \u201cantiderivative\u201d at the endpoints of the curve and subtracting, just as for single-variable integrals.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term255\" data-type=\"term\">Fundamental Theorem for Line Integrals.<\/span><\/strong>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: the fundamental theorem for line integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794043104\">Let [latex]C[\/latex] be a piecewise smooth curve with parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]f[\/latex] be a function of two or three variables with first-order partial derivatives that exist and are continuous on [latex]C[\/latex]. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=f({\\bf{r}}(b))-f({\\bf{r}}(a))}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1167793948209\">By\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt[\/latex].<\/p>\r\n<p id=\"fs-id1167793828355\">By the chain rule,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}(f({\\bf{r}}(t))=\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)[\/latex].<\/p>\r\n<p id=\"fs-id1167793396439\">Therefore, by the Fundamental Theorem of Calculus,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}&amp;=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_a^b\\frac{d}{dt}(f({\\bf{r}}(t)) \\ dt \\\\\r\n&amp;=[{f}({\\bf{r}}(t))]_{t=a}^{t=b} \\\\\r\n&amp;={f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1167794022012\">We know that if\u00a0[latex]\\bf{F}[\/latex]\u00a0is a conservative vector field, there are potential functions [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]. Therefore [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}}={f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))[\/latex]. In other words, just as with the Fundamental Theorem of Calculus, computing the line integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, is a two-step process: (1) find a potential function (\u201cantiderivative\u201d) [latex]f[\/latex] for\u00a0[latex]\\bf{F}[\/latex]\u00a0and (2) compute the value of [latex]f[\/latex] at the endpoints of [latex]C[\/latex] and calculate their difference [latex]{f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))[\/latex]. Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals.\u00a0<em data-effect=\"italics\">A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.<\/em><\/p>\r\n\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: applying the fundamental theorem<\/h3>\r\n<p id=\"fs-id1167794029002\">Calculate integral [latex]\\displaystyle\\int_c{\\bf{f}}.d{\\bf{r}}[\/latex], where [latex]{\\bf{F}}(x,y,z)=\\left\\langle2x\\ln{y},\\frac{x^2}y+z^2,2yz\\right\\rangle[\/latex] and [latex]C[\/latex] is a curve with parameterization [latex]{\\bf{r}}(t)=\\langle{t^2},t,t\\rangle[\/latex],\u00a0[latex]1\\leq{t}\\leq1{e}[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1167793292432\" type=\"a\">\r\n \t<li>without using the Fundamental Theorem of Line Integrals and<\/li>\r\n \t<li>using the Fundamental Theorem of Line Integrals.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"847519662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847519662\"]\r\n<ol id=\"fs-id1167793240974\" type=\"a\">\r\n \t<li>First, let\u2019s calculate the integral without the Fundamental Theorem for Line Integrals and instead use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793912263\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&amp;=\\displaystyle\\int_1^e{\\bf{F}}(r(t)).r^\\prime(t) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_1^e\\left\\langle2t^2\\ln{t},\\frac{t^4}t+t^2,2t^2\\right\\rangle. \\langle2t,1,1\\rangle dt \\\\\r\n&amp;=\\displaystyle\\int_1^e(4t^3\\ln{t}+t^3+3t^2) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt +\\displaystyle\\int_1^e(t^3+3t^2) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt+\\left[\\frac{t^4}4+t^3\\right]_1^e \\\\\r\n&amp;=2\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54.\r\n\\end{align}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Integral [latex]\\displaystyle\\int_1^et^3\\ln{t}[\/latex] requires integration by parts. Let [latex]u=\\ln{t}[\/latex] and [latex]dv=t^3[\/latex]. Then [latex]u=\\ln{t}[\/latex],\u00a0[latex]dv=t^3[\/latex],<span data-type=\"newline\">\r\n<\/span>and<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793620013\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]du=\\frac{1}t dt[\/latex],[latex]v=\\frac{t^4}4[\/latex].<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Therefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793977267\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_1^et^3\\ln{t}dt&amp;=\\left[\\frac{t^4}4\\ln{t}\\right]_1^e-\\frac14\\displaystyle\\int_1^et^3 \\ dt \\\\\r\n&amp;=\\frac{e^4}4-\\frac1r\\left(\\frac{e^4}4-\\frac14\\right).\r\n\\end{align}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Thus,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793419427\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&amp;=4\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&amp;=4\\left(\\frac{e^4}4-\\frac14\\left(\\frac{e^4}4-\\frac14\\right)\\right)+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&amp;=e^4-\\frac{e^4}4+\\frac14+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&amp;=e^4+e^3-1\r\n\\end{align}[\/latex].<\/div>\r\n<ul>\r\n \t<li>Given that [latex]f(x,y,z)=x^2\\ln{y}+yz^2[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex], let\u2019s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793828376\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&amp;=\\displaystyle\\int_C\\nabla{f}.d{\\bf{r}} \\\\\r\n&amp;=f({\\bf{r}}(e))-f({\\bf{r}}(1)) \\\\\r\n&amp;=f(e^2,e,e)-f(1,1,1) \\\\\r\n&amp;=e^4+e^3-1\r\n\\end{align}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nExample \"Applying the Fundamental Theorem\"\u00a0illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nGiven that [latex]f(x,y)=(x-1)^2y+(y+1)^2x[\/latex] is a potential function for [latex]{\\bf{F}}=\\langle2xy-2y+(y+1)^2,(x-1)^2+2yx+2x\\rangle[\/latex], calculate integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]C[\/latex] is the lower half of the unit circle oriented counterclockwise.\r\n\r\n<img class=\"aligncenter size-full wp-image-5252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223444\/6.26.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/9f28a5ea7739065108e57a778883047758bdf7f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.