{"id":5500,"date":"2022-06-02T18:54:31","date_gmt":"2022-06-02T18:54:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5500"},"modified":"2022-11-01T05:25:30","modified_gmt":"2022-11-01T05:25:30","slug":"curl","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/curl\/","title":{"raw":"Curl","rendered":"Curl"},"content":{"raw":"<div data-type=\"note\">\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\">Determine curl from the formula for a given vector field.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167794332441\">The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of a fluid. Then, the curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at point [latex]P[\/latex] is a vector that measures the tendency of particles near [latex]P[\/latex] to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at [latex]P[\/latex] measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field\u2019s \u201cspin\u201d at that point. Visually, imagine placing a paddlewheel into a fluid at [latex]P[\/latex], with the axis of the paddlewheel aligned with the curl vector (Figure 1). The curl measures the tendency of the paddlewheel to rotate.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5330\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-5330\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31233532\/6.54.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/c85c11c2efce444d0adf8685a16f0f99c3fd3b6c&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram of a small paddlewheel in water. Arrows are drawn surrounding the center in a counterclockwise circle. At the center, the height is labeled n.&quot; id=&quot;16&quot;&gt;\" width=\"487\" height=\"411\" \/> Figure 1. To visualize curl at a point, imagine placing a small paddlewheel into the vector field at a point.[\/caption]\r\n<p id=\"fs-id1167794290442\">Consider the vector fields in\u00a0Figure 1 from the <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/divergence\/\">Divergence<\/a> learning objective. In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is <span style=\"white-space: nowrap;\">[latex]2{\\bf{k}}[\/latex]). <\/span><\/p>\r\n<p id=\"fs-id1167793931994\">To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn\u2019t rotate as it moves through the fluid.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794025989\">If [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^3[\/latex], and [latex]P_y[\/latex], [latex]P_z[\/latex],\u00a0[latex]Q_x[\/latex], [latex]Q_z[\/latex],\u00a0[latex]R_x[\/latex], and [latex]R_y[\/latex] all exist, then the\u00a0<span id=\"6f668e5d-300d-4e6c-aec3-70b59f71fe69_term267\" data-type=\"term\">curl<\/span>\u00a0of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\text{curl }{\\bf{F}}&amp;=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\\r\n&amp;=\\left(\\frac{\\partial{R}}{\\partial{y}}-\\frac{\\partial{Q}}{\\partial{z}}\\right){\\bf{i}}+\\left(\\frac{\\partial{P}}{\\partial{z}}-\\frac{\\partial{R}}{\\partial{x}}\\right){\\bf{j}}+\\left(\\frac{\\partial{Q}}{\\partial{x}}-\\frac{\\partial{P}}{\\partial{y}}\\right){\\bf{k}}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793931192\">Note that the curl of a vector field is a vector field, in contrast to divergence.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793831500\">The definition of curl can be difficult to remember. To help with remembering, we use the notation [latex]\\nabla\\times{\\bf{F}}[\/latex] to stand for a \u201cdeterminant\u201d that gives the curl formula:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{vmatrix}\r\n{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\\r\n\\frac{\\partial}{\\partial{x}}&amp;\\frac{\\partial}{\\partial{y}}&amp;\\frac{\\partial}{\\partial{z}} \\\\\r\nP&amp;Q&amp;R\r\n\\end{vmatrix}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793965705\">The determinant of this matrix is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{(R_y-Q_z){\\bf{i}}-(R_x-P_z){\\bf{j}}+(Q_x-P_y){\\bf{k}}=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}=\\text{curl }{\\bf{F}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794170488\">Thus, this matrix is a way to help remember the formula for curl. Keep in mind, though, that the word\u00a0<em data-effect=\"italics\">determinant<\/em>\u00a0is used very loosely. A determinant is not really defined on a matrix with entries that are three vectors, three operators, and three functions.