{"id":760,"date":"2025-10-09T15:37:05","date_gmt":"2025-10-09T15:37:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=760"},"modified":"2025-11-13T20:18:40","modified_gmt":"2025-11-13T20:18:40","slug":"greens-theorem-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/greens-theorem-fresh-take\/","title":{"raw":"Green\u2019s Theorem: Fresh Take","rendered":"Green\u2019s Theorem: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Apply the circulation form of Green's theorem<\/li>\r\n \t<li>Apply the flux form of Green's theorem<\/li>\r\n \t<li>Calculate circulation and flux on general regions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Circulation Form of Green\u2019s Theorem<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Green's Theorem is a powerful bridge between line integrals and double integrals. It transforms the circulation of a vector field around a closed curve into a double integral over the region inside that curve\u2014often making difficult calculations much simpler.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a vector field [latex]\\mathbf{F} = \\langle P, Q \\rangle[\/latex] and a simple closed curve [latex]C[\/latex] (oriented counterclockwise) enclosing region [latex]D[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_C \\mathbf{F} \\cdot d\\mathbf{r} = \\oint_C P\\,dx + Q\\,dy = \\iint_D (Q_x - P_y)\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The left side is a line integral around the boundary. The right side is a double integral over the interior. The expression [latex]Q_x - P_y[\/latex] measures how the vector field \"curls\" in the region.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>When to Use Green's Theorem<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The theorem requires:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">A <strong>two-dimensional<\/strong> vector field (not three-dimensional)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">An <strong>open, simply connected<\/strong> region [latex]D[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">A <strong>simple closed curve<\/strong> [latex]C[\/latex] that's piecewise smooth and oriented counterclockwise<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Component functions with <strong>continuous partial derivatives<\/strong> on [latex]D[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Strategic Thinking<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Green's Theorem is a tool for choosing the easier calculation. Sometimes the line integral is simpler to compute directly. Other times\u2014especially when [latex]C[\/latex] has multiple pieces or the integrand is complex\u2014converting to a double integral saves significant work. The theorem can also work in reverse: you can use a line integral to find area by choosing [latex]\\mathbf{F}[\/latex] strategically so that [latex]Q_x - P_y = 1[\/latex].<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why It Works for Conservative Fields<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]\\mathbf{F}[\/latex] is conservative, then [latex]P_y = Q_x[\/latex], making [latex]Q_x - P_y = 0[\/latex]. Both sides of Green's Theorem equal zero, confirming that circulation around a closed curve in a conservative field is always zero.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">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].[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]<\/section>\r\n<h2>Flux Form of Green\u2019s Theorem<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">While the circulation form of Green's Theorem relates circulation around a curve to a double integral, the flux form relates flux across a curve to a double integral. This version is particularly useful when you need to calculate how much of a vector field flows across a boundary\u2014often easier to compute as a double integral than as a flux line integral.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a vector field [latex]\\mathbf{F} = \\langle P, Q \\rangle[\/latex] and a simple closed curve [latex]C[\/latex] (oriented counterclockwise) with outward normal vector [latex]\\mathbf{N}[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_C \\mathbf{F} \\cdot \\mathbf{N}\\,ds = \\iint_D (P_x + Q_y)\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The left side measures flux across the boundary. The right side is a double integral over region [latex]D[\/latex]. The expression [latex]P_x + Q_y[\/latex] is called the <strong>divergence<\/strong> of [latex]\\mathbf{F}[\/latex]\u2014it measures how much the field \"spreads out\" from each point.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Source-Free Vector Fields<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A vector field is <strong>source-free<\/strong> if it has zero flux across any closed curve. On a simply connected domain, [latex]\\mathbf{F}[\/latex] is source-free if and only if [latex]P_x + Q_y = 0[\/latex]. Source-free fields have <strong>stream functions<\/strong> [latex]g[\/latex] where [latex]P = g_y[\/latex] and [latex]Q = -g_x[\/latex]. Stream functions work like potential functions: they let you understand the field through a single scalar function.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Conservative AND Source-Free Fields<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When a vector field is both conservative and source-free, any potential function [latex]f[\/latex] satisfies <strong>Laplace's equation<\/strong>: [latex]f_{xx} + f_{yy} = 0[\/latex]. Such functions are called <strong>harmonic<\/strong> and appear throughout physics\u2014modeling gravitational potentials, electromagnetic fields, and fluid flow.