&quot; id=&quot;15&quot;&gt;\" width=\"642\" height=\"459\" \/>\r\n\r\n[reveal-answer q=\"964357240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"964357240\"]\r\n\r\n[latex]2[\/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=8250317&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nfZUeb79cSM&amp;video_target=tpm-plugin-srmwxh3x-nfZUeb79cSM\" 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\/CP6.26_transcript.html\">transcript for \u201cCP 6.26\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167794291540\">The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and [latex]C[\/latex] is a closed curve, then the circulation of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] is zero\u2014that is, [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=0[\/latex]. To see why this is true, let [latex]f[\/latex] be a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Since [latex]C[\/latex] is a closed curve, the terminal point [latex]{\\bf{r}}(b)[\/latex] of [latex]C[\/latex] is the same as the initial point [latex]{\\bf{r}}(a)[\/latex] of [latex]C[\/latex]\u2014that is, [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex]. Therefore, by the Fundamental Theorem for Line Integrals,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\oint_C\\nabla{f}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=f({\\bf{r}}(b))-f({\\bf{r}}(a)) \\\\\r\n&amp;=f({\\bf{r}}(b))-f({\\bf{r}}(b)) \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793618332\">Recall that the reason a conservative vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is called \u201cconservative\u201d is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0in integral [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] as a gravitational field, then the equation [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=0[\/latex] follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.<\/p>\r\n<p id=\"fs-id1167793945028\">The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path\u2014meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.<\/p>\r\n\r\n<div id=\"fs-id1167793510046\" 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\nLet\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field with domain [latex]D[\/latex]. The vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term256\" data-type=\"term\">independent of path<\/span><\/strong>\u00a0(or\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term257\" data-type=\"term\">path independent<\/span><\/strong>) if [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] for any paths [latex]C_1[\/latex] and [latex]C_2[\/latex] in [latex]D[\/latex] with the same initial and terminal points.\r\n\r\n<\/div>\r\nThe second consequence is stated formally in the following theorem.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: path independence of conservative fields<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a conservative vector field, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path.\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1167793340652\">Let [latex]D[\/latex] denote the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0and let [latex]C_1[\/latex] and [latex]C_2[\/latex] be two paths in [latex]D[\/latex] with the same initial and terminal points (Figure 5). Call the initial point [latex]P_1[\/latex] and the terminal point [latex]P_2[\/latex]. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, there is a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. By the Fundamental Theorem for Line Integrals,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=f({\\bf{P}}_2)-f({\\bf{P}}_1)=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793918318\">Therefore, [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n[caption id=\"attachment_5253\" align=\"aligncenter\" width=\"655\"]<img class=\"size-full wp-image-5253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223535\/6.29.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/db3b7b07dc4a4243dcbf4dd01addbc41609ec1eb&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows are shorter the closer to the x axis and line x=1.5 they become. The arrows point up, converging around x=1.5 in quadrant 1. That line is approached from the left and from the right. Below, in quadrant 4, the arrows in the rough interval [1,2.5] curve out, away from the given line x=1.5, but do turn back in and converge to x=1.5 above the x axis. Outside of that interval, the arrows go to the left and right horizontally for x values less than 1 and greater than 2.5, respectively. A line is drawn from P_1 at the origin to P_2 at (3,.75) and labeled C_2. C_1 is a simple curve that connects the given endpoints above C_2, C_3 is a simple curve that connects the given endpoints below C_2.&quot; id=&quot;19&quot;&gt;\" width=\"655\" height=\"352\" \/> Figure 5. The vector field is conservative, and therefore independent of path.[\/caption]<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div class=\"ui-has-child-title\" data-type=\"note\">\r\n<p id=\"fs-id1167794125835\">To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.<\/p>\r\n<p id=\"fs-id1167793287896\">We have shown that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path. It turns out that if the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then the converse is also true. That is, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path and the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: The path independence test of conservative fields<\/h3>\r\n\r\n<hr \/>\r\n\r\nIf\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a continuous vector field that is independent of path and the domain [latex]D[\/latex] of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1167793394250\">We prove the theorem for vector fields in [latex]\\mathbb{R}^2[\/latex]. The proof for vector fields in [latex]\\mathbb{R}^3[\/latex] is similar. To show that [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is conservative, we must find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. To that end, let [latex]X[\/latex] be a fixed point in [latex]D[\/latex]. For any point [latex](x, y)[\/latex] in [latex]D[\/latex], let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex]. Define [latex]f(x, y)[\/latex] by [latex]f(x,y)=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. (Note that this definition of [latex]f[\/latex] makes sense only because\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was not independent of path, then it might be possible to find another path [latex]C'[\/latex] from [latex]X[\/latex] to [latex](x, y)[\/latex] such that [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], and in such a case [latex]f(x, y)[\/latex] would not be a function.) We want to show that [latex]f[\/latex] has the property [latex]\\nabla{f}={\\bf{F}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794126268\">Since domain [latex]D[\/latex] is open, it is possible to find a disk centered at [latex](x, y)[\/latex] such that the disk is contained entirely inside [latex]D[\/latex]. Let [latex](a, y)[\/latex] with [latex]a&lt;x[\/latex] be a point in that disk. Let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex] that consists of two pieces: [latex]C_1[\/latex] and [latex]C_2[\/latex]. The first piece, [latex]C_1[\/latex], is any path from [latex]X[\/latex] to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex]; [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x ,y)[\/latex] (Figure 6). Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f(x,y)=\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793426381\">The first integral does not depend on [latex]x[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac{\\partial}{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793959035\">If we parameterize [latex]C_2[\/latex] by [latex]{\\bf{r}}(t)=\\langle{t},y\\rangle[\/latex], [latex]a\\leq{t}\\leq{x}[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nf_x&amp;=\\frac\\partial{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\\r\n&amp;=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\frac{d}{dt}(\\langle{t},y\\rangle \\ dt \\\\\r\n&amp;=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\langle1,0\\rangle \\ dt \\\\\r\n&amp;=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793881085\">By the Fundamental Theorem of Calculus (part 1),<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt=P(x,y)}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5255\" align=\"aligncenter\" width=\"428\"]<img class=\"size-full wp-image-5255\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223631\/6.30.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/96903a4e93cc5b2a37d8748493ad2684ea71c218&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram of a region D in the rough shape of a backwards C. It is a simply connected region formed by a closed curve. Another curve C_1 is drawn inside D from point X to (a,y). C_2 is a horizontal line segment drawn from (a,y) to (x,y). Arrowheads point to (a,y) on C_1 and to (x,y) on C_2.&quot; id=&quot;21&quot;&gt;\" width=\"428\" height=\"511\" \/> Figure 6. Here, [latex]C_1[\/latex] is any path from [latex]X[\/latex]; to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex], and [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x, y)[\/latex].[\/caption]<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\">\r\n<p id=\"fs-id1167794211104\">A similar argument using a vertical line segment rather than a horizontal line segment shows that [latex]f_y= Q(x, y)[\/latex].<\/p>\r\n<p id=\"fs-id1167793609990\">Therefore [latex]\\nabla{f}={\\bf{F}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1167793469703\">We have spent a lot of time discussing and proving\u00a0Path Independence of Conservative Fields Theorem\u00a0and\u00a0The Path Independence Test for Conservative Fields Theorem, but we can summarize them simply: a vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: showing that a vector field is not conservative<\/h3>\r\nUse path independence to show that vector field [latex]{\\bf{F}}(x,y)=\\langle{x^2}y,y+5\\rangle[\/latex] is not conservative.\r\n\r\n[reveal-answer q=\"093720434\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"093720434\"]\r\n<p id=\"fs-id1167794210536\">We can indicate that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative by showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not path independent. We do so by giving two different paths, [latex]C_1[\/latex] and [latex]C_2[\/latex], that both start at [latex](0, 0)[\/latex] and end at [latex](1, 1)[\/latex], and yet [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793311592\">Let [latex]C_1[\/latex] be the curve with parameterization [latex]r_1(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] and let [latex]C_2[\/latex] be the curve with parameterization [latex]r_2(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] (Figure 7). Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_0^1{\\bf{F}}(r_1(t))\\cdot{r}_1^\\prime(t) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^1\\langle{t^3},t+5\\rangle\\cdot\\langle1,1\\rangle \\ dt =\\displaystyle\\int_0^1(t^3+t+5) \\ dt \\\\\r\n&amp;=\\left[\\frac{t^4}4+\\frac{t^2}2+5t\\right]_0^1=\\frac{23}4\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793925125\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_0^1{\\bf{F}}(r_2(t))\\cdot{r}_2^\\prime(t) \\ dt \\\\\r\n&amp;=\\displaystyle\\int_0^1\\langle{t^4},t^2+5\\rangle\\cdot\\langle1,2t\\rangle \\ dt =\\displaystyle\\int_0^1(t^4+2t^2+10t) \\ dt \\\\\r\n&amp;=\\left[\\frac{t^5}5+\\frac{t^4}2+5t^2\\right]_0^1=\\frac{57}{10}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793498806\">Since [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], the value of a line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0depends on the path between two given points. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not independent of path, and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative.<\/p>\r\n\r\n[caption id=\"attachment_5261\" align=\"aligncenter\" width=\"340\"]<img class=\"size-full wp-image-5261\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31224359\/6.31.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/f04c50a036101a7fa1172a0b89c787ab72e1aa7c&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field drawn in two dimensions. The arrows are roughly the same length. They point directly up but tend to shift to the right in the upper right portion of quadrant 1. Curves C_1 and C_2 connect the origin to point (1,1). They are both simple curves, and their arrowheads point to (1,1).&quot; id=&quot;23&quot;&gt;\" width=\"340\" height=\"348\" \/> Figure 7. Curves [latex]C_1[\/latex] and [latex]C_2[\/latex] are both oriented from left to right.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nShow that [latex]{\\bf{F}}(x,y)=\\langle{x}y,x^2y^2\\rangle[\/latex] is not path independent by considering the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] and the piece of the graph of [latex]y=\\frac{x^2}2[\/latex] that goes from [latex](0, 0)[\/latex] to\u00a0[latex](2, 2)[\/latex].\r\n\r\n[reveal-answer q=\"483725098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483725098\"]\r\n\r\nIf [latex]C_1[\/latex] and [latex]C_2[\/latex] represent the two curves, then [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\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\">Describe simple and closed curves; define connected and simply connected regions.