{"id":2042,"date":"2025-11-06T13:39:15","date_gmt":"2025-11-06T13:39:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=2042"},"modified":"2025-11-06T14:37:58","modified_gmt":"2025-11-06T14:37:58","slug":"greens-theorem-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/greens-theorem-learn-it-4\/","title":{"raw":"Green\u2019s Theorem: Learn It 4","rendered":"Green\u2019s Theorem: Learn It 4"},"content":{"raw":"<h2 data-type=\"title\">Green\u2019s Theorem on General Regions<\/h2>\r\nGreen's theorem, as stated, applies only to regions that are simply connected\u2014that is, Green's theorem cannot handle regions with holes. Here, we extend Green's theorem so that it does work on regions with finitely many holes (Figure 1).\r\n\r\n[caption id=\"attachment_1416\" align=\"aligncenter\" width=\"273\"]<img class=\"size-full wp-image-1416\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145733\/7407ed159946b7804a92fac2aaf626ae545a69db.webp\" alt=\"A nonsimply connected, oval-shaped region with three circular holes.\" width=\"273\" height=\"148\" \/> Figure 1. Green\u2019s theorem, as stated, does not apply to a nonsimply connected region with three holes like this one.[\/caption]\r\n<p class=\"whitespace-normal break-words\">Before discussing extensions of Green's theorem, we need to go over some terminology regarding the boundary of a region. Let [latex]D[\/latex] be a region and let [latex]C[\/latex] be a component of the boundary of [latex]D[\/latex].<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\">We say that [latex]C[\/latex] is <strong>positively oriented<\/strong> if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our left. The counterclockwise orientation of the boundary of a disk is a positive orientation, for example.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Curve [latex]C[\/latex] is <strong>negatively oriented<\/strong> if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Let [latex]D[\/latex] be a region with finitely many holes (so that [latex]D[\/latex] has finitely many boundary curves), and denote the boundary of [latex]D[\/latex] by [latex]\\partial D[\/latex] (Figure 2). To extend Green's theorem so it can handle [latex]D[\/latex], we divide region [latex]D[\/latex] into two regions, [latex]D_1[\/latex] and [latex]D_2[\/latex] (with respective boundaries [latex]\\partial D_1[\/latex] and [latex]\\partial D_2[\/latex]), in such a way that [latex]D=D_1\\cup D_2[\/latex] and neither [latex]D_1[\/latex] nor [latex]D_2[\/latex] has any holes (Figure 2).<\/p>\r\n\r\n[caption id=\"attachment_1417\" align=\"aligncenter\" width=\"652\"]<img class=\"size-full wp-image-1417\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc.webp\" alt=\"Two regions. The first region D is oval-shaped with three circular holes in it. Its oriented boundary is counterclockwise. The second region is region D split horizontally down the middle into two simply connected regions with no holes. It still has a boundary oriented counterclockwise.\" width=\"652\" height=\"302\" \/> Figure 2. (a) Region [latex]D[\/latex] with an oriented boundary has three holes. (b) Region [latex]D[\/latex] split into two simply connected regions has no holes.[\/caption]Assume the boundary of [latex]D[\/latex] is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region [latex]D_1[\/latex] and [latex]D_2[\/latex] is positively oriented. If [latex]\\mathbf{F}[\/latex] is a vector field defined on [latex]D[\/latex], then Green's theorem says that:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\oint_{\\partial{D_1}}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\oint_{\\partial{D_2}}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\iint_{D_1}Q_x-P_ydA+\\displaystyle\\iint_{D_2}Q_x-P_ydA \\\\\r\n&amp;=\\displaystyle\\iint_{D}(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167794206293\">Therefore, Green\u2019s theorem still works on a region with holes.<\/p>\r\n<p class=\"whitespace-normal break-words\">To see how this works in practice, consider annulus [latex]D[\/latex] in Figure 3 and suppose that [latex]\\mathbf{F}=\\langle P,Q\\rangle[\/latex] is a vector field defined on this annulus. Region [latex]D[\/latex] has a hole, so it is not simply connected.