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>When to Use This Form<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Use the flux form when calculating how much fluid crosses a boundary, when working with problems involving divergence, or when the flux integral looks complicated but [latex]P_x + Q_y[\/latex] is simple.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the flux of [latex]{\\bf{F}}(x,y)=\\langle{x}^3,y^3\\rangle[\/latex]\u00a0 across a unit circle oriented counterclockwise.[reveal-answer q=\"457134886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457134886\"][latex]\\frac{3\\pi}2[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250321&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=6YIilpzQIhI&amp;video_target=tpm-plugin-zx9nr5sz-6YIilpzQIhI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.36_transcript.html\">transcript for \u201cCP 6.36\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Find a stream function for vector field [latex]{\\bf{F}}(x,y)=\\langle{x}\\sin{y},\\cos{y}\\rangle[\/latex].[reveal-answer q=\"938299357\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"938299357\"][latex]g(x,y)=-x\\cos{y}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Is the function [latex]f(x,y)=e^{x+5y}[\/latex] harmonic?[reveal-answer q=\"394592875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394592875\"]No.[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Green\u2019s Theorem on General Regions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">As originally stated, Green's Theorem only works on simply connected regions\u2014regions without holes. But many real-world problems involve regions with holes (like annuli or domains with obstacles). The good news: Green's Theorem can be extended to handle regions with finitely many holes.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding Orientation<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When a region has holes, proper orientation matters:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Positively oriented<\/strong>: As you walk along the boundary in the direction of orientation, the region is on your left (counterclockwise for outer boundaries)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Negatively oriented<\/strong>: The region is on your right (clockwise for inner holes)<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For regions with holes, the outer boundary is oriented counterclockwise and each inner boundary (hole) is oriented clockwise.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Key Strategy<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">To apply Green's Theorem to a region [latex]D[\/latex] with holes, divide [latex]D[\/latex] into simply connected pieces [latex]D_1, D_2, \\ldots[\/latex] that have no holes. When you apply Green's Theorem to each piece and add the results, something important happens: the line integrals over the shared internal boundaries cancel out (they're traversed in opposite directions). What remains is just the integral over the actual boundary of [latex]D[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_{\\partial D} \\mathbf{F} \\cdot d\\mathbf{r} = \\iint_D (Q_x - P_y)\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The division creates internal boundaries that get traversed twice\u2014once in each direction\u2014so their contributions cancel. Only the true outer and inner boundaries of [latex]D[\/latex] remain. This means Green's Theorem extends naturally to regions with holes without changing the formula.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Impact<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This extension is crucial for fields with singularities (like [latex]\\mathbf{F} = \\left\\langle \\frac{y}{x^2+y^2}, \\frac{-x}{x^2+y^2} \\right\\rangle[\/latex] near the origin). Whether the line integral is zero or nonzero depends entirely on whether the curve encloses the singularity\u2014not on the curve's specific shape.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate integral [latex]\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]2\\leq{r}\\leq5,0\\leq\\theta\\leq2\\pi[\/latex], and [latex]{\\bf{F}}(x,y)=\\langle{x}^3,5x+e^y\\sin{y}\\rangle[\/latex].[reveal-answer q=\"743192057\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"743192057\"][latex]105\\pi[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250322&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pT4YCxdUh7s&amp;video_target=tpm-plugin-13jbef4f-pT4YCxdUh7s\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.39_transcript.html\">transcript for \u201cCP 6.39\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Apply the circulation form of Green&#8217;s theorem<\/li>\n<li>Apply the flux form of Green&#8217;s theorem<\/li>\n<li>Calculate circulation and flux on general regions<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Circulation Form of Green\u2019s Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Green&#8217;s Theorem is a powerful bridge between line integrals and double integrals. It transforms the circulation of a vector field around a closed curve into a double integral over the region inside that curve\u2014often making difficult calculations much simpler.