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167794160226\" data-depth=\"1\">\n<h2 data-type=\"title\">Curves and Regions<\/h2>\n<p id=\"fs-id1167793876161\">Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.<\/p>\n<p id=\"fs-id1167793516148\">We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure 1).<\/p>\n<div id=\"fs-id1167793928878\" 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>Curve [latex]C[\/latex] is a\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term251\" data-type=\"term\">closed curve<\/span><\/strong>\u00a0if there is a parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] of [latex]C[\/latex] such that the parameterization traverses the curve exactly once and [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex]. Curve [latex]C[\/latex] is a\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term252\" data-type=\"term\">simple curve<\/span><\/strong>\u00a0if [latex]C[\/latex] does not cross itself. That is, [latex]C[\/latex] is simple if there exists a parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] of [latex]C[\/latex] such that\u00a0[latex]{\\bf{r}}[\/latex]\u00a0is one-to-one over [latex](a, b)[\/latex]. It is possible for [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex], meaning that the simple curve is also closed.<\/p>\n<\/div>\n<div id=\"attachment_5246\" style=\"width: 939px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5246\" class=\"size-full wp-image-5246\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222810\/6.25.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/6b7ab47eeaf8153b1a5df319d257a4e13e034e1d&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;An image showing eight curves and their types. The first curve is neither simple nor closed; it has two endpoints and crosses itself twice. The second curve is simple but not closed; it does not cross itself and has two endpoints. The third curve is closed but is not simple; it crosses itself a few times. The fourth is a simple closed curve; it does not cross itself and has no endpoints. The fifth is a simple, not closed curve; it does not cross itself, but it has endpoints. The sixth is a simple, closed curve; it does not cross itself and has no endpoints. The seventh is closed but not a simple curve; it crosses itself but has no endpoints. The last is not simple and not closed; it crosses itself and has endpoints.&quot; id=&quot;3&quot;&gt;\" width=\"929\" height=\"379\" \/><\/p>\n<p id=\"caption-attachment-5246\" class=\"wp-caption-text\">Figure 1. Types of curves that are simple or not simple and closed or not closed.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: determining whether a curve is simple and closed<\/h3>\n<p>Is the curve with parameterization [latex]{\\bf{r}}(t)=\\left\\langle\\cos{t},\\frac{\\sin{(2t)}}2\\right\\rangle, \\ 0\\leq{t}\\leq2\\pi[\/latex] a simple closed curve?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834725833\">Show Solution<\/span><\/p>\n<div id=\"q834725833\" class=\"hidden-answer\" style=\"display: none\">\n<p>Note that [latex]{\\bf{r}}(0)=\\langle0,1\\rangle[\/latex]; therefore, the curve is closed. The curve is not simple, however. To see this, note that [latex]{\\bf{r}}\\left(\\frac{\\pi}2\\right)=\\langle0,0\\rangle={\\bf{r}}\\left(\\frac{3\\pi}2\\right)[\/latex], and therefore the curve crosses itself at the origin (Figure 2).<\/p>\n<div id=\"attachment_5709\" style=\"width: 652px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5709\" class=\"wp-image-5709 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2022\/06\/03210940\/6.26.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220509.174553\/resources\/e3c0f528ccac9907f5cd1fba1bcfa807ed43ebe7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram in the (x,y) coordinate plane that shows a closed but not simple curve. It looks like a horizontal figure eight with the crossing point at the origin.&quot; id=&quot;5&quot;&gt;\" width=\"642\" height=\"371\" \/><\/p>\n<p id=\"caption-attachment-5709\" class=\"wp-caption-text\">Figure 2. A curve that is closed but not simple.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Is the curve given by parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},3\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex] a simple closed curve?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q814582267\">Show Solution<\/span><\/p>\n<div id=\"q814582267\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793361387\">Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region\u2019s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.<\/p>\n<div id=\"fs-id1167793847738\" 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>A region [latex]D[\/latex] is a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term253\" data-type=\"term\">connected region<\/span>\u00a0if, for any two points [latex]P_1[\/latex] and [latex]P_2[\/latex], there is a path from [latex]P_1[\/latex] to [latex]P_2[\/latex] with a trace contained entirely inside [latex]D[\/latex]. A region [latex]D[\/latex] is a\u00a0<span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term254\" data-type=\"term\">simply connected region<\/span>\u00a0if\u00a0<em data-effect=\"italics\">D<\/em>\u00a0is connected for any simple closed curve [latex]C[\/latex] that lies inside [latex]D[\/latex], and curve [latex]C[\/latex] can be shrunk continuously to a point while staying entirely inside [latex]D[\/latex]. In two dimensions, a region is simply connected if it is connected and has no holes.<\/p>\n<\/div>\n<p>All simply connected regions are connected, but not all connected regions are simply connected (Figure 3).<\/p>\n<div id=\"attachment_5248\" style=\"width: 558px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5248\" class=\"size-full wp-image-5248\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31222905\/6.27.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/8d0eb3f345f768a2c295aa0f1991aaf19d3bcca2&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram showing simply connected, connected, and not connected regions. The simply connected regions have no holes. The connected regions may have holes, but a path can still be found between any two points in the region. The not connected region has some points that cannot be connected by a path in the region. Here, this is illustrated by showing two circular shapes that are defined as part of region D1 but are separated by white space.