<\/p>\r\n<p class=\"whitespace-normal break-words\">Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure 3) so that, when we divide the region into [latex]D_1[\/latex] and [latex]D_2[\/latex], we are able to keep the region on our left as we walk along a path that traverses the boundary. Let [latex]D_1[\/latex] be the upper half of the annulus and [latex]D_2[\/latex] be the lower half. Neither of these regions has holes, so we have divided [latex]D[\/latex] into two simply connected regions.<\/p>\r\n<p class=\"whitespace-normal break-words\">We label each piece of these new boundaries as [latex]P_i[\/latex] for some [latex]i[\/latex], as in Figure 3. If we begin at [latex]P[\/latex] and travel along the oriented boundary, the first segment is [latex]P_1[\/latex], then [latex]P_2[\/latex], [latex]P_3[\/latex], and [latex]P_4[\/latex]. Now we have traversed [latex]D_1[\/latex] and returned to [latex]P[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Next, we start at [latex]P[\/latex] again and traverse [latex]D_2[\/latex]. Since the first piece of the boundary is the same as [latex]P_4[\/latex] in [latex]D_1[\/latex], but oriented in the opposite direction, the first piece of [latex]D_2[\/latex] is [latex]-P_4[\/latex]. Next, we have [latex]P_5[\/latex], then [latex]-P_2[\/latex], and finally [latex]P_6[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1418\" align=\"aligncenter\" width=\"794\"]<img class=\"size-full wp-image-1418\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be.webp\" alt=\"A diagram of an annulus \u2013 a circular region with a hole in in like a donut. Its boundary is oriented counterclockwise. One point P on the outer boundary is labeled. It is the right endpoint of the horizontal diameter. The annulus is split horizontally down the middle into two separate regions that are each simply connected. Point P is labeled on both of these regions, D1 and D2. Each region has boundaries oriented counterclockwise. The upper curve of D1 is labeled P1, the left flat side is P2, the lower curve is P3, and the right flat side is P4. The lower curve of D2 is P6, the left flat side is \u2013P2, the upper curve is P5, and the right flat side is \u2013P4.\" width=\"794\" height=\"407\" \/> Figure 3. Breaking the annulus into two separate regions gives us two simply connected regions. The line integrals over the common boundaries cancel out.[\/caption]\r\n<p class=\"whitespace-normal break-words\">Figure 3 shows a path that traverses the boundary of [latex]D[\/latex]. Notice that this path traverses the boundary of region [latex]D_1[\/latex], returns to the starting point, and then traverses the boundary of region [latex]D_2[\/latex]. Furthermore, as we walk along the path, the region is always on our left.<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that this traversal of the [latex]P_i[\/latex] paths covers the entire boundary of region [latex]D[\/latex]. If we had only traversed one portion of the boundary of [latex]D[\/latex], then we cannot apply Green's theorem to [latex]D[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The boundary of the upper half of the annulus, therefore, is [latex]P_1\\cup P_2\\cup P_3\\cup P_4[\/latex] and the boundary of the lower half of the annulus is [latex]-P_4\\cup P_5\\cup -P_2\\cup P_6[\/latex]. Then, Green's theorem implies:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{-P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\int_{\\partial{D}_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{\\partial{D}_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&amp;=\\displaystyle\\iint_{D_1}(Q_x-P_y)dA+\\displaystyle\\iint_{D_2}(Q_x-P_y)dA \\\\\r\n&amp;=\\displaystyle\\iint_{D}(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, we arrive at the equation found in Green's theorem:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_D(Q_z-P_y)dA[\/latex].<\/p>\r\n<p id=\"fs-id1167793928375\">The same logic implies that the flux form of Green's theorem can also be extended to a region with finitely many holes:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_{C}{\\bf{F}}\\cdot{\\bf{D}}ds=\\displaystyle\\iint_D(P_x+Q_y)dA[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793925125\">Calculate integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy}[\/latex],<\/p>\r\n<p id=\"fs-id1167794054408\">where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]1\\leq{\\bf{r}}\\leq2,0\\leq \\theta\\leq2\\pi[\/latex].