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a vector field [latex]\\mathbf{F} = \\langle P, Q \\rangle[\/latex] and a simple closed curve [latex]C[\/latex] (oriented counterclockwise) enclosing region [latex]D[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_C \\mathbf{F} \\cdot d\\mathbf{r} = \\oint_C P\\,dx + Q\\,dy = \\iint_D (Q_x - P_y)\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The left side is a line integral around the boundary. The right side is a double integral over the interior. The expression [latex]Q_x - P_y[\/latex] measures how the vector field &#8220;curls&#8221; in the region.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>When to Use Green&#8217;s Theorem<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The theorem requires:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">A <strong>two-dimensional<\/strong> vector field (not three-dimensional)<\/li>\n<li class=\"whitespace-normal break-words\">An <strong>open, simply connected<\/strong> region [latex]D[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">A <strong>simple closed curve<\/strong> [latex]C[\/latex] that&#8217;s piecewise smooth and oriented counterclockwise<\/li>\n<li class=\"whitespace-normal break-words\">Component functions with <strong>continuous partial derivatives<\/strong> on [latex]D[\/latex]<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Strategic Thinking<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Green&#8217;s Theorem is a tool for choosing the easier calculation. Sometimes the line integral is simpler to compute directly. Other times\u2014especially when [latex]C[\/latex] has multiple pieces or the integrand is complex\u2014converting to a double integral saves significant work. The theorem can also work in reverse: you can use a line integral to find area by choosing [latex]\\mathbf{F}[\/latex] strategically so that [latex]Q_x - P_y = 1[\/latex].<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why It Works for Conservative Fields<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]\\mathbf{F}[\/latex] is conservative, then [latex]P_y = Q_x[\/latex], making [latex]Q_x - P_y = 0[\/latex]. Both sides of Green&#8217;s Theorem equal zero, confirming that circulation around a closed curve in a conservative field is always zero.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q304250970\">Show Solution<\/button><\/p>\n<div id=\"q304250970\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{45}2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<figure id=\"attachment_5285\" aria-describedby=\"caption-attachment-5285\" style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q487592847\">Show Solution<\/button><\/p>\n<div id=\"q487592847\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac43[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2>Flux Form of Green\u2019s Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">While the circulation form of Green&#8217;s Theorem relates circulation around a curve to a double integral, the flux form relates flux across a curve to a double integral. This version is particularly useful when you need to calculate how much of a vector field flows across a boundary\u2014often easier to compute as a double integral than as a flux line integral.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Formula<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a vector field [latex]\\mathbf{F} = \\langle P, Q \\rangle[\/latex] and a simple closed curve [latex]C[\/latex] (oriented counterclockwise) with outward normal vector [latex]\\mathbf{N}[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_C \\mathbf{F} \\cdot \\mathbf{N}\\,ds = \\iint_D (P_x + Q_y)\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The left side measures flux across the boundary. The right side is a double integral over region [latex]D[\/latex]. The expression [latex]P_x + Q_y[\/latex] is called the <strong>divergence<\/strong> of [latex]\\mathbf{F}[\/latex]\u2014it measures how much the field &#8220;spreads out&#8221; from each point.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Source-Free Vector Fields<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A vector field is <strong>source-free<\/strong> if it has zero flux across any closed curve. On a simply connected domain, [latex]\\mathbf{F}[\/latex] is source-free if and only if [latex]P_x + Q_y = 0[\/latex]. Source-free fields have <strong>stream functions<\/strong> [latex]g[\/latex] where [latex]P = g_y[\/latex] and [latex]Q = -g_x[\/latex]. Stream functions work like potential functions: they let you understand the field through a single scalar function.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Conservative AND Source-Free Fields<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When a vector field is both conservative and source-free, any potential function [latex]f[\/latex] satisfies <strong>Laplace&#8217;s equation<\/strong>: [latex]f_{xx} + f_{yy} = 0[\/latex]. Such functions are called <strong>harmonic<\/strong> and appear throughout physics\u2014modeling gravitational potentials, electromagnetic fields, and fluid flow.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>When to Use This Form<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Use the flux form when calculating how much fluid crosses a boundary, when working with problems involving divergence, or when the flux integral looks complicated but [latex]P_x + Q_y[\/latex] is simple.