&quot; id=&quot;8&quot;&gt;\" width=\"548\" height=\"657\" \/><\/p>\n<p id=\"caption-attachment-5248\" class=\"wp-caption-text\">Figure 3. Not all connected regions are simply connected. (a) Simply connected regions have no holes. (b) Connected regions that are not simply connected may have holes but you can still find a path in the region between any two points. (c) A region that is not connected has some points that cannot be connected by a path in the region.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Is the region in the below image connected? Is the region simply connected?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5249\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223004\/6.25a.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/4fa7f7d2717cf20dc660c0bca880c7c7de267432&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A shaded circle with an open space in the shape of a circle inside it but very close to the boundary.&quot; id=&quot;9&quot;&gt;\" width=\"254\" height=\"254\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q977374235\">Show Solution<\/span><\/p>\n<div id=\"q977374235\" class=\"hidden-answer\" style=\"display: none\">\n<p>The region in the figure is connected. The region in the figure is not simply connected.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm253265\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=253265&theme=oea&iframe_resize_id=ohm253265&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Fundamental Theorem for Line Integrals<\/h2>\n<p id=\"fs-id1167793786300\">Now that we understand some basic curves and regions, let\u2019s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function [latex]f[\/latex] has an antiderivative\u00a0<em data-effect=\"italics\">F<\/em>, then the integral of [latex]f[\/latex] from\u00a0<em data-effect=\"italics\">a<\/em>\u00a0to\u00a0<em data-effect=\"italics\">b<\/em>\u00a0depends only on the values of\u00a0<em data-effect=\"italics\">F<\/em>\u00a0at\u00a0<em data-effect=\"italics\">a<\/em>\u00a0and at\u00a0<em data-effect=\"italics\">b<\/em>\u2014that is,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_a^bf(x) \\ dx =F(b)-F(a)}[\/latex].<\/p>\n<p id=\"fs-id1167794292575\">If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a line integral and the Antiderivatives of the endpoints<\/h3>\n<p>Let [latex]{\\bf{F}}(x,y)=\\langle2x,4y\\rangle[\/latex]. Calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex], where [latex]C[\/latex] is the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] (Figure 4).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q243750312\">Show Solution<\/span><\/p>\n<div id=\"q243750312\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794118593\">We use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]\u00a0to calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex]. Curve [latex]C[\/latex] can be parameterized by [latex]{\\bf{r}}(t)=\\langle2t,2t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle4t,8t\\rangle[\/latex] and [latex]{\\bf{r}}^\\prime(t)=\\langle2,2\\rangle[\/latex], which implies that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1\\langle4t,8t\\rangle\\cdot\\langle2,2\\rangle \\ dt \\\\<\/p>\n<p>  &=\\displaystyle\\int_0^1(8t+16t)dt=\\displaystyle\\int_0^124t \\ dt \\\\    &=[12t^2]_0^1=12.    \\end{aligned}[\/latex]<\/p>\n<div id=\"attachment_5250\" style=\"width: 465px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5250\" class=\"size-full wp-image-5250\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223147\/6.28.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/b801957dd45b11865bb430466bc4718add1ba07a&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows are longer the further away from the origin they are. They stretch out from the origin, forming a rectangular pattern. A line segment is drawn from P_0 at (0,0) to P_1 at (2,2).&quot; id=&quot;12&quot;&gt;\" width=\"455\" height=\"385\" \/><\/p>\n<p id=\"caption-attachment-5250\" class=\"wp-caption-text\">Figure 4. The value of line integral [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex] depends only on the value of the potential function of [latex]\\bf{F}[\/latex] at the endpoints of the curve.<\/p>\n<\/div>\n<p id=\"fs-id1167794128287\">Notice that [latex]F=\\nabla{f}[\/latex], where [latex]f(x, y)=x^{2}+2y^{2}[\/latex]. If we think of the gradient as a derivative, then [latex]f[\/latex] is an \u201cantiderivative\u201d of\u00a0[latex]\\bf{F}[\/latex]. In the case of single-variable integrals, the integral of derivative [latex]g'(x)[\/latex] is [latex]g(b)-g(a)[\/latex], where\u00a0<em data-effect=\"italics\">a<\/em>\u00a0is the start point of the interval of integration and\u00a0<em data-effect=\"italics\">b<\/em>\u00a0is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral\u00a0[latex]\\bf{F}[\/latex]\u00a0to be [latex]f(P_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of the curve of integration and [latex]P_0[\/latex] is the start point. Notice that this is the case for this example:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=12}[\/latex]<\/p>\n<p id=\"fs-id1167794335010\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f(2,2)-f(0,0)=4+8-0=12}[\/latex].<\/p>\n<p id=\"fs-id1167793832103\">In other words, the integral of a \u201cderivative\u201d can be calculated by evaluating an \u201cantiderivative\u201d at the endpoints of the curve and subtracting, just as for single-variable integrals.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term255\" data-type=\"term\">Fundamental Theorem for Line Integrals.<\/span><\/strong><\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: the fundamental theorem for line integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1167794043104\">Let [latex]C[\/latex] be a piecewise smooth curve with parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]f[\/latex] be a function of two or three variables with first-order partial derivatives that exist and are continuous on [latex]C[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=f({\\bf{r}}(b))-f({\\bf{r}}(a))}[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1167793948209\">By\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt[\/latex].<\/p>\n<p id=\"fs-id1167793828355\">By the chain rule,<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}(f({\\bf{r}}(t))=\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)[\/latex].