<\/p>\r\n<p id=\"fs-id1167793925808\">If [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^2[\/latex], then the curl of\u00a0[latex]{\\bf{F}}[\/latex], by definition, is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=(Q_x-P_y){\\bf{k}}=\\left(\\frac{\\partial{Q}}{\\partial{x}}-\\frac{\\partial{P}}{\\partial{y}}\\right){\\bf{k}}}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167794026966\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the curl of a three-dimensional vector field<\/h3>\r\nFind the curl of [latex]{\\bf{F}}(P,Q,R)=\\langle{x}^2z,e^y+xz,xyz\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"275847941\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"275847941\"]\r\n<p id=\"fs-id1167793959778\">The curl is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\text{curl }{\\bf{F}}&amp;=\\nabla\\times{\\bf{F}} \\\\\r\n&amp;=\\begin{vmatrix}\r\n{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\\r\n\\frac{\\partial}{\\partial{x}}&amp;\\frac{\\partial}{\\partial{y}}&amp;\\frac{\\partial}{\\partial{z}} \\\\\r\nP&amp;Q&amp;R\r\n\\end{vmatrix} \\\\\r\n&amp;=(R_y-Q)z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\\r\n&amp;=(xz-z){\\bf{i}}+(x^2-yz){\\bf{j}}+z{\\bf{k}}\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the curl of [latex]{\\bf{F}}=\\langle\\sin{x}\\cos{z},\\sin{y}\\sin{z},\\cos{x}\\cos{y}\\rangle[\/latex] at point [latex]\\left(0,\\frac{\\pi}2,\\frac{\\pi}2\\right)[\/latex].\r\n\r\n[reveal-answer q=\"995421768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995421768\"]\r\n\r\n[latex]-{\\bf{i}}[\/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=8250324&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11v52S_m5wc&amp;video_target=tpm-plugin-i4jt3djg-11v52S_m5wc\" 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\/CP6.43_transcript.html\">transcript for \u201cCP 6.43\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6225[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the curl of a two-dimensional vector field<\/h3>\r\nFind the curl of [latex]{\\bf{F}}=\\langle{P},Q\\rangle=\\langle{y},0\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"569401235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569401235\"]\r\n<p id=\"fs-id1167793489271\">Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the\u00a0<em data-effect=\"italics\">x<\/em>-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=(Q_x-P_t){\\bf{k}}=-{\\bf{k}}\\ne0}[\/latex].<\/p>\r\n<p id=\"fs-id1167793640854\">Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (Figure 2). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.<\/p>\r\n\r\n[caption id=\"attachment_5331\" align=\"aligncenter\" width=\"492\"]<img class=\"size-full wp-image-5331\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31233628\/6.55.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0e80978e3662ede11c0d539b929a8dd5be857b40&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;Two vector field diagrams consisting of vectors that are all parallel. The closer they are to the x axis, the shorter the arrows are. Above the x axis, the arrows point to the right, and below the x axis, the arrows point to the left.&quot; id=&quot;21&quot;&gt;\" width=\"492\" height=\"497\" \/> Figure 2. Vector field [latex]{\\bf{F}}(x,y)=\\langle{y},0\\rangle[\/latex] consists of vectors that are all parallel.[\/caption][\/hidden-answer]<\/div>\r\n<p id=\"fs-id1167793245268\">Note that if [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field in a plane, then [latex]\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}=(Q_x-P_y){\\bf{k}}\\cdot{\\bf{k}}=Q_x-P_y[\/latex]. Therefore, the circulation form of Green\u2019s theorem is sometimes written as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_D\\text{curl }{\\bf{}}\\cdot{\\bf{k}}dA}[\/latex],<\/p>\r\n<p id=\"fs-id1167793420776\">where [latex]C[\/latex] is a simple closed curve and [latex]D[\/latex] is the region enclosed by [latex]C[\/latex]. Therefore, the circulation form of Green\u2019s theorem can be written in terms of the curl. If we think of curl as a derivative of sorts, then Green\u2019s theorem says that the \u201cderivative\u201d of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on a region can be translated into a line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function [latex]f[\/latex] on line segment [latex][a,b][\/latex] can be translated into a statement about [latex]f[\/latex] on the boundary of [latex][a,b][\/latex]. Using curl, we can see the circulation form of Green\u2019s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.<\/p>\r\n<p id=\"fs-id1167793426386\">We can now use what we have learned about curl to show that gravitational fields have no \u201cspin.