{"id":790,"date":"2025-10-09T15:39:29","date_gmt":"2025-10-09T15:39:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=790"},"modified":"2025-11-13T20:32:59","modified_gmt":"2025-11-13T20:32:59","slug":"surface-integrals-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/surface-integrals-fresh-take\/","title":{"raw":"Surface Integrals: Fresh Take","rendered":"Surface Integrals: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Understand and find parametric representations of surfaces<\/li>\r\n \t<li>Calculate surface integrals of scalar functions<\/li>\r\n \t<li>Calculate surface integrals of vector fields<\/li>\r\n \t<li>Use surface integrals to solve real-world problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Parametric Surfaces<\/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\">Just as we parameterize curves using one parameter to trace out a path, we parameterize surfaces using two parameters to trace out a two-dimensional object in space. This is the foundation for surface integrals\u2014the higher-dimensional analog of line integrals.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parametric surface is given by [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex], where parameters [latex]u[\/latex] and [latex]v[\/latex] vary over a <strong>parameter domain<\/strong> (a region in the [latex]uv[\/latex]-plane). Each choice of [latex](u,v)[\/latex] produces a point on the surface, and varying [latex]u[\/latex] and [latex]v[\/latex] over their entire domain traces out the complete surface.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Visualizing Surfaces<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">To understand a parametric surface, hold one parameter constant and vary the other. This creates two families of curves on the surface\u2014a \"skeleton\" that reveals its shape. For example, a cylinder [latex]x^2 + y^2 = R^2[\/latex] has circles (constant [latex]v[\/latex]) and vertical lines (constant [latex]u[\/latex]).<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Common Parameterizations<\/strong><\/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>Cylinders<\/strong>: [latex]\\mathbf{r}(u,v) = \\langle R\\cos u, R\\sin u, v \\rangle[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Spheres<\/strong>: [latex]\\mathbf{r}(\\phi,\\theta) = \\langle \\rho\\cos\\theta\\sin\\phi, \\rho\\sin\\theta\\sin\\phi, \\rho\\cos\\phi \\rangle[\/latex] (using spherical coordinates)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Graphs of functions<\/strong>: [latex]\\mathbf{r}(x,y) = \\langle x, y, f(x,y) \\rangle[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Regular and Smooth Parameterizations<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parameterization is <strong>regular<\/strong> if [latex]\\mathbf{r}_u \\times \\mathbf{r}_v \\neq \\mathbf{0}[\/latex] everywhere in the parameter domain. This ensures the image is genuinely two-dimensional (an actual surface, not a curve or point).<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parameterization is <strong>smooth<\/strong> if [latex]\\mathbf{r}_u \\times \\mathbf{r}_v[\/latex] is continuous and never zero\u2014meaning the surface has no sharp corners. A surface is <strong>piecewise smooth<\/strong> if it has smooth faces but sharp edges where derivatives don't exist (like a pyramid).<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Describe the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle2\\cos{u},2\\sin{u},v\\rangle,0\\leq{u}\\leq2\\pi,-\\infty &lt; v &lt; \\infty[\/latex].[reveal-answer q=\"537294031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"537294031\"]Cylinder\u00a0[latex]x^{2}+y^{2}=4[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Describe the surface parameterized by [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle,-\\infty &lt; u,\\infty,0\\leq{v}\\leq2\\pi[\/latex].[reveal-answer q=\"759828461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759828461\"]Cone [latex]x^{2}+y^{2}=z^{2}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Give a parameterization for the portion of cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying in the first octant.[reveal-answer q=\"893249821\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"893249821\"][latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle,0 &lt; u &lt; \\infty,0\\leq{v} &lt; \\frac{\\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=8250327&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=odVIoTzG2b4&amp;video_target=tpm-plugin-sgncyqxs-odVIoTzG2b4\" 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\/CP6.49_transcript.html\">transcript for \u201cCP 6.49\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Is the surface parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}^{2v},v+1,\\sin{u}\\rangle[\/latex], [latex]0\\leq{u}\\leq2, \\ 0\\leq{v}\\leq3[\/latex] smooth?[reveal-answer q=\"957983471\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"957983471\"]Yes.