{"id":5488,"date":"2022-06-02T18:49:57","date_gmt":"2022-06-02T18:49:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5488"},"modified":"2022-11-01T05:17:39","modified_gmt":"2022-11-01T05:17:39","slug":"circulation-form-of-greens-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/circulation-form-of-greens-theorem\/","title":{"raw":"Circulation Form of Green\u2019s Theorem","rendered":"Circulation Form of Green\u2019s Theorem"},"content":{"raw":"<div data-type=\"example\">\r\n<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\">Apply the circulation form of Green\u2019s theorem.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Extending the Fundamental Theorem of Calculus<\/h2>\r\n<p id=\"fs-id1167794000131\">Recall that the Fundamental Theorem of Calculus says that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_a^b{F}^\\prime(x)dx=F(b)-F(a)}[\/latex].<\/p>\r\n<p id=\"fs-id1167793361383\">As a geometric statement, this equation says that the integral over the region below the graph of [latex]F^\\prime(x)[\/latex] and above the line segment [latex][a,b][\/latex] depends only on the value of [latex]F[\/latex] at the endpoints\u00a0<em data-effect=\"italics\">a<\/em>\u00a0and\u00a0<em data-effect=\"italics\">b<\/em>\u00a0of that segment. Since the numbers [latex]a[\/latex]\u00a0and\u00a0[latex]b[\/latex]\u00a0are the boundary of the line segment [latex][a,b][\/latex], the theorem says we can calculate integral [latex]\\displaystyle\\int_a^b{F}^\\prime(x)dx[\/latex] based on information about the boundary of line segment [latex][a,b][\/latex] (Figure 1). The same idea is true of the Fundamental Theorem for Line Integrals:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}}=f({\\bf{r}}(b))-f({\\bf{r}}(a))}[\/latex].<\/p>\r\n<p id=\"fs-id1167793831466\">When we have a potential function (an \u201cantiderivative\u201d), we can calculate the line integral based solely on information about the boundary of curve [latex]C[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5268\" align=\"aligncenter\" width=\"454\"]<img class=\"size-full wp-image-5268\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225241\/6.32.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/1fd7145a16a0b05f17857a2d38eae6e520680739&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A graph in quadrant 1 of a generic function f(x). It is an increasing concave up function for the first quarter, an increasing concave down function for the second quarter, a decreasing concave down function for the third quarter, and an increasing concave down function for the last quarter. In the second quarter, a point a is marked on the x axis, and in the third quarter, a point b is marked on the x axis. The area under the curve and between a and b is shaded. This area is labeled the integral from a to b of f(x) dx.&quot; id=&quot;2&quot;&gt;\" width=\"454\" height=\"275\" \/> Figure 1. The Fundamental Theorem of Calculus says that the integral over line segment [latex][a,b][\/latex] depends only on the values of the antiderivative at the endpoints of [latex][a,b][\/latex].[\/caption]<section id=\"fs-id1167794163661\" data-depth=\"1\">\r\n<p id=\"fs-id1167794022133\"><strong><span id=\"b15b6c19-717d-45d1-877b-d79c5b8518b9_term259\" data-type=\"term\">Green\u2019s theorem<\/span><\/strong>\u00a0takes this idea and extends it to calculating double integrals. Green\u2019s theorem says that we can calculate a double integral over region [latex]D[\/latex] based solely on information about the boundary of [latex]D[\/latex] Green\u2019s theorem also says we can calculate a line integral over a simple closed curve [latex]C[\/latex] based solely on information about the region that [latex]C[\/latex] encloses. In particular, Green\u2019s theorem connects a double integral over region [latex]D[\/latex] to a line integral around the boundary of [latex]D[\/latex].<\/p>\r\n\r\n<\/section><section id=\"fs-id1167793472730\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Circulation Form of Green\u2019s Theorem<\/h2>\r\n<p id=\"fs-id1167793804583\">The first form of Green\u2019s theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve [latex]C[\/latex] to a double integral over the region enclosed by [latex]C[\/latex]. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.<\/p>\r\n\r\n<\/section><\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: green's theorem, circulation form<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793879614\">Let [latex]D[\/latex] be an open, simply connected region with a boundary curve [latex]C[\/latex] that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 2). Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives on [latex]D[\/latex]. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\oint_CPdx+Qdy=\\displaystyle\\iint_D(Q_x-P_y)dA[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"attachment_5271\" align=\"aligncenter\" width=\"493\"]<img class=\"size-full wp-image-5271\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225421\/6.33.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/36fc30dd3c8b4bbbf13c91ec20e65b9d6aa4a4f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions with all of the arrows pointing up and to the right. A curve C oriented counterclockwise sections off a region D around the origin. It is a simple, closed region.&quot; id=&quot;4&quot;&gt;\" width=\"493\" height=\"349\" \/> Figure 2. The circulation form of Green\u2019s theorem relates a line integral over curve [latex]C[\/latex] to a double integral over region [latex]D[\/latex].