<\/p>\n<p id=\"fs-id1167793396439\">Therefore, by the Fundamental Theorem of Calculus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}&=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\  &=\\displaystyle\\int_a^b\\frac{d}{dt}(f({\\bf{r}}(t)) \\ dt \\\\  &=[{f}({\\bf{r}}(t))]_{t=a}^{t=b} \\\\  &={f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))  \\end{aligned}[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167794022012\">We know that if\u00a0[latex]\\bf{F}[\/latex]\u00a0is a conservative vector field, there are potential functions [latex]f[\/latex] such that [latex]\\nabla{f}={\\bf{F}}[\/latex]. Therefore [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}}={f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))[\/latex]. In other words, just as with the Fundamental Theorem of Calculus, computing the line integral [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative, is a two-step process: (1) find a potential function (\u201cantiderivative\u201d) [latex]f[\/latex] for\u00a0[latex]\\bf{F}[\/latex]\u00a0and (2) compute the value of [latex]f[\/latex] at the endpoints of [latex]C[\/latex] and calculate their difference [latex]{f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))[\/latex]. Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals.\u00a0<em data-effect=\"italics\">A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.<\/em><\/p>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: applying the fundamental theorem<\/h3>\n<p id=\"fs-id1167794029002\">Calculate integral [latex]\\displaystyle\\int_c{\\bf{f}}.d{\\bf{r}}[\/latex], where [latex]{\\bf{F}}(x,y,z)=\\left\\langle2x\\ln{y},\\frac{x^2}y+z^2,2yz\\right\\rangle[\/latex] and [latex]C[\/latex] is a curve with parameterization [latex]{\\bf{r}}(t)=\\langle{t^2},t,t\\rangle[\/latex],\u00a0[latex]1\\leq{t}\\leq1{e}[\/latex]<\/p>\n<ol id=\"fs-id1167793292432\" type=\"a\">\n<li>without using the Fundamental Theorem of Line Integrals and<\/li>\n<li>using the Fundamental Theorem of Line Integrals.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q847519662\">Show Solution<\/span><\/p>\n<div id=\"q847519662\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793240974\" type=\"a\">\n<li>First, let\u2019s calculate the integral without the Fundamental Theorem for Line Integrals and instead use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793912263\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}  \\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=\\displaystyle\\int_1^e{\\bf{F}}(r(t)).r^\\prime(t) \\ dt \\\\  &=\\displaystyle\\int_1^e\\left\\langle2t^2\\ln{t},\\frac{t^4}t+t^2,2t^2\\right\\rangle. \\langle2t,1,1\\rangle dt \\\\  &=\\displaystyle\\int_1^e(4t^3\\ln{t}+t^3+3t^2) \\ dt \\\\  &=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt +\\displaystyle\\int_1^e(t^3+3t^2) \\ dt \\\\  &=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt+\\left[\\frac{t^4}4+t^3\\right]_1^e \\\\  &=2\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54.  \\end{align}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Integral [latex]\\displaystyle\\int_1^et^3\\ln{t}[\/latex] requires integration by parts. Let [latex]u=\\ln{t}[\/latex] and [latex]dv=t^3[\/latex]. Then [latex]u=\\ln{t}[\/latex],\u00a0[latex]dv=t^3[\/latex],<span data-type=\"newline\"><br \/>\n<\/span>and<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793620013\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]du=\\frac{1}t dt[\/latex],[latex]v=\\frac{t^4}4[\/latex].<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793977267\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}  \\displaystyle\\int_1^et^3\\ln{t}dt&=\\left[\\frac{t^4}4\\ln{t}\\right]_1^e-\\frac14\\displaystyle\\int_1^et^3 \\ dt \\\\  &=\\frac{e^4}4-\\frac1r\\left(\\frac{e^4}4-\\frac14\\right).  \\end{align}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Thus,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793419427\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}  \\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=4\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54 \\\\  &=4\\left(\\frac{e^4}4-\\frac14\\left(\\frac{e^4}4-\\frac14\\right)\\right)+\\frac{e^4}4+e^3-\\frac54 \\\\  &=e^4-\\frac{e^4}4+\\frac14+\\frac{e^4}4+e^3-\\frac54 \\\\  &=e^4+e^3-1  \\end{align}[\/latex].<\/div>\n<ul>\n<li>Given that [latex]f(x,y,z)=x^2\\ln{y}+yz^2[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex], let\u2019s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793828376\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\begin{align}  \\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=\\displaystyle\\int_C\\nabla{f}.d{\\bf{r}} \\\\  &=f({\\bf{r}}(e))-f({\\bf{r}}(1)) \\\\  &=f(e^2,e,e)-f(1,1,1) \\\\  &=e^4+e^3-1  \\end{align}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p>Example &#8220;Applying the Fundamental Theorem&#8221;\u00a0illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Given that [latex]f(x,y)=(x-1)^2y+(y+1)^2x[\/latex] is a potential function for [latex]{\\bf{F}}=\\langle2xy-2y+(y+1)^2,(x-1)^2+2yx+2x\\rangle[\/latex], calculate integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]C[\/latex] is the lower half of the unit circle oriented counterclockwise.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223444\/6.26.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/9f28a5ea7739065108e57a778883047758bdf7f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.&quot; id=&quot;15&quot;&gt;\" width=\"642\" height=\"459\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q964357240\">Show Solution<\/span><\/p>\n<div id=\"q964357240\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2[\/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=8250317&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nfZUeb79cSM&amp;video_target=tpm-plugin-srmwxh3x-nfZUeb79cSM\" 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\/CP6.26_transcript.html\">transcript for \u201cCP 6.26\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167794291540\">The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative and [latex]C[\/latex] is a closed curve, then the circulation of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] is zero\u2014that is, [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=0[\/latex]. To see why this is true, let [latex]f[\/latex] be a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Since [latex]C[\/latex] is a closed curve, the terminal point [latex]{\\bf{r}}(b)[\/latex] of [latex]C[\/latex] is the same as the initial point [latex]{\\bf{r}}(a)[\/latex] of [latex]C[\/latex]\u2014that is, [latex]{\\bf{r}}(a)={\\bf{r}}(b)[\/latex]. Therefore, by the Fundamental Theorem for Line Integrals,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\oint_C\\nabla{f}\\cdot{d}{\\bf{r}} \\\\  &=f({\\bf{r}}(b))-f({\\bf{r}}(a)) \\\\  &=f({\\bf{r}}(b))-f({\\bf{r}}(b)) \\\\  &=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793618332\">Recall that the reason a conservative vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is called \u201cconservative\u201d is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0in integral [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] as a gravitational field, then the equation [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=0[\/latex] follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.