[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Surface Area of a Parametric Surface<\/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\">To find the surface area of a parametric surface, we extend the idea of arc length for curves to two dimensions. Just as arc length uses the derivative of a curve, surface area uses the partial derivatives of a surface parameterization.<\/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 smooth surface [latex]S[\/latex] parameterized by [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] over parameter domain [latex]D[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{Surface Area} = \\iint_D ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">where [latex]\\mathbf{t}_u = \\mathbf{r}_u[\/latex] and [latex]\\mathbf{t}_v = \\mathbf{r}_v[\/latex] are the tangent vectors along the grid curves.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding the Approach<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Divide the parameter domain into small rectangles of size [latex]\\Delta u \\times \\Delta v[\/latex]. Each rectangle maps to a curved patch on the surface. At any point, the tangent vectors [latex]\\mathbf{t}_u[\/latex] and [latex]\\mathbf{t}_v[\/latex] span a parallelogram in the tangent plane that approximates the curved patch. The area of this parallelogram is [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||\\Delta u\\Delta v[\/latex]. Summing over all patches and taking a limit gives the surface area integral.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The cross product [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] captures how the parameterization stretches and distorts the surface. Its magnitude measures the \"scaling factor\" between area in the parameter domain and area on the actual surface.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Common Applications<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This formula recovers familiar results: the lateral surface area of a cone is [latex]\\pi r\\sqrt{h^2 + r^2}[\/latex], and the surface area of a sphere is [latex]4\\pi r^2[\/latex]. For surfaces of revolution generated by rotating [latex]y = f(x)[\/latex] around the [latex]x[\/latex]-axis, use parameterization [latex]\\mathbf{r}(x,\\theta) = \\langle x, f(x)\\cos\\theta, f(x)\\sin\\theta \\rangle[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the surface area of the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}+v,u^2,2v\\rangle[\/latex], [latex]0\\leq{u}\\leq3[\/latex], [latex]0\\leq{v}\\leq2[\/latex].[reveal-answer q=\"847590824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847590824\"][latex]\\approx43.02[\/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=8250326&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=BJGW8e2nxH0&amp;video_target=tpm-plugin-h4f59i0l-BJGW8e2nxH0\" 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\/CP6.51_transcript.html\">transcript for \u201cCP 6.51\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Show that the surface area of cylinder [latex]x^2+y^2=r^2[\/latex], [latex]0\\leq{z}\\leq{h}[\/latex] is [latex]2\\pi{r}h[\/latex]. Notice that this cylinder does not include the top and bottom circles.[reveal-answer q=\"948375921\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"948375921\"]With the standard parameterization of a cylinder, The Surface Area Equation\u00a0shows that the surface area is [latex]2\\pi{r}h[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Use The Surface Area Equation\u00a0to find the area of the surface of revolution obtained by rotating curve [latex]y=\\sin{x}, \\ 0\\leq{x}\\leq\\pi[\/latex] about the [latex]x[\/latex]-axis.[reveal-answer q=\"253747231\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"253747231\"][latex]2\\pi(\\sqrt2+\\sin{h}^{-1}(1))[\/latex][\/hidden-answer]<\/section>\r\n<h2>Surface Integral of a Scalar-Valued Function<\/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\">Surface integrals extend the concept of line integrals from one-dimensional curves to two-dimensional surfaces. Just as line integrals sum function values along a curve, surface integrals sum function values over a surface\u2014weighted by the surface area.<\/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 function [latex]f(x,y,z)[\/latex] over a surface [latex]S[\/latex] parameterized by [latex]\\mathbf{r}(u,v)[\/latex] with parameter domain [latex]D[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S f(x,y,z)\\,dS = \\iint_D f(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The notation [latex]dS[\/latex] represents an infinitesimal piece of surface area, and [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||[\/latex] converts area in the parameter domain to area on the actual surface.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding the Process<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The definition mirrors scalar line integrals: divide the surface into tiny pieces [latex]S_{ij}[\/latex], evaluate [latex]f[\/latex] at a sample point in each piece, multiply by the area [latex]\\Delta S_{ij}[\/latex], and sum. Taking the limit as pieces shrink to zero gives the integral. The formula transforms this into a double integral by approximating each [latex]\\Delta S_{ij}[\/latex] using the tangent plane, giving [latex]\\Delta S_{ij} \\approx ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\Delta u\\Delta v[\/latex].