[\/caption]\r\n<p id=\"fs-id1167793422976\">Notice that Green\u2019s theorem can be used only for a two-dimensional vector field\u00a0[latex]{\\bf{F}}[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a three-dimensional field, then Green\u2019s theorem does not apply. Since<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_CPdx+Qdy=\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{T}}ds}[\/latex],<\/p>\r\n<p id=\"fs-id1167793430412\">this version of Green\u2019s theorem is sometimes referred to as the\u00a0<span id=\"b15b6c19-717d-45d1-877b-d79c5b8518b9_term260\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">tangential form<\/em>\u00a0of Green\u2019s theorem<\/span>.<\/p>\r\n<p id=\"fs-id1167794222761\">The proof of Green\u2019s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that [latex]D[\/latex] is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which [latex]{\\bf{F}}\\langle{P},Q\\rangle[\/latex] is conservative. In this case,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_CPdx+Qdy=0}[\/latex]<\/p>\r\n<p id=\"fs-id1167793569904\">because the circulation is zero in conservative vector fields. By\u00a0Cross-Partial Property of Conservative Fields Theorem,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial condition, so [latex]P_y=Q_x[\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D0dA=0=\\displaystyle\\oint_CPdx+Qdy}[\/latex],<\/p>\r\n<p id=\"fs-id1167794098233\">which confirms Green\u2019s theorem in the case of conservative vector fields.<\/p>\r\n\r\n<section id=\"fs-id1167794144755\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1167793370605\">Let\u2019s now prove that the circulation form of Green\u2019s theorem is true when the region [latex]D[\/latex] is a rectangle. Let [latex]D[\/latex] be the rectangle [latex][a,b]\\times[c,d][\/latex] oriented counterclockwise. Then, the boundary [latex]C[\/latex] of [latex]D[\/latex] consists of four piecewise smooth pieces [latex]C_1[\/latex], [latex]C_2[\/latex], [latex]C_3[\/latex], and [latex]C_4[\/latex] (Figure 3). We parameterize each side of [latex]D[\/latex] as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nC_1:{\\bf{r}}_1(t)&amp;=\\langle{t},c\\rangle,a\\leq{t}\\leq{b} \\\\\r\nC_2:{\\bf{r}}_2(t)&amp;=\\langle{b},t\\rangle,c\\leq{t}\\leq{d} \\\\\r\n-C_3:{\\bf{r}}_3(t)&amp;=\\langle{t},d\\rangle,a\\leq{t}\\leq{b} \\\\\r\n-C_4:{\\bf{r}}_4(t)&amp;=\\langle{a},t\\rangle,c\\leq{t}\\leq{d}\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/section><\/div>\r\n<div>[caption id=\"attachment_5273\" align=\"aligncenter\" width=\"337\"]<img class=\"size-full wp-image-5273\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225540\/6.34.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0177a6156983d7186594e82b72926d649a2e18f1&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram in quadrant 1. Rectangle D is oriented counterclockwise. Points a and b are on the x axis, and points c and d are on the y axis with b &gt; a and d &gt; c. The sides of the rectangle are side c1 with endpoints at (a,c) and (b,c), side c2 with endpoints at (b,c) and (b,d), side c3 with endpoints at (b,d) and (a,d), and side c4 with endpoints at (a,d) and (a,c).&quot; id=&quot;5&quot;&gt;\" width=\"337\" height=\"280\" \/> Figure 3. Rectangle [latex]D[\/latex] is oriented counterclockwise.[\/caption]<\/div>\r\n<div data-type=\"example\">\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">Then,<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}&amp;=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_1(t))\\bullet{\\bf{r}}_1(t)dt+\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_2(t))\\bullet{\\bf{r}}_2(t)dt-\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_3(t))\\bullet{\\bf{r}}_3(t)dt-\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_4(t))\\bullet{\\bf{r}}_4(t)dt \\\\\r\n&amp;=\\displaystyle\\int_a^bP(t,c)dt+\\displaystyle\\int_c^dQ(b,t)dt-\\displaystyle\\int_a^bP(t,d)dt-\\displaystyle\\int_c^dQ(a,t)dt \\\\\r\n&amp;=\\displaystyle\\int_a^b(P(t,c)-P(t,d))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\\r\n&amp;=-\\displaystyle\\int_a^b(P(t,d)-P(t,c))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793636189\">By the Fundamental Theorem of Calculus,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{P(t,d)-P(t,c)=\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dy\\text{ and }Q(b,t)-Q(a,t)=\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167793931876\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n&amp;-\\displaystyle\\int_a^b(P(t,d)-P(t,c)dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\\r\n&amp;=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793918518\">But,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt&amp;=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(x,y)dydx+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdy \\\\\r\n&amp;=\\displaystyle\\int_a^b\\displaystyle\\int_c^d(Q_x-P_y)dydx \\\\\r\n&amp;=\\displaystyle\\int \\ \\displaystyle\\int_D(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793637686\">Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\int\\int_D(Q_{x}-P_{y})dA[\/latex] and we have proved Green\u2019s theorem in the case of a rectangle.<\/p>\r\n<p id=\"fs-id1167794076825\">To prove Green\u2019s theorem over a general region [latex]D[\/latex], we can decompose [latex]D[\/latex] into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: applying Green's Theorem over a rectangle<\/h3>\r\n<p id=\"fs-id1167793778221\">Calculate the line integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_Cx^2ydx+(y-3)dy[\/latex]<\/p>\r\n<p id=\"fs-id1167793771032\">where [latex]C[\/latex] is a rectangle with vertices [latex](1, 1)[\/latex], [latex](4, 1)[\/latex], [latex](4, 5)[\/latex], and [latex](1, 5)[\/latex] oriented counterclockwise.<\/p>\r\n[reveal-answer q=\"384752098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384752098\"]\r\n<h2>Solution<\/h2>\r\n<p id=\"fs-id1167793610999\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{x}^2y,y-3\\rangle[\/latex]. Then, [latex]Q_x=0[\/latex] and [latex]P_y=x^{2}[\/latex]. Therefore, [latex]W_x-P_y=-x^{2}[\/latex].