<\/p>\r\n[reveal-answer q=\"437584636\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437584636\"]\r\n<p id=\"fs-id1167793322337\">Although [latex]D[\/latex] is not simply connected, we can use the extended form of Green\u2019s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy&amp;=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&amp;=\\displaystyle\\iint_D(x^2+y^2)dA \\\\\r\n&amp;=\\displaystyle\\int_0^\r\n{2\\pi}\\displaystyle\\int_1^2r^3drd\\theta=\\displaystyle\\int_0^{2\\pi}\\frac{15}4d\\theta \\\\\r\n&amp;=\\frac{15\\pi}2\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle=\\langle\\frac{y}{x^2+y^2},-\\frac{y}{x^2+y^2}[\/latex] and let [latex]C[\/latex] be any simple closed curve in a plane oriented counterclockwise. What are the possible values of [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]?[reveal-answer q=\"329492936\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"329492936\"]\r\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">We use the extended form of Green\u2019s theorem to show that [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is either [latex]0[\/latex] or [latex]-2\\pi[\/latex]\u2013that is, no matter how crazy curve [latex]C[\/latex] is, the line integral of [latex]{\\bf{F}}[\/latex] along [latex]C[\/latex] can have only one of two possible values. We consider two cases: the case when [latex]C[\/latex] encompasses the origin and the case when [latex]C[\/latex] does not encompass the origin.<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n\r\n<strong>Case 1: [latex]C [\/latex] Does Not Encompass the Origin<\/strong>\r\n\r\n<section id=\"fs-id1167793886542\" data-depth=\"1\">\r\n<p id=\"fs-id1167793383039\">In this case, the region enclosed by [latex]C[\/latex] is simply connected because the only hole in the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is at the origin. We showed in our discussion of cross-partials that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial condition. If we restrict the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0just to [latex]C[\/latex] and the region it encloses, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0with this restricted domain is now defined on a simply connected domain. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property on its restricted domain, the field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative on this simply connected region and hence the circulation [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is zero.<\/p>\r\n<strong>Case 2: [latex]C[\/latex] Does Encompass the Origin<\/strong>\r\n\r\n<\/section><section id=\"fs-id1167793913142\" data-depth=\"1\">\r\n<p id=\"fs-id1167793913152\">In this case, the region enclosed by [latex]C[\/latex] is not simply connected because this region contains a hole at the origin. Let [latex]C_1[\/latex] be a circle of radius\u00a0<em data-effect=\"italics\">a<\/em>\u00a0centered at the origin so that [latex]C_1[\/latex] is entirely inside the region enclosed by [latex]C[\/latex] (Figure 4). Give [latex]C_1[\/latex] a clockwise orientation.<\/p>\r\n\r\n[caption id=\"attachment_2058\" align=\"aligncenter\" width=\"454\"]<img class=\"size-full wp-image-2058\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9.webp\" alt=\"A diagram in two dimensions. A circle C1 oriented clockwise is centered at the origin completely inside a generic curve C that is in all four quadrants. Curve C is oriented counterclockwise.\" width=\"454\" height=\"309\" \/> Figure 4. Choose circle [latex]C_1[\/latex] centered at the origin that is contained entirely inside [latex]C[\/latex].[\/caption]<\/section><section id=\"fs-id1167793913142\" data-depth=\"1\">\r\n<p id=\"fs-id1167793244246\">Let [latex]D[\/latex] be the region between [latex]C_1[\/latex] and [latex]C[\/latex], and [latex]C[\/latex] is orientated counterclockwise. By the extended version of Green\u2019s theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\iint_DQ_x-P_ydA \\\\\r\n&amp;=\\displaystyle\\iint_D-\\frac{y^2-x^2}{(x^2+y^2)^2}+\\frac{y^2-x^2}{(x^2+y^2)^2}dA \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex],<\/p>\r\n<p id=\"fs-id1167793951778\">and therefore<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=-\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793546872\">Since [latex]C_1[\/latex] is a specific curve, we can evaluate [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. Let<\/p>\r\n<p style=\"text-align: center;\">[latex]x=a\\cos{t},y=a\\sin{t},0\\leq{t}\\leq2\\pi[\/latex]<\/p>\r\n<p id=\"fs-id1167793470618\">be a parameterization of [latex]C_1[\/latex]. 