<\/p>\n<p id=\"fs-id1167793945028\">The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path\u2014meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.<\/p>\n<div id=\"fs-id1167793510046\" 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>Let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field with domain [latex]D[\/latex]. The vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term256\" data-type=\"term\">independent of path<\/span><\/strong>\u00a0(or\u00a0<strong><span id=\"47dae7f0-fc5d-4d6c-a882-d19c748319d9_term257\" data-type=\"term\">path independent<\/span><\/strong>) if [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] for any paths [latex]C_1[\/latex] and [latex]C_2[\/latex] in [latex]D[\/latex] with the same initial and terminal points.<\/p>\n<\/div>\n<p>The second consequence is stated formally in the following theorem.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: path independence of conservative fields<\/h3>\n<hr \/>\n<p>If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a conservative vector field, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path.<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1167793340652\">Let [latex]D[\/latex] denote the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0and let [latex]C_1[\/latex] and [latex]C_2[\/latex] be two paths in [latex]D[\/latex] with the same initial and terminal points (Figure 5). Call the initial point [latex]P_1[\/latex] and the terminal point [latex]P_2[\/latex]. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, there is a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. By the Fundamental Theorem for Line Integrals,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=f({\\bf{P}}_2)-f({\\bf{P}}_1)=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\n<p id=\"fs-id1167793918318\">Therefore, [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"attachment_5253\" style=\"width: 665px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5253\" class=\"size-full wp-image-5253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223535\/6.29.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/db3b7b07dc4a4243dcbf4dd01addbc41609ec1eb&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows are shorter the closer to the x axis and line x=1.5 they become. The arrows point up, converging around x=1.5 in quadrant 1. That line is approached from the left and from the right. Below, in quadrant 4, the arrows in the rough interval [1,2.5] curve out, away from the given line x=1.5, but do turn back in and converge to x=1.5 above the x axis. Outside of that interval, the arrows go to the left and right horizontally for x values less than 1 and greater than 2.5, respectively. A line is drawn from P_1 at the origin to P_2 at (3,.75) and labeled C_2. C_1 is a simple curve that connects the given endpoints above C_2, C_3 is a simple curve that connects the given endpoints below C_2.&quot; id=&quot;19&quot;&gt;\" width=\"655\" height=\"352\" \/><\/p>\n<p id=\"caption-attachment-5253\" class=\"wp-caption-text\">Figure 5. The vector field is conservative, and therefore independent of path.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div class=\"ui-has-child-title\" data-type=\"note\">\n<p id=\"fs-id1167794125835\">To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.<\/p>\n<p id=\"fs-id1167793287896\">We have shown that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path. It turns out that if the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then the converse is also true. That is, if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path and the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: The path independence test of conservative fields<\/h3>\n<hr \/>\n<p>If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a continuous vector field that is independent of path and the domain [latex]D[\/latex] of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is open and connected, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1167793394250\">We prove the theorem for vector fields in [latex]\\mathbb{R}^2[\/latex]. The proof for vector fields in [latex]\\mathbb{R}^3[\/latex] is similar. To show that [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is conservative, we must find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. To that end, let [latex]X[\/latex] be a fixed point in [latex]D[\/latex]. For any point [latex](x, y)[\/latex] in [latex]D[\/latex], let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex]. Define [latex]f(x, y)[\/latex] by [latex]f(x,y)=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. (Note that this definition of [latex]f[\/latex] makes sense only because\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was not independent of path, then it might be possible to find another path [latex]C'[\/latex] from [latex]X[\/latex] to [latex](x, y)[\/latex] such that [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], and in such a case [latex]f(x, y)[\/latex] would not be a function.) We want to show that [latex]f[\/latex] has the property [latex]\\nabla{f}={\\bf{F}}[\/latex].<\/p>\n<p id=\"fs-id1167794126268\">Since domain [latex]D[\/latex] is open, it is possible to find a disk centered at [latex](x, y)[\/latex] such that the disk is contained entirely inside [latex]D[\/latex]. Let [latex](a, y)[\/latex] with [latex]a<x[\/latex] be a point in that disk. Let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex] that consists of two pieces: [latex]C_1[\/latex] and [latex]C_2[\/latex]. The first piece, [latex]C_1[\/latex], is any path from [latex]X[\/latex] to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex]; [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x ,y)[\/latex] (Figure 6). Then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f(x,y)=\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\n<p id=\"fs-id1167793426381\">The first integral does not depend on [latex]x[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac{\\partial}{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\n<p id=\"fs-id1167793959035\">If we parameterize [latex]C_2[\/latex] by [latex]{\\bf{r}}(t)=\\langle{t},y\\rangle[\/latex], [latex]a\\leq{t}\\leq{x}[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  f_x&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\  &=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\frac{d}{dt}(\\langle{t},y\\rangle \\ dt \\\\  &=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\langle1,0\\rangle \\ dt \\\\  &=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793881085\">By the Fundamental Theorem of Calculus (part 1),<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f_x=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt=P(x,y)}[\/latex].