<\/p>\r\n<p id=\"fs-id1167794050031\">Let [latex]D[\/latex] be the rectangular region enclosed by [latex]C[\/latex] (Figure 4). By Green\u2019s theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_Cx^2ydx+(y-3)dy&amp;=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&amp;=\\displaystyle\\int \\ \\displaystyle\\int_D-x^2dA=\\displaystyle\\int_1^5 \\ \\displaystyle\\int_1^4-x^2dxdy \\\\\r\n&amp;=\\displaystyle\\int_1^5-21dy = -84\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_5278\" align=\"aligncenter\" width=\"719\"]<img class=\"size-full wp-image-5278\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230043\/6.35.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/e1e134b9025c929cea27bc975695f84f13acd178&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions with focus on quadrant 1. The arrows near the origin are short, and the arrows further away from the origin are longer. A rectangle has endpoints at (1,1), (4,1), (4,5), and (1,5). The arrows in quadrant 3 are pointing to the right. At the y axis, they split at y = 3. Arrows above that line curve up at the y axis and shift until they are horizontally pointing to the right in quadrant 1. Arrows below that line and above the x axis curve down at the y axis and shift until they are horizontally pointing to the right. Arrows below the x axis point to the left and down, pointing back to the y axis.&quot; id=&quot;7&quot;&gt;\" width=\"719\" height=\"646\" \/> Figure 4. The line integral over the boundary of the rectangle can be transformed into a double integral over the rectangle.[\/caption]\r\n<h2 data-type=\"commentary-title\">[\/hidden-answer]<\/h2>\r\n<h2 id=\"8\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\r\n<p id=\"fs-id1167793340651\">If we were to evaluate this line integral without using Green\u2019s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-line-integrals\/\" data-page-slug=\"6-2-line-integrals\" data-page-uuid=\"34b090a5-96d8-48dc-97e5-4aab175adbb5\" data-page-fragment=\"page_34b090a5-96d8-48dc-97e5-4aab175adbb5\">Line Integrals<\/a>\u00a0to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green\u2019s theorem makes the calculation much simpler.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_16_04_004\" class=\"os-figure\"><\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying Green's Theorem to Calculate work<\/h3>\r\n<p id=\"fs-id1167793948916\">Calculate the work done on a particle by force field<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{F}}(x,y)=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]<\/p>\r\n<p id=\"fs-id1167793416833\">as the particle traverses circle [latex]x^2+y^2=4[\/latex] exactly once in the counterclockwise direction, starting and ending at point [latex](2, 0)[\/latex].<\/p>\r\n[reveal-answer q=\"187491994\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"187491994\"]\r\n<p id=\"fs-id1167793443552\">Let [latex]C[\/latex] denote the circle and let [latex]D[\/latex] be the disk enclosed by [latex]C[\/latex]. The work done on the particle is<\/p>\r\n<p style=\"text-align: center;\">[latex]W=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy[\/latex].<\/p>\r\n<p id=\"fs-id1167794117883\">As with\u00a0Example \"Applying Green\u2019s Theorem over a Rectangle\", this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green\u2019s theorem (Figure 5).<\/p>\r\n<p id=\"fs-id1167793579556\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]. Then, [latex]Q_x=-1\\text{ and }P_y=1[\/latex]. Therefore, [latex]Q_x-P_y=-2[\/latex].<\/p>\r\n<p id=\"fs-id1167794144052\">By Green\u2019s theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nW&amp;=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy \\\\\r\n&amp;=\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D-2dA \\\\\r\n&amp;=-2(\\text{area}(D))=-2\\pi(2^2)=-8\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_5280\" align=\"aligncenter\" width=\"717\"]<img class=\"size-full wp-image-5280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230135\/6.36.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/628e498ab257314c7470cab521111c2e02af3ce1&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows further away from the origin are much longer than those near the origin. The arrows curve out from about (.5,.5) in a clockwise spiral pattern.&quot; id=&quot;10&quot;&gt;\" width=\"717\" height=\"572\" \/> Figure 5. The line integral over the boundary circle can be transformed into a double integral over the disk enclosed by the circle.[\/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\n<p id=\"fs-id1167794246701\">Use Green\u2019s theorem to calculate line integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_C\\sin(x^2)dx+(3x-y)dy[\/latex],<\/p>\r\n<p id=\"fs-id1167793433362\">where [latex]C[\/latex] is a right triangle with vertices [latex](-1, 2)[\/latex], [latex](4, 2)[\/latex], and [latex](4, 5)[\/latex] oriented counterclockwise.<\/p>\r\n[reveal-answer q=\"304250970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"304250970\"]\r\n\r\n[latex]\\frac{45}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=8250320&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ULOBy2zIJqA&amp;video_target=tpm-plugin-scbmiged-ULOBy2zIJqA\" 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\/CP6.34_transcript.html\">transcript for \u201cCP 6.34\u201d here (opens in new window).<\/a><\/center>In the preceding two examples, the double integral in Green\u2019s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green\u2019s theorem to translate a double integral into a line integral.\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: applying green's theorem over an ellipse<\/h3>\r\nCalculate the area enclosed by ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] (Figure 6).\r\n\r\n[caption id=\"attachment_5283\" align=\"aligncenter\" width=\"349\"]<img class=\"size-full wp-image-5283\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230231\/6.37.