\u201d Suppose there is an object at the origin with mass [latex]m_1[\/latex] at the origin and an object with mass [latex]m_2[\/latex]. Recall that the gravitational force that object 1 exerts on object 2 is given by field<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y,z)=-Gm_1m_2\\left\\langle\\frac{x}{(x^2+y^2+z^2)^{3\/2}},\\frac{y}{(x^2+y^2+z^2)^{3\/2}},\\frac{z}{(x^2+y^2+z^2)^{3\/2}}\\right\\rangle}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167793505725\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: determining the spin of a gravitational field<\/h3>\r\nShow that a gravitational field has no spin.\r\n\r\n[reveal-answer q=\"864371936\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"864371936\"]\r\n<p id=\"fs-id1167793355047\">To show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has no spin, we calculate its curl. Let [latex]P(x,y,z)=\\frac{x}{(x^2+y^2+z^2)^{3\/2}}[\/latex], [latex]Q(x,y,z)=\\frac{y}{(x^2+y^2+z^2)^{3\/2}}[\/latex], and [latex]R(x,y,z)=\\frac{z}{(x^2+y^2+z^2)^{3\/2}}[\/latex]. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\text{curl }{\\bf{F}}&amp;=-Gm_1m_2[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\\r\n&amp;=-Gm_1m_2\r\n\\left[\\begin{aligned}\r\n&amp;\\left(\\frac{-3yz}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3yz}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{i}} \\\\\r\n&amp;+\\left(\\frac{-3xz}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3xz}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{j}} \\\\\r\n&amp;+\\left(\\frac{-3xy}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3xy}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{k}}\r\n\\end{aligned}\\right]\\\\\r\n&amp;=0\\\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793319476\">Since the curl of the gravitational field is zero, the field has no spin.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nField [latex]{\\bf{v}}(x,y)=\\left\\langle-\\frac{y}{x^2+y^2},\\frac{x}{x^2+y^2}\\right\\rangle[\/latex] models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.\r\n\r\n[reveal-answer q=\"935016173\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"935016173\"]\r\n\r\ncurl\u00a0[latex]{\\bf{v}}=0[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div data-type=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Determine curl from the formula for a given vector field.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167794332441\">The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of a fluid. Then, the curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at point [latex]P[\/latex] is a vector that measures the tendency of particles near [latex]P[\/latex] to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at [latex]P[\/latex] measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field\u2019s \u201cspin\u201d at that point. Visually, imagine placing a paddlewheel into a fluid at [latex]P[\/latex], with the axis of the paddlewheel aligned with the curl vector (Figure 1). The curl measures the tendency of the paddlewheel to rotate.<\/p>\n<div id=\"attachment_5330\" style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5330\" class=\"size-full wp-image-5330\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31233532\/6.54.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/c85c11c2efce444d0adf8685a16f0f99c3fd3b6c&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;A diagram of a small paddlewheel in water. Arrows are drawn surrounding the center in a counterclockwise circle. At the center, the height is labeled n.&quot; id=&quot;16&quot;&gt;\" width=\"487\" height=\"411\" \/><\/p>\n<p id=\"caption-attachment-5330\" class=\"wp-caption-text\">Figure 1. To visualize curl at a point, imagine placing a small paddlewheel into the vector field at a point.<\/p>\n<\/div>\n<p id=\"fs-id1167794290442\">Consider the vector fields in\u00a0Figure 1 from the <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/divergence\/\">Divergence<\/a> learning objective. In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is <span style=\"white-space: nowrap;\">[latex]2{\\bf{k}}[\/latex]). <\/span><\/p>\n<p id=\"fs-id1167793931994\">To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn\u2019t rotate as it moves through the fluid.<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167794025989\">If [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^3[\/latex], and [latex]P_y[\/latex], [latex]P_z[\/latex],\u00a0[latex]Q_x[\/latex], [latex]Q_z[\/latex],\u00a0[latex]R_x[\/latex], and [latex]R_y[\/latex] all exist, then the\u00a0<span id=\"6f668e5d-300d-4e6c-aec3-70b59f71fe69_term267\" data-type=\"term\">curl<\/span>\u00a0of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is defined by<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\text{curl }{\\bf{F}}&=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\  &=\\left(\\frac{\\partial{R}}{\\partial{y}}-\\frac{\\partial{Q}}{\\partial{z}}\\right){\\bf{i}}+\\left(\\frac{\\partial{P}}{\\partial{z}}-\\frac{\\partial{R}}{\\partial{x}}\\right){\\bf{j}}+\\left(\\frac{\\partial{Q}}{\\partial{x}}-\\frac{\\partial{P}}{\\partial{y}}\\right){\\bf{k}}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793931192\">Note that the curl of a vector field is a vector field, in contrast to divergence.