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the flux of [latex]{\\bf{F}}(x,y)=\\langle{x}^3,y^3\\rangle[\/latex]\u00a0 across a unit circle oriented counterclockwise.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q457134886\">Show Solution<\/button><\/p>\n<div id=\"q457134886\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{3\\pi}2[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250321&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=6YIilpzQIhI&amp;video_target=tpm-plugin-zx9nr5sz-6YIilpzQIhI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP6.36_transcript.html\">transcript for \u201cCP 6.36\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find a stream function for vector field [latex]{\\bf{F}}(x,y)=\\langle{x}\\sin{y},\\cos{y}\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q938299357\">Show Solution<\/button><\/p>\n<div id=\"q938299357\" class=\"hidden-answer\" style=\"display: none\">[latex]g(x,y)=-x\\cos{y}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Is the function [latex]f(x,y)=e^{x+5y}[\/latex] harmonic?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q394592875\">Show Solution<\/button><\/p>\n<div id=\"q394592875\" class=\"hidden-answer\" style=\"display: none\">No.<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Green\u2019s Theorem on General Regions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">As originally stated, Green&#8217;s Theorem only works on simply connected regions\u2014regions without holes. But many real-world problems involve regions with holes (like annuli or domains with obstacles). The good news: Green&#8217;s Theorem can be extended to handle regions with finitely many holes.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding Orientation<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When a region has holes, proper orientation matters:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Positively oriented<\/strong>: As you walk along the boundary in the direction of orientation, the region is on your left (counterclockwise for outer boundaries)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Negatively oriented<\/strong>: The region is on your right (clockwise for inner holes)<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For regions with holes, the outer boundary is oriented counterclockwise and each inner boundary (hole) is oriented clockwise.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Key Strategy<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">To apply Green&#8217;s Theorem to a region [latex]D[\/latex] with holes, divide [latex]D[\/latex] into simply connected pieces [latex]D_1, D_2, \\ldots[\/latex] that have no holes. When you apply Green&#8217;s Theorem to each piece and add the results, something important happens: the line integrals over the shared internal boundaries cancel out (they&#8217;re traversed in opposite directions). What remains is just the integral over the actual boundary of [latex]D[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\oint_{\\partial D} \\mathbf{F} \\cdot d\\mathbf{r} = \\iint_D (Q_x - P_y)\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The division creates internal boundaries that get traversed twice\u2014once in each direction\u2014so their contributions cancel. Only the true outer and inner boundaries of [latex]D[\/latex] remain. This means Green&#8217;s Theorem extends naturally to regions with holes without changing the formula.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Impact<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This extension is crucial for fields with singularities (like [latex]\\mathbf{F} = \\left\\langle \\frac{y}{x^2+y^2}, \\frac{-x}{x^2+y^2} \\right\\rangle[\/latex] near the origin). Whether the line integral is zero or nonzero depends entirely on whether the curve encloses the singularity\u2014not on the curve&#8217;s specific shape.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate integral [latex]\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]2\\leq{r}\\leq5,0\\leq\\theta\\leq2\\pi[\/latex], and [latex]{\\bf{F}}(x,y)=\\langle{x}^3,5x+e^y\\sin{y}\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q743192057\">Show Solution<\/button><\/p>\n<div id=\"q743192057\" class=\"hidden-answer\" style=\"display: none\">[latex]105\\pi[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250322&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pT4YCxdUh7s&amp;video_target=tpm-plugin-13jbef4f-pT4YCxdUh7s\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP6.39_transcript.html\">transcript for \u201cCP 6.39\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":27,"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-760","chapter","type-chapter","status-publish","hentry"],"part":436,"module-header":"fresh_take","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\/760","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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/760\/revisions"}],"predecessor-version":[{"id":2236,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/760\/revisions\/2236"}],"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\/760\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=760"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=760"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=760"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=760"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}