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/d7d97e8443810811ba4304de3fe2d855f57a3cae&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A horizontal ellipse graphed in two dimensions. It has vertices at (-a, 0), (0, -b), (a, 0), and (0, b), where the absolute value of a is between 2.5 and 5 and the absolute value of b is between 0 and 2.5.&quot; id=&quot;13&quot;&gt;\" width=\"349\" height=\"197\" \/> Figure 6. Ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] is denoted by [latex]C[\/latex].[\/caption][reveal-answer q=\"837599384\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"837599384\"]\r\n<p id=\"fs-id1167793378251\">Let [latex]C[\/latex] denote the ellipse and let [latex]D[\/latex] be the region enclosed by [latex]C[\/latex]. Recall that ellipse [latex]C[\/latex] can be parameterized by<\/p>\r\n<p style=\"text-align: center;\">[latex]x=a\\cos{t},y=b\\sin{t},0\\leq{t}\\leq2\\pi[\/latex].<\/p>\r\n<p id=\"fs-id1167793915906\">Calculating the area of [latex]D[\/latex] is equivalent to computing double integral [latex]\\displaystyle\\iint_DdA[\/latex]. To calculate this integral without Green\u2019s theorem, we would need to divide [latex]D[\/latex] into two regions: the region above the\u00a0<em data-effect=\"italics\">x<\/em>-axis and the region below. The area of the ellipse is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{-a}^a\\displaystyle\\int_0^{\\sqrt{b^2-(bx\/a)^2}}dydx+\\displaystyle\\int_{-a}^a\\displaystyle\\int_{\\sqrt{b^2-(bx\/a)^2}}^0dydx[\/latex].<\/p>\r\n<p id=\"fs-id1167793932632\">These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green\u2019s theorem to transform [latex]\\displaystyle\\iint_DdA[\/latex] into a line integral around the boundary [latex]C[\/latex].<\/p>\r\n<p id=\"fs-id1167793626809\">Consider vector field<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1167793720113\">Then, [latex]Q_x=\\frac12[\/latex] and [latex]P_y=-\\frac12[\/latex], and therefore [latex]Q_x-P_y=1[\/latex]. Notice that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was chosen to have the property that [latex]Q_x-P_y=1[\/latex]. Since this is the case, Green\u2019s theorem transforms the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over [latex]C[\/latex] into the double integral of 1 over [latex]D[\/latex].<\/p>\r\n<p id=\"fs-id1167793521511\">By Green\u2019s theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_DdA&amp;=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&amp;=\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\frac12\\displaystyle\\int_C-ydx+xdy \\\\\r\n&amp;=\\frac12\\displaystyle\\int_0^{2\\pi}-b\\sin{t}(-a\\sin{t})+a(\\cos{t})b\\cos{t}dt \\\\\r\n&amp;=\\frac12\\displaystyle\\int_0^{2\\pi}ab\\cos^2t+ab\\sin^2tdt=\\frac12\\displaystyle\\int_0^{2\\pi}abdt=\\pi{a}b\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167794334808\">Therefore, the area of the ellipse is\u00a0[latex]\\pi{a}b[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794147373\">In\u00a0Example \"Applying Green\u2019s Theorem over an Ellipse\", we used vector field [latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex] to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region [latex]D[\/latex]. Let [latex]D[\/latex] be any region with a boundary that is a simple closed curve [latex]C[\/latex] oriented counterclockwise. If [latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex]. Therefore, by the same logic as in\u00a0Example \"Applying Green\u2019s Theorem over an Ellipse\",<\/p>\r\n<p style=\"text-align: center;\"><span style=\"text-align: center; white-space: nowrap; word-spacing: normal;\">area of [latex]D=\\displaystyle\\iint_DdA=\\frac12\\displaystyle\\oint_C-ydx+xdy[\/latex].<\/span><\/p>\r\n<p id=\"fs-id1167794094247\">It\u2019s worth noting that if [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is any vector field with [latex]Q_x-P_y=1[\/latex], then the logic of the previous paragraph works. So, the equation above\u00a0is not the only equation that uses a vector field\u2019s mixed partials to get the area of a region.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the area of the region enclosed by the curve with parameterization\u00a0[latex]{\\bf{r}}(t)=\\langle\\sin{t}\\cos{t},\\sin{t}\\rangle,\\text{ }0\\leq{t}\\leq\\pi[\/latex].\r\n\r\n[caption id=\"attachment_5285\" align=\"aligncenter\" width=\"343\"]<img class=\"size-full wp-image-5285\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230402\/6.35c.jpg\" alt=\"An image of a curve in quadrants 1 and 2. The curve begins at the origin, curves up and to the right until about (.5, .8), curves to the left nearly horizontally, goes through (0,1), continues until about (-1, .7), and then curves down and to the right until it hits the origin again.\" width=\"343\" height=\"234\" \/> Figure 7. This region is enclosed by the curve with parameterization\u00a0[latex]{\\bf{r}}(t)=\\langle\\sin{t}\\cos{t},\\sin{t}\\rangle,\\text{ }0\\leq{t}\\leq\\pi[\/latex].[\/caption][reveal-answer q=\"487592847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"487592847\"][latex]\\frac43[\/latex][\/hidden-answer]<\/div>\r\n<\/div>","rendered":"<div data-type=\"example\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Apply the circulation form of Green\u2019s theorem.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Extending the Fundamental Theorem of Calculus<\/h2>\n<p id=\"fs-id1167794000131\">Recall that the Fundamental Theorem of Calculus says that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_a^b{F}^\\prime(x)dx=F(b)-F(a)}[\/latex].