<\/p>\n<\/div>\n<p id=\"fs-id1167793831500\">The definition of curl can be difficult to remember. To help with remembering, we use the notation [latex]\\nabla\\times{\\bf{F}}[\/latex] to stand for a \u201cdeterminant\u201d that gives the curl formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{vmatrix}  {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\  \\frac{\\partial}{\\partial{x}}&\\frac{\\partial}{\\partial{y}}&\\frac{\\partial}{\\partial{z}} \\\\  P&Q&R  \\end{vmatrix}}[\/latex].<\/p>\n<p id=\"fs-id1167793965705\">The determinant of this matrix is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{(R_y-Q_z){\\bf{i}}-(R_x-P_z){\\bf{j}}+(Q_x-P_y){\\bf{k}}=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}=\\text{curl }{\\bf{F}}}[\/latex].<\/p>\n<p id=\"fs-id1167794170488\">Thus, this matrix is a way to help remember the formula for curl. Keep in mind, though, that the word\u00a0<em data-effect=\"italics\">determinant<\/em>\u00a0is used very loosely. A determinant is not really defined on a matrix with entries that are three vectors, three operators, and three functions.<\/p>\n<p id=\"fs-id1167793925808\">If [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field in [latex]\\mathbb{R}^2[\/latex], then the curl of\u00a0[latex]{\\bf{F}}[\/latex], by definition, is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=(Q_x-P_y){\\bf{k}}=\\left(\\frac{\\partial{Q}}{\\partial{x}}-\\frac{\\partial{P}}{\\partial{y}}\\right){\\bf{k}}}[\/latex].<\/p>\n<div id=\"fs-id1167794026966\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding the curl of a three-dimensional vector field<\/h3>\n<p>Find the curl of [latex]{\\bf{F}}(P,Q,R)=\\langle{x}^2z,e^y+xz,xyz\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q275847941\">Show Solution<\/span><\/p>\n<div id=\"q275847941\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793959778\">The curl is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\text{curl }{\\bf{F}}&=\\nabla\\times{\\bf{F}} \\\\  &=\\begin{vmatrix}  {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\  \\frac{\\partial}{\\partial{x}}&\\frac{\\partial}{\\partial{y}}&\\frac{\\partial}{\\partial{z}} \\\\  P&Q&R  \\end{vmatrix} \\\\  &=(R_y-Q)z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\  &=(xz-z){\\bf{i}}+(x^2-yz){\\bf{j}}+z{\\bf{k}}  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the curl of [latex]{\\bf{F}}=\\langle\\sin{x}\\cos{z},\\sin{y}\\sin{z},\\cos{x}\\cos{y}\\rangle[\/latex] at point [latex]\\left(0,\\frac{\\pi}2,\\frac{\\pi}2\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q995421768\">Show Solution<\/span><\/p>\n<div id=\"q995421768\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-{\\bf{i}}[\/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=8250324&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=11v52S_m5wc&amp;video_target=tpm-plugin-i4jt3djg-11v52S_m5wc\" 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\/CP6.43_transcript.html\">transcript for \u201cCP 6.43\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6225\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6225&theme=oea&iframe_resize_id=ohm6225&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the curl of a two-dimensional vector field<\/h3>\n<p>Find the curl of [latex]{\\bf{F}}=\\langle{P},Q\\rangle=\\langle{y},0\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q569401235\">Show Solution<\/span><\/p>\n<div id=\"q569401235\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793489271\">Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the\u00a0<em data-effect=\"italics\">x<\/em>-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=(Q_x-P_t){\\bf{k}}=-{\\bf{k}}\\ne0}[\/latex].<\/p>\n<p id=\"fs-id1167793640854\">Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (Figure 2). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.<\/p>\n<div id=\"attachment_5331\" style=\"width: 502px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5331\" class=\"size-full wp-image-5331\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31233628\/6.55.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/0e80978e3662ede11c0d539b929a8dd5be857b40&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;Two vector field diagrams consisting of vectors that are all parallel. The closer they are to the x axis, the shorter the arrows are. Above the x axis, the arrows point to the right, and below the x axis, the arrows point to the left.