<\/p>\n<div id=\"attachment_5255\" style=\"width: 438px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5255\" class=\"size-full wp-image-5255\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31223631\/6.30.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/96903a4e93cc5b2a37d8748493ad2684ea71c218&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram of a region D in the rough shape of a backwards C. It is a simply connected region formed by a closed curve. Another curve C_1 is drawn inside D from point X to (a,y). C_2 is a horizontal line segment drawn from (a,y) to (x,y). Arrowheads point to (a,y) on C_1 and to (x,y) on C_2.&quot; id=&quot;21&quot;&gt;\" width=\"428\" height=\"511\" \/><\/p>\n<p id=\"caption-attachment-5255\" class=\"wp-caption-text\">Figure 6. Here, [latex]C_1[\/latex] is any path from [latex]X[\/latex]; to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex], and [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x, y)[\/latex].<\/p>\n<\/div>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\">\n<p id=\"fs-id1167794211104\">A similar argument using a vertical line segment rather than a horizontal line segment shows that [latex]f_y= Q(x, y)[\/latex].<\/p>\n<p id=\"fs-id1167793609990\">Therefore [latex]\\nabla{f}={\\bf{F}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167793469703\">We have spent a lot of time discussing and proving\u00a0Path Independence of Conservative Fields Theorem\u00a0and\u00a0The Path Independence Test for Conservative Fields Theorem, but we can summarize them simply: a vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: showing that a vector field is not conservative<\/h3>\n<p>Use path independence to show that vector field [latex]{\\bf{F}}(x,y)=\\langle{x^2}y,y+5\\rangle[\/latex] is not conservative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q093720434\">Show Solution<\/span><\/p>\n<div id=\"q093720434\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794210536\">We can indicate that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative by showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not path independent. We do so by giving two different paths, [latex]C_1[\/latex] and [latex]C_2[\/latex], that both start at [latex](0, 0)[\/latex] and end at [latex](1, 1)[\/latex], and yet [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\n<p id=\"fs-id1167793311592\">Let [latex]C_1[\/latex] be the curve with parameterization [latex]r_1(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] and let [latex]C_2[\/latex] be the curve with parameterization [latex]r_2(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] (Figure 7). Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1{\\bf{F}}(r_1(t))\\cdot{r}_1^\\prime(t) \\ dt \\\\  &=\\displaystyle\\int_0^1\\langle{t^3},t+5\\rangle\\cdot\\langle1,1\\rangle \\ dt =\\displaystyle\\int_0^1(t^3+t+5) \\ dt \\\\  &=\\left[\\frac{t^4}4+\\frac{t^2}2+5t\\right]_0^1=\\frac{23}4  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793925125\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1{\\bf{F}}(r_2(t))\\cdot{r}_2^\\prime(t) \\ dt \\\\  &=\\displaystyle\\int_0^1\\langle{t^4},t^2+5\\rangle\\cdot\\langle1,2t\\rangle \\ dt =\\displaystyle\\int_0^1(t^4+2t^2+10t) \\ dt \\\\  &=\\left[\\frac{t^5}5+\\frac{t^4}2+5t^2\\right]_0^1=\\frac{57}{10}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793498806\">Since [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], the value of a line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0depends on the path between two given points. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not independent of path, and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative.<\/p>\n<div id=\"attachment_5261\" style=\"width: 350px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5261\" class=\"size-full wp-image-5261\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31224359\/6.31.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/f04c50a036101a7fa1172a0b89c787ab72e1aa7c&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field drawn in two dimensions. The arrows are roughly the same length. They point directly up but tend to shift to the right in the upper right portion of quadrant 1. Curves C_1 and C_2 connect the origin to point (1,1). They are both simple curves, and their arrowheads point to (1,1).&quot; id=&quot;23&quot;&gt;\" width=\"340\" height=\"348\" \/><\/p>\n<p id=\"caption-attachment-5261\" class=\"wp-caption-text\">Figure 7. Curves [latex]C_1[\/latex] and [latex]C_2[\/latex] are both oriented from left to right.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Show that [latex]{\\bf{F}}(x,y)=\\langle{x}y,x^2y^2\\rangle[\/latex] is not path independent by considering the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] and the piece of the graph of [latex]y=\\frac{x^2}2[\/latex] that goes from [latex](0, 0)[\/latex] to\u00a0[latex](2, 2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483725098\">Show Solution<\/span><\/p>\n<div id=\"q483725098\" class=\"hidden-answer\" style=\"display: none\">\n<p>If [latex]C_1[\/latex] and [latex]C_2[\/latex] represent the two curves, then [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\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-5485\">\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 6.26. <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":12,"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 6.26\",\"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-5485","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5485","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\/349141"}],"version-history":[{"count":28,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5485\/revisions"}],"predecessor-version":[{"id":6459,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5485\/revisions\/6459"}],"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\/5485\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5485"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5485"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5485"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5485"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}