<\/p>\n<p id=\"fs-id1167793361383\">As a geometric statement, this equation says that the integral over the region below the graph of [latex]F^\\prime(x)[\/latex] and above the line segment [latex][a,b][\/latex] depends only on the value of [latex]F[\/latex] at the endpoints\u00a0<em data-effect=\"italics\">a<\/em>\u00a0and\u00a0<em data-effect=\"italics\">b<\/em>\u00a0of that segment. Since the numbers [latex]a[\/latex]\u00a0and\u00a0[latex]b[\/latex]\u00a0are the boundary of the line segment [latex][a,b][\/latex], the theorem says we can calculate integral [latex]\\displaystyle\\int_a^b{F}^\\prime(x)dx[\/latex] based on information about the boundary of line segment [latex][a,b][\/latex] (Figure 1). The same idea is true of the Fundamental Theorem for Line Integrals:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_C\\nabla{f}\\cdot{d}{\\bf{r}}=f({\\bf{r}}(b))-f({\\bf{r}}(a))}[\/latex].<\/p>\n<p id=\"fs-id1167793831466\">When we have a potential function (an \u201cantiderivative\u201d), we can calculate the line integral based solely on information about the boundary of curve [latex]C[\/latex].<\/p>\n<div id=\"attachment_5268\" style=\"width: 464px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5268\" class=\"size-full wp-image-5268\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225241\/6.32.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/1fd7145a16a0b05f17857a2d38eae6e520680739&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A graph in quadrant 1 of a generic function f(x). It is an increasing concave up function for the first quarter, an increasing concave down function for the second quarter, a decreasing concave down function for the third quarter, and an increasing concave down function for the last quarter. In the second quarter, a point a is marked on the x axis, and in the third quarter, a point b is marked on the x axis. The area under the curve and between a and b is shaded. This area is labeled the integral from a to b of f(x) dx.&quot; id=&quot;2&quot;&gt;\" width=\"454\" height=\"275\" \/><\/p>\n<p id=\"caption-attachment-5268\" class=\"wp-caption-text\">Figure 1. The Fundamental Theorem of Calculus says that the integral over line segment [latex][a,b][\/latex] depends only on the values of the antiderivative at the endpoints of [latex][a,b][\/latex].<\/p>\n<\/div>\n<section id=\"fs-id1167794163661\" data-depth=\"1\">\n<p id=\"fs-id1167794022133\"><strong><span id=\"b15b6c19-717d-45d1-877b-d79c5b8518b9_term259\" data-type=\"term\">Green\u2019s theorem<\/span><\/strong>\u00a0takes this idea and extends it to calculating double integrals. Green\u2019s theorem says that we can calculate a double integral over region [latex]D[\/latex] based solely on information about the boundary of [latex]D[\/latex] Green\u2019s theorem also says we can calculate a line integral over a simple closed curve [latex]C[\/latex] based solely on information about the region that [latex]C[\/latex] encloses. In particular, Green\u2019s theorem connects a double integral over region [latex]D[\/latex] to a line integral around the boundary of [latex]D[\/latex].<\/p>\n<\/section>\n<section id=\"fs-id1167793472730\" data-depth=\"1\">\n<h2 data-type=\"title\">Circulation Form of Green\u2019s Theorem<\/h2>\n<p id=\"fs-id1167793804583\">The first form of Green\u2019s theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve [latex]C[\/latex] to a double integral over the region enclosed by [latex]C[\/latex]. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.<\/p>\n<\/section>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: green&#8217;s theorem, circulation form<\/h3>\n<hr \/>\n<p id=\"fs-id1167793879614\">Let [latex]D[\/latex] be an open, simply connected region with a boundary curve [latex]C[\/latex] that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 2). Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives on [latex]D[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\oint_CPdx+Qdy=\\displaystyle\\iint_D(Q_x-P_y)dA[\/latex].<\/p>\n<\/div>\n<div id=\"attachment_5271\" style=\"width: 503px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5271\" class=\"size-full wp-image-5271\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225421\/6.33.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/36fc30dd3c8b4bbbf13c91ec20e65b9d6aa4a4f7&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions with all of the arrows pointing up and to the right. A curve C oriented counterclockwise sections off a region D around the origin. It is a simple, closed region.&quot; id=&quot;4&quot;&gt;\" width=\"493\" height=\"349\" \/><\/p>\n<p id=\"caption-attachment-5271\" class=\"wp-caption-text\">Figure 2. The circulation form of Green\u2019s theorem relates a line integral over curve [latex]C[\/latex] to a double integral over region [latex]D[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793422976\">Notice that Green\u2019s theorem can be used only for a two-dimensional vector field\u00a0[latex]{\\bf{F}}[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a three-dimensional field, then Green\u2019s theorem does not apply. Since<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_CPdx+Qdy=\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{T}}ds}[\/latex],<\/p>\n<p id=\"fs-id1167793430412\">this version of Green\u2019s theorem is sometimes referred to as the\u00a0<span id=\"b15b6c19-717d-45d1-877b-d79c5b8518b9_term260\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">tangential form<\/em>\u00a0of Green\u2019s theorem<\/span>.<\/p>\n<p id=\"fs-id1167794222761\">The proof of Green\u2019s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that [latex]D[\/latex] is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which [latex]{\\bf{F}}\\langle{P},Q\\rangle[\/latex] is conservative. In this case,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_CPdx+Qdy=0}[\/latex]<\/p>\n<p id=\"fs-id1167793569904\">because the circulation is zero in conservative vector fields. By\u00a0Cross-Partial Property of Conservative Fields Theorem,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial condition, so [latex]P_y=Q_x[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D0dA=0=\\displaystyle\\oint_CPdx+Qdy}[\/latex],<\/p>\n<p id=\"fs-id1167794098233\">which confirms Green\u2019s theorem in the case of conservative vector fields.