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^{2\\pi}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}\\left\\langle-\\frac{\\sin{(t)}}a,-\\frac{\\cos{(t)}}1\\right\\rangle\\cdot\\langle-a\\sin{(t)},-\\cos{(t)}\\rangle dt \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}\\sin^2(t)+\\cos^2(t)dt=\\displaystyle\\int_0^{2\\pi}dt=2\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793725941\" style=\"text-align: center;\">Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}s=-2\\pi[\/latex].<\/p>\r\n\r\n<\/section><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2 data-type=\"title\">Green\u2019s Theorem on General Regions<\/h2>\n<p>Green&#8217;s theorem, as stated, applies only to regions that are simply connected\u2014that is, Green&#8217;s theorem cannot handle regions with holes. Here, we extend Green&#8217;s theorem so that it does work on regions with finitely many holes (Figure 1).<\/p>\n<figure id=\"attachment_1416\" aria-describedby=\"caption-attachment-1416\" style=\"width: 273px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1416\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145733\/7407ed159946b7804a92fac2aaf626ae545a69db.webp\" alt=\"A nonsimply connected, oval-shaped region with three circular holes.\" width=\"273\" height=\"148\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145733\/7407ed159946b7804a92fac2aaf626ae545a69db.webp 273w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145733\/7407ed159946b7804a92fac2aaf626ae545a69db-65x35.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145733\/7407ed159946b7804a92fac2aaf626ae545a69db-225x122.webp 225w\" sizes=\"auto, (max-width: 273px) 100vw, 273px\" \/><figcaption id=\"caption-attachment-1416\" class=\"wp-caption-text\">Figure 1. Green\u2019s theorem, as stated, does not apply to a nonsimply connected region with three holes like this one.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">Before discussing extensions of Green&#8217;s theorem, we need to go over some terminology regarding the boundary of a region. Let [latex]D[\/latex] be a region and let [latex]C[\/latex] be a component of the boundary of [latex]D[\/latex].<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">We say that [latex]C[\/latex] is <strong>positively oriented<\/strong> if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our left. The counterclockwise orientation of the boundary of a disk is a positive orientation, for example.<\/li>\n<li class=\"whitespace-normal break-words\">Curve [latex]C[\/latex] is <strong>negatively oriented<\/strong> if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Let [latex]D[\/latex] be a region with finitely many holes (so that [latex]D[\/latex] has finitely many boundary curves), and denote the boundary of [latex]D[\/latex] by [latex]\\partial D[\/latex] (Figure 2). To extend Green&#8217;s theorem so it can handle [latex]D[\/latex], we divide region [latex]D[\/latex] into two regions, [latex]D_1[\/latex] and [latex]D_2[\/latex] (with respective boundaries [latex]\\partial D_1[\/latex] and [latex]\\partial D_2[\/latex]), in such a way that [latex]D=D_1\\cup D_2[\/latex] and neither [latex]D_1[\/latex] nor [latex]D_2[\/latex] has any holes (Figure 2).<\/p>\n<figure id=\"attachment_1417\" aria-describedby=\"caption-attachment-1417\" style=\"width: 652px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1417\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc.webp\" alt=\"Two regions. The first region D is oval-shaped with three circular holes in it. Its oriented boundary is counterclockwise. The second region is region D split horizontally down the middle into two simply connected regions with no holes. It still has a boundary oriented counterclockwise.\" width=\"652\" height=\"302\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc.webp 652w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc-300x139.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc-65x30.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc-225x104.