&quot; id=&quot;21&quot;&gt;\" width=\"492\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-5331\" class=\"wp-caption-text\">Figure 2. Vector field [latex]{\\bf{F}}(x,y)=\\langle{y},0\\rangle[\/latex] consists of vectors that are all parallel.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793245268\">Note that if [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field in a plane, then [latex]\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}=(Q_x-P_y){\\bf{k}}\\cdot{\\bf{k}}=Q_x-P_y[\/latex]. Therefore, the circulation form of Green\u2019s theorem is sometimes written as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_D\\text{curl }{\\bf{}}\\cdot{\\bf{k}}dA}[\/latex],<\/p>\n<p id=\"fs-id1167793420776\">where [latex]C[\/latex] is a simple closed curve and [latex]D[\/latex] is the region enclosed by [latex]C[\/latex]. Therefore, the circulation form of Green\u2019s theorem can be written in terms of the curl. If we think of curl as a derivative of sorts, then Green\u2019s theorem says that the \u201cderivative\u201d of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0on a region can be translated into a line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function [latex]f[\/latex] on line segment [latex][a,b][\/latex] can be translated into a statement about [latex]f[\/latex] on the boundary of [latex][a,b][\/latex]. Using curl, we can see the circulation form of Green\u2019s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.<\/p>\n<p id=\"fs-id1167793426386\">We can now use what we have learned about curl to show that gravitational fields have no \u201cspin.\u201d Suppose there is an object at the origin with mass [latex]m_1[\/latex] at the origin and an object with mass [latex]m_2[\/latex]. Recall that the gravitational force that object 1 exerts on object 2 is given by field<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}(x,y,z)=-Gm_1m_2\\left\\langle\\frac{x}{(x^2+y^2+z^2)^{3\/2}},\\frac{y}{(x^2+y^2+z^2)^{3\/2}},\\frac{z}{(x^2+y^2+z^2)^{3\/2}}\\right\\rangle}[\/latex].<\/p>\n<div id=\"fs-id1167793505725\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: determining the spin of a gravitational field<\/h3>\n<p>Show that a gravitational field has no spin.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864371936\">Show Solution<\/span><\/p>\n<div id=\"q864371936\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793355047\">To show that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has no spin, we calculate its curl. Let [latex]P(x,y,z)=\\frac{x}{(x^2+y^2+z^2)^{3\/2}}[\/latex], [latex]Q(x,y,z)=\\frac{y}{(x^2+y^2+z^2)^{3\/2}}[\/latex], and [latex]R(x,y,z)=\\frac{z}{(x^2+y^2+z^2)^{3\/2}}[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\text{curl }{\\bf{F}}&=-Gm_1m_2[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\  &=-Gm_1m_2  \\left[\\begin{aligned}  &\\left(\\frac{-3yz}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3yz}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{i}} \\\\  &+\\left(\\frac{-3xz}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3xz}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{j}} \\\\  &+\\left(\\frac{-3xy}{(x^2+y^2+z^2)^{5\/2}}-\\left(\\frac{-3xy}{(x^2+y^2+z^2)^{5\/2}}\\right)\\right){\\bf{k}}  \\end{aligned}\\right]\\\\  &=0\\  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793319476\">Since the curl of the gravitational field is zero, the field has no spin.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Field [latex]{\\bf{v}}(x,y)=\\left\\langle-\\frac{y}{x^2+y^2},\\frac{x}{x^2+y^2}\\right\\rangle[\/latex] models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q935016173\">Show Solution<\/span><\/p>\n<div id=\"q935016173\" class=\"hidden-answer\" style=\"display: none\">\n<p>curl\u00a0[latex]{\\bf{v}}=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5500\">\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 6.43. <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":22,"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 6.43\",\"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-5500","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5500","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":11,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5500\/revisions"}],"predecessor-version":[{"id":6116,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5500\/revisions\/6116"}],"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\/5500\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5500"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5500"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5500"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5500"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}