<\/p>\n<section id=\"fs-id1167794144755\" data-depth=\"2\">\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1167793370605\">Let\u2019s now prove that the circulation form of Green\u2019s theorem is true when the region [latex]D[\/latex] is a rectangle. Let [latex]D[\/latex] be the rectangle [latex][a,b]\\times[c,d][\/latex] oriented counterclockwise. Then, the boundary [latex]C[\/latex] of [latex]D[\/latex] consists of four piecewise smooth pieces [latex]C_1[\/latex], [latex]C_2[\/latex], [latex]C_3[\/latex], and [latex]C_4[\/latex] (Figure 3). We parameterize each side of [latex]D[\/latex] as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  C_1:{\\bf{r}}_1(t)&=\\langle{t},c\\rangle,a\\leq{t}\\leq{b} \\\\  C_2:{\\bf{r}}_2(t)&=\\langle{b},t\\rangle,c\\leq{t}\\leq{d} \\\\  -C_3:{\\bf{r}}_3(t)&=\\langle{t},d\\rangle,a\\leq{t}\\leq{b} \\\\  -C_4:{\\bf{r}}_4(t)&=\\langle{a},t\\rangle,c\\leq{t}\\leq{d}  \\end{aligned}[\/latex].<\/p>\n<\/section>\n<\/div>\n<div>\n<div id=\"attachment_5273\" style=\"width: 347px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5273\" class=\"size-full wp-image-5273\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31225540\/6.34.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0177a6156983d7186594e82b72926d649a2e18f1&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram in quadrant 1. Rectangle D is oriented counterclockwise. Points a and b are on the x axis, and points c and d are on the y axis with b &gt; a and d &gt; c. The sides of the rectangle are side c1 with endpoints at (a,c) and (b,c), side c2 with endpoints at (b,c) and (b,d), side c3 with endpoints at (b,d) and (a,d), and side c4 with endpoints at (a,d) and (a,c).&quot; id=&quot;5&quot;&gt;\" width=\"337\" height=\"280\" \/><\/p>\n<p id=\"caption-attachment-5273\" class=\"wp-caption-text\">Figure 3. Rectangle [latex]D[\/latex] is oriented counterclockwise.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">Then,<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}&=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\  &=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\  &=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_1(t))\\bullet{\\bf{r}}_1(t)dt+\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_2(t))\\bullet{\\bf{r}}_2(t)dt-\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_3(t))\\bullet{\\bf{r}}_3(t)dt-\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_4(t))\\bullet{\\bf{r}}_4(t)dt \\\\  &=\\displaystyle\\int_a^bP(t,c)dt+\\displaystyle\\int_c^dQ(b,t)dt-\\displaystyle\\int_a^bP(t,d)dt-\\displaystyle\\int_c^dQ(a,t)dt \\\\  &=\\displaystyle\\int_a^b(P(t,c)-P(t,d))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\  &=-\\displaystyle\\int_a^b(P(t,d)-P(t,c))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793636189\">By the Fundamental Theorem of Calculus,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{P(t,d)-P(t,c)=\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dy\\text{ and }Q(b,t)-Q(a,t)=\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dx}[\/latex].<\/p>\n<p id=\"fs-id1167793931876\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  &-\\displaystyle\\int_a^b(P(t,d)-P(t,c)dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\  &=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793918518\">But,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  -\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt&=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(x,y)dydx+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdy \\\\  &=\\displaystyle\\int_a^b\\displaystyle\\int_c^d(Q_x-P_y)dydx \\\\  &=\\displaystyle\\int \\ \\displaystyle\\int_D(Q_x-P_y)dA  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793637686\">Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\int\\int_D(Q_{x}-P_{y})dA[\/latex] and we have proved Green\u2019s theorem in the case of a rectangle.<\/p>\n<p id=\"fs-id1167794076825\">To prove Green\u2019s theorem over a general region [latex]D[\/latex], we can decompose [latex]D[\/latex] into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: applying Green&#8217;s Theorem over a rectangle<\/h3>\n<p id=\"fs-id1167793778221\">Calculate the line integral<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_Cx^2ydx+(y-3)dy[\/latex]<\/p>\n<p id=\"fs-id1167793771032\">where [latex]C[\/latex] is a rectangle with vertices [latex](1, 1)[\/latex], [latex](4, 1)[\/latex], [latex](4, 5)[\/latex], and [latex](1, 5)[\/latex] oriented counterclockwise.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384752098\">Show Solution<\/span><\/p>\n<div id=\"q384752098\" class=\"hidden-answer\" style=\"display: none\">\n<h2>Solution<\/h2>\n<p id=\"fs-id1167793610999\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{x}^2y,y-3\\rangle[\/latex]. Then, [latex]Q_x=0[\/latex] and [latex]P_y=x^{2}[\/latex]. Therefore, [latex]W_x-P_y=-x^{2}[\/latex].<\/p>\n<p id=\"fs-id1167794050031\">Let [latex]D[\/latex] be the rectangular region enclosed by [latex]C[\/latex] (Figure 4). By Green\u2019s theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_Cx^2ydx+(y-3)dy&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\  &=\\displaystyle\\int \\ \\displaystyle\\int_D-x^2dA=\\displaystyle\\int_1^5 \\ \\displaystyle\\int_1^4-x^2dxdy \\\\  &=\\displaystyle\\int_1^5-21dy = -84  \\end{aligned}[\/latex].<\/p>\n<div id=\"attachment_5278\" style=\"width: 729px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5278\" class=\"size-full wp-image-5278\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230043\/6.35.