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145753\/bc0655a25bdd4acc3350193625c05d6479da73cc-350x162.webp 350w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><figcaption id=\"caption-attachment-1417\" class=\"wp-caption-text\">Figure 2. (a) Region [latex]D[\/latex] with an oriented boundary has three holes. (b) Region [latex]D[\/latex] split into two simply connected regions has no holes.<\/figcaption><\/figure>\n<p>Assume the boundary of [latex]D[\/latex] is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region [latex]D_1[\/latex] and [latex]D_2[\/latex] is positively oriented. If [latex]\\mathbf{F}[\/latex] is a vector field defined on [latex]D[\/latex], then Green&#8217;s theorem says that:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\oint_{\\partial{D_1}}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\oint_{\\partial{D_2}}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\displaystyle\\iint_{D_1}Q_x-P_ydA+\\displaystyle\\iint_{D_2}Q_x-P_ydA \\\\  &=\\displaystyle\\iint_{D}(Q_x-P_y)dA  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167794206293\">Therefore, Green\u2019s theorem still works on a region with holes.<\/p>\n<p class=\"whitespace-normal break-words\">To see how this works in practice, consider annulus [latex]D[\/latex] in Figure 3 and suppose that [latex]\\mathbf{F}=\\langle P,Q\\rangle[\/latex] is a vector field defined on this annulus. Region [latex]D[\/latex] has a hole, so it is not simply connected.<\/p>\n<p class=\"whitespace-normal break-words\">Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure 3) so that, when we divide the region into [latex]D_1[\/latex] and [latex]D_2[\/latex], we are able to keep the region on our left as we walk along a path that traverses the boundary. Let [latex]D_1[\/latex] be the upper half of the annulus and [latex]D_2[\/latex] be the lower half. Neither of these regions has holes, so we have divided [latex]D[\/latex] into two simply connected regions.<\/p>\n<p class=\"whitespace-normal break-words\">We label each piece of these new boundaries as [latex]P_i[\/latex] for some [latex]i[\/latex], as in Figure 3. If we begin at [latex]P[\/latex] and travel along the oriented boundary, the first segment is [latex]P_1[\/latex], then [latex]P_2[\/latex], [latex]P_3[\/latex], and [latex]P_4[\/latex]. Now we have traversed [latex]D_1[\/latex] and returned to [latex]P[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Next, we start at [latex]P[\/latex] again and traverse [latex]D_2[\/latex]. Since the first piece of the boundary is the same as [latex]P_4[\/latex] in [latex]D_1[\/latex], but oriented in the opposite direction, the first piece of [latex]D_2[\/latex] is [latex]-P_4[\/latex]. Next, we have [latex]P_5[\/latex], then [latex]-P_2[\/latex], and finally [latex]P_6[\/latex].<\/p>\n<figure id=\"attachment_1418\" aria-describedby=\"caption-attachment-1418\" style=\"width: 794px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1418\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be.webp\" alt=\"A diagram of an annulus \u2013 a circular region with a hole in in like a donut. Its boundary is oriented counterclockwise. One point P on the outer boundary is labeled. It is the right endpoint of the horizontal diameter. The annulus is split horizontally down the middle into two separate regions that are each simply connected. Point P is labeled on both of these regions, D1 and D2. Each region has boundaries oriented counterclockwise. The upper curve of D1 is labeled P1, the left flat side is P2, the lower curve is P3, and the right flat side is P4. The lower curve of D2 is P6, the left flat side is \u2013P2, the upper curve is P5, and the right flat side is \u2013P4.\" width=\"794\" height=\"407\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be.webp 794w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be-300x154.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be-768x394.webp 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be-65x33.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be-225x115.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17145817\/dc0143617fb2ea61001615439154fd3c0f28e0be-350x179.webp 350w\" sizes=\"auto, (max-width: 794px) 100vw, 794px\" \/><figcaption id=\"caption-attachment-1418\" class=\"wp-caption-text\">Figure 3. Breaking the annulus into two separate regions gives us two simply connected regions. The line integrals over the common boundaries cancel out.