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/e1e134b9025c929cea27bc975695f84f13acd178&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions with focus on quadrant 1. The arrows near the origin are short, and the arrows further away from the origin are longer. A rectangle has endpoints at (1,1), (4,1), (4,5), and (1,5). The arrows in quadrant 3 are pointing to the right. At the y axis, they split at y = 3. Arrows above that line curve up at the y axis and shift until they are horizontally pointing to the right in quadrant 1. Arrows below that line and above the x axis curve down at the y axis and shift until they are horizontally pointing to the right. Arrows below the x axis point to the left and down, pointing back to the y axis.&quot; id=&quot;7&quot;&gt;\" width=\"719\" height=\"646\" \/><\/p>\n<p id=\"caption-attachment-5278\" class=\"wp-caption-text\">Figure 4. The line integral over the boundary of the rectangle can be transformed into a double integral over the rectangle.<\/p>\n<\/div>\n<h2 data-type=\"commentary-title\"><\/div>\n<\/div>\n<\/h2>\n<h2 id=\"8\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\n<p id=\"fs-id1167793340651\">If we were to evaluate this line integral without using Green\u2019s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-line-integrals\/\" data-page-slug=\"6-2-line-integrals\" data-page-uuid=\"34b090a5-96d8-48dc-97e5-4aab175adbb5\" data-page-fragment=\"page_34b090a5-96d8-48dc-97e5-4aab175adbb5\">Line Integrals<\/a>\u00a0to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green\u2019s theorem makes the calculation much simpler.<\/p>\n<div id=\"CNX_Calc_Figure_16_04_004\" class=\"os-figure\"><\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying Green&#8217;s Theorem to Calculate work<\/h3>\n<p id=\"fs-id1167793948916\">Calculate the work done on a particle by force field<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{F}}(x,y)=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]<\/p>\n<p id=\"fs-id1167793416833\">as the particle traverses circle [latex]x^2+y^2=4[\/latex] exactly once in the counterclockwise direction, starting and ending at point [latex](2, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q187491994\">Show Solution<\/span><\/p>\n<div id=\"q187491994\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793443552\">Let [latex]C[\/latex] denote the circle and let [latex]D[\/latex] be the disk enclosed by [latex]C[\/latex]. The work done on the particle is<\/p>\n<p style=\"text-align: center;\">[latex]W=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy[\/latex].<\/p>\n<p id=\"fs-id1167794117883\">As with\u00a0Example &#8220;Applying Green\u2019s Theorem over a Rectangle&#8221;, this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green\u2019s theorem (Figure 5).<\/p>\n<p id=\"fs-id1167793579556\">Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]. Then, [latex]Q_x=-1\\text{ and }P_y=1[\/latex]. Therefore, [latex]Q_x-P_y=-2[\/latex].<\/p>\n<p id=\"fs-id1167794144052\">By Green\u2019s theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  W&=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy \\\\  &=\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D-2dA \\\\  &=-2(\\text{area}(D))=-2\\pi(2^2)=-8\\pi  \\end{aligned}[\/latex].<\/p>\n<div id=\"attachment_5280\" style=\"width: 727px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5280\" class=\"size-full wp-image-5280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230135\/6.36.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/628e498ab257314c7470cab521111c2e02af3ce1&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A vector field in two dimensions. The arrows further away from the origin are much longer than those near the origin. The arrows curve out from about (.5,.5) in a clockwise spiral pattern.&quot; id=&quot;10&quot;&gt;\" width=\"717\" height=\"572\" \/><\/p>\n<p id=\"caption-attachment-5280\" class=\"wp-caption-text\">Figure 5. The line integral over the boundary circle can be transformed into a double integral over the disk enclosed by the circle.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1167794246701\">Use Green\u2019s theorem to calculate line integral<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_C\\sin(x^2)dx+(3x-y)dy[\/latex],<\/p>\n<p id=\"fs-id1167793433362\">where [latex]C[\/latex] is a right triangle with vertices [latex](-1, 2)[\/latex], [latex](4, 2)[\/latex], and [latex](4, 5)[\/latex] oriented counterclockwise.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q304250970\">Show Solution<\/span><\/p>\n<div id=\"q304250970\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{45}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=8250320&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ULOBy2zIJqA&amp;video_target=tpm-plugin-scbmiged-ULOBy2zIJqA\" 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\/CP6.34_transcript.html\">transcript for \u201cCP 6.34\u201d here (opens in new window).<\/a><\/div>\n<p>In the preceding two examples, the double integral in Green\u2019s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green\u2019s theorem to translate a double integral into a line integral.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: applying green&#8217;s theorem over an ellipse<\/h3>\n<p>Calculate the area enclosed by ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] (Figure 6).<\/p>\n<div id=\"attachment_5283\" style=\"width: 359px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5283\" class=\"size-full wp-image-5283\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230231\/6.37.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/d7d97e8443810811ba4304de3fe2d855f57a3cae&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A horizontal ellipse graphed in two dimensions. It has vertices at (-a, 0), (0, -b), (a, 0), and (0, b), where the absolute value of a is between 2.5 and 5 and the absolute value of b is between 0 and 2.5.