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">Figure 3 shows a path that traverses the boundary of [latex]D[\/latex]. Notice that this path traverses the boundary of region [latex]D_1[\/latex], returns to the starting point, and then traverses the boundary of region [latex]D_2[\/latex]. Furthermore, as we walk along the path, the region is always on our left.<\/p>\n<p class=\"whitespace-normal break-words\">Notice that this traversal of the [latex]P_i[\/latex] paths covers the entire boundary of region [latex]D[\/latex]. If we had only traversed one portion of the boundary of [latex]D[\/latex], then we cannot apply Green&#8217;s theorem to [latex]D[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The boundary of the upper half of the annulus, therefore, is [latex]P_1\\cup P_2\\cup P_3\\cup P_4[\/latex] and the boundary of the lower half of the annulus is [latex]-P_4\\cup P_5\\cup -P_2\\cup P_6[\/latex]. Then, Green&#8217;s theorem implies:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{-P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\displaystyle\\int_{\\partial{D}_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{\\partial{D}_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\  &=\\displaystyle\\iint_{D_1}(Q_x-P_y)dA+\\displaystyle\\iint_{D_2}(Q_x-P_y)dA \\\\  &=\\displaystyle\\iint_{D}(Q_x-P_y)dA  \\end{aligned}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, we arrive at the equation found in Green&#8217;s theorem:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_D(Q_z-P_y)dA[\/latex].<\/p>\n<p id=\"fs-id1167793928375\">The same logic implies that the flux form of Green&#8217;s theorem can also be extended to a region with finitely many holes:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\oint_{C}{\\bf{F}}\\cdot{\\bf{D}}ds=\\displaystyle\\iint_D(P_x+Q_y)dA[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793925125\">Calculate integral<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy}[\/latex],<\/p>\n<p id=\"fs-id1167794054408\">where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]1\\leq{\\bf{r}}\\leq2,0\\leq \\theta\\leq2\\pi[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q437584636\">Show Solution<\/button><\/p>\n<div id=\"q437584636\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793322337\">Although [latex]D[\/latex] is not simply connected, we can use the extended form of Green\u2019s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\  &=\\displaystyle\\iint_D(x^2+y^2)dA \\\\  &=\\displaystyle\\int_0^  {2\\pi}\\displaystyle\\int_1^2r^3drd\\theta=\\displaystyle\\int_0^{2\\pi}\\frac{15}4d\\theta \\\\  &=\\frac{15\\pi}2  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle=\\langle\\frac{y}{x^2+y^2},-\\frac{y}{x^2+y^2}[\/latex] and let [latex]C[\/latex] be any simple closed curve in a plane oriented counterclockwise. What are the possible values of [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q329492936\">Show Solution<\/button><\/p>\n<div id=\"q329492936\" class=\"hidden-answer\" style=\"display: none\">\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">We use the extended form of Green\u2019s theorem to show that [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is either [latex]0[\/latex] or [latex]-2\\pi[\/latex]\u2013that is, no matter how crazy curve [latex]C[\/latex] is, the line integral of [latex]{\\bf{F}}[\/latex] along [latex]C[\/latex] can have only one of two possible values. We consider two cases: the case when [latex]C[\/latex] encompasses the origin and the case when [latex]C[\/latex] does not encompass the origin.<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p><strong>Case 1: [latex]C[\/latex] Does Not Encompass the Origin<\/strong><\/p>\n<section id=\"fs-id1167793886542\" data-depth=\"1\">\n<p id=\"fs-id1167793383039\">In this case, the region enclosed by [latex]C[\/latex] is simply connected because the only hole in the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is at the origin. We showed in our discussion of cross-partials that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial condition. If we restrict the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0just to [latex]C[\/latex] and the region it encloses, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0with this restricted domain is now defined on a simply connected domain. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property on its restricted domain, the field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative on this simply connected region and hence the circulation [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is zero.<\/p>\n<p><strong>Case 2: [latex]C[\/latex] Does Encompass the Origin<\/strong><\/p>\n<\/section>\n<section id=\"fs-id1167793913142\" data-depth=\"1\">\n<p id=\"fs-id1167793913152\">In this case, the region enclosed by [latex]C[\/latex] is not simply connected because this region contains a hole at the origin. Let [latex]C_1[\/latex] be a circle of radius\u00a0<em data-effect=\"italics\">a<\/em>\u00a0centered at the origin so that [latex]C_1[\/latex] is entirely inside the region enclosed by [latex]C[\/latex] (Figure 4). Give [latex]C_1[\/latex] a clockwise orientation.<\/p>\n<figure id=\"attachment_2058\" aria-describedby=\"caption-attachment-2058\" style=\"width: 454px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-2058\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9.webp\" alt=\"A diagram in two dimensions. A circle C1 oriented clockwise is centered at the origin completely inside a generic curve C that is in all four quadrants. Curve C is oriented counterclockwise.\" width=\"454\" height=\"309\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9.webp 454w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9-300x204.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9-65x44.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9-225x153.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/11\/06143702\/bff0e1965585abf3138aa8fef022ae3971f5dfe9-350x238.webp 350w\" sizes=\"auto, (max-width: 454px) 100vw, 454px\" \/><figcaption id=\"caption-attachment-2058\" class=\"wp-caption-text\">Figure 4. Choose circle [latex]C_1[\/latex] centered at the origin that is contained entirely inside [latex]C[\/latex].<\/figcaption><\/figure>\n<\/section>\n<section data-depth=\"1\">\n<p id=\"fs-id1167793244246\">Let [latex]D[\/latex] be the region between [latex]C_1[\/latex] and [latex]C[\/latex], and [latex]C[\/latex] is orientated counterclockwise. By the extended version of Green\u2019s theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\iint_DQ_x-P_ydA \\\\  &=\\displaystyle\\iint_D-\\frac{y^2-x^2}{(x^2+y^2)^2}+\\frac{y^2-x^2}{(x^2+y^2)^2}dA \\\\  &=0  \\end{aligned}[\/latex],<\/p>\n<p id=\"fs-id1167793951778\">and therefore<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=-\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\n<p id=\"fs-id1167793546872\">Since [latex]C_1[\/latex] is a specific curve, we can evaluate [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. Let<\/p>\n<p style=\"text-align: center;\">[latex]x=a\\cos{t},y=a\\sin{t},0\\leq{t}\\leq2\\pi[\/latex]<\/p>\n<p id=\"fs-id1167793470618\">be a parameterization of [latex]C_1[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^{2\\pi}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)dt \\\\  &=\\displaystyle\\int_0^{2\\pi}\\left\\langle-\\frac{\\sin{(t)}}a,-\\frac{\\cos{(t)}}1\\right\\rangle\\cdot\\langle-a\\sin{(t)},-\\cos{(t)}\\rangle dt \\\\  &=\\displaystyle\\int_0^{2\\pi}\\sin^2(t)+\\cos^2(t)dt=\\displaystyle\\int_0^{2\\pi}dt=2\\pi  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793725941\" style=\"text-align: center;\">Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}s=-2\\pi[\/latex].<\/p>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":25,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2042","chapter","type-chapter","status-publish","hentry"],"part":436,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2042","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2042\/revisions"}],"predecessor-version":[{"id":2060,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2042\/revisions\/2060"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/436"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/2042\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=2042"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=2042"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=2042"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=2042"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}