&quot; id=&quot;13&quot;&gt;\" width=\"349\" height=\"197\" \/><\/p>\n<p id=\"caption-attachment-5283\" class=\"wp-caption-text\">Figure 6. Ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] is denoted by [latex]C[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q837599384\">Show Solution<\/span><\/p>\n<div id=\"q837599384\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793378251\">Let [latex]C[\/latex] denote the ellipse and let [latex]D[\/latex] be the region enclosed by [latex]C[\/latex]. Recall that ellipse [latex]C[\/latex] can be parameterized by<\/p>\n<p style=\"text-align: center;\">[latex]x=a\\cos{t},y=b\\sin{t},0\\leq{t}\\leq2\\pi[\/latex].<\/p>\n<p id=\"fs-id1167793915906\">Calculating the area of [latex]D[\/latex] is equivalent to computing double integral [latex]\\displaystyle\\iint_DdA[\/latex]. To calculate this integral without Green\u2019s theorem, we would need to divide [latex]D[\/latex] into two regions: the region above the\u00a0<em data-effect=\"italics\">x<\/em>-axis and the region below. The area of the ellipse is<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_{-a}^a\\displaystyle\\int_0^{\\sqrt{b^2-(bx\/a)^2}}dydx+\\displaystyle\\int_{-a}^a\\displaystyle\\int_{\\sqrt{b^2-(bx\/a)^2}}^0dydx[\/latex].<\/p>\n<p id=\"fs-id1167793932632\">These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green\u2019s theorem to transform [latex]\\displaystyle\\iint_DdA[\/latex] into a line integral around the boundary [latex]C[\/latex].<\/p>\n<p id=\"fs-id1167793626809\">Consider vector field<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex].<\/p>\n<p id=\"fs-id1167793720113\">Then, [latex]Q_x=\\frac12[\/latex] and [latex]P_y=-\\frac12[\/latex], and therefore [latex]Q_x-P_y=1[\/latex]. Notice that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was chosen to have the property that [latex]Q_x-P_y=1[\/latex]. Since this is the case, Green\u2019s theorem transforms the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over [latex]C[\/latex] into the double integral of 1 over [latex]D[\/latex].<\/p>\n<p id=\"fs-id1167793521511\">By Green\u2019s theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_DdA&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\  &=\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\frac12\\displaystyle\\int_C-ydx+xdy \\\\  &=\\frac12\\displaystyle\\int_0^{2\\pi}-b\\sin{t}(-a\\sin{t})+a(\\cos{t})b\\cos{t}dt \\\\  &=\\frac12\\displaystyle\\int_0^{2\\pi}ab\\cos^2t+ab\\sin^2tdt=\\frac12\\displaystyle\\int_0^{2\\pi}abdt=\\pi{a}b  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167794334808\">Therefore, the area of the ellipse is\u00a0[latex]\\pi{a}b[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167794147373\">In\u00a0Example &#8220;Applying Green\u2019s Theorem over an Ellipse&#8221;, we used vector field [latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex] to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region [latex]D[\/latex]. Let [latex]D[\/latex] be any region with a boundary that is a simple closed curve [latex]C[\/latex] oriented counterclockwise. If [latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex]. Therefore, by the same logic as in\u00a0Example &#8220;Applying Green\u2019s Theorem over an Ellipse&#8221;,<\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center; white-space: nowrap; word-spacing: normal;\">area of [latex]D=\\displaystyle\\iint_DdA=\\frac12\\displaystyle\\oint_C-ydx+xdy[\/latex].<\/span><\/p>\n<p id=\"fs-id1167794094247\">It\u2019s worth noting that if [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is any vector field with [latex]Q_x-P_y=1[\/latex], then the logic of the previous paragraph works. So, the equation above\u00a0is not the only equation that uses a vector field\u2019s mixed partials to get the area of a region.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the area of the region enclosed by the curve with parameterization\u00a0[latex]{\\bf{r}}(t)=\\langle\\sin{t}\\cos{t},\\sin{t}\\rangle,\\text{ }0\\leq{t}\\leq\\pi[\/latex].<\/p>\n<div id=\"attachment_5285\" style=\"width: 353px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5285\" class=\"size-full wp-image-5285\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31230402\/6.35c.jpg\" alt=\"An image of a curve in quadrants 1 and 2. The curve begins at the origin, curves up and to the right until about (.5, .8), curves to the left nearly horizontally, goes through (0,1), continues until about (-1, .7), and then curves down and to the right until it hits the origin again.\" width=\"343\" height=\"234\" \/><\/p>\n<p id=\"caption-attachment-5285\" class=\"wp-caption-text\">Figure 7. This region is enclosed by the curve with parameterization\u00a0[latex]{\\bf{r}}(t)=\\langle\\sin{t}\\cos{t},\\sin{t}\\rangle,\\text{ }0\\leq{t}\\leq\\pi[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q487592847\">Show Solution<\/span><\/p>\n<div id=\"q487592847\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac43[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5488\">\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 6.34. <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":16,"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 6.34\",\"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-5488","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5488","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":17,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5488\/revisions"}],"predecessor-version":[{"id":6102,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5488\/revisions\/6102"}],"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\/5488\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5488"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5488"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5488"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5488"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}