<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Comparison<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For line integrals: [latex]\\int_C f,ds = \\int_a^b f(\\mathbf{r}(t)) ||\\mathbf{r}'(t)||\\,dt[\/latex] (derivative is tangent to curve)<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For surface integrals: [latex]\\iint_S f,dS = \\iint_D f(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex] (cross product is perpendicular to surface)<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Applications<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]S[\/latex] represents a thin metal sheet and [latex]\\rho(x,y,z)[\/latex] is its density, then [latex]m = \\iint_S \\rho(x,y,z)\\,dS[\/latex] gives the total mass. For piecewise smooth surfaces, break the surface into smooth pieces and add the integrals.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]\\displaystyle\\iint_S(x^2-z)dS[\/latex], where [latex]S[\/latex] is the surface with parameterization\u00a0[latex]{\\bf{r}}(u,v)=\\langle{u},u^2+v^2,1\\rangle[\/latex], [latex]0\\leq{u}\\leq2, \\ 0\\leq{v}\\leq3[\/latex].[reveal-answer q=\"834661790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834661790\"][latex]24[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate line integral [latex]\\displaystyle\\iint_S(x-y)dS[\/latex], where [latex]S[\/latex] is cylinder [latex]x^2+y^2=1[\/latex],[latex]0\\leq{z}\\leq2[\/latex], including the circular top and bottom.[reveal-answer q=\"843749219\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843749219\"][latex]0[\/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=8250328&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=RpZfvx5TarU&amp;video_target=tpm-plugin-15l3ck1c-RpZfvx5TarU\" 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\/CP6.55_transcript.html\">transcript for \u201cCP 6.55\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">A piece of metal has a shape that is modeled by paraboloid [latex]z=x^2+y^2[\/latex], [latex]0\\leq{z}\\leq4[\/latex], and the density of the metal is given by [latex]\\rho(x,y,z)=z+1[\/latex]. Find the mass of the piece of metal.[reveal-answer q=\"682307653\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"682307653\"][latex]38.401\\pi\\approx120.640[\/latex].[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Orientation of a Surface<\/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\">Just as vector line integrals require an oriented curve (a direction of travel), surface integrals of vector fields require an oriented surface. An orientation specifies which side of the surface is \"positive\"\u2014think of it as choosing an \"outward\" vs. \"inward\" or \"upward\" vs. \"downward\" side.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>What Makes a Surface Orientable?<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A smooth surface [latex]S[\/latex] is <strong>orientable<\/strong> if you can choose a unit normal vector [latex]\\mathbf{N}[\/latex] at every point on [latex]S[\/latex] such that [latex]\\mathbf{N}[\/latex] varies continuously across the entire surface. This continuous field of normal vectors defines the orientation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a closed surface like a sphere, the outward-pointing normal gives the <strong>positive orientation<\/strong>, while the inward-pointing normal gives the <strong>negative orientation<\/strong>. For graphs of functions [latex]z = f(x,y)[\/latex], the upward-pointing normal (positive [latex]z[\/latex]-component) typically defines the positive orientation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding an Orientation from a Parameterization<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]S[\/latex] is parameterized by [latex]\\mathbf{r}(u,v)[\/latex], the cross product [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] is perpendicular to the surface at each point. The unit normal vector:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{N} = \\frac{\\mathbf{t}_u \\times \\mathbf{t}_v}{||\\mathbf{t}_u \\times \\mathbf{t}_v||}[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">gives an orientation of the surface. The direction depends on the order of the cross product\u2014reversing [latex]u[\/latex] and [latex]v[\/latex] flips the orientation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Nonorientable Surfaces<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Not all surfaces can be oriented. The famous <strong>M\u00f6bius strip<\/strong>\u2014formed by giving a rectangular strip a half-twist before connecting the ends\u2014has no consistent \"inside\" or \"outside.\" If you place a normal vector on the strip and move it continuously around the surface, it points in the opposite direction when it returns to its starting point. Such surfaces have only one side and cannot support vector surface integrals.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Orientation is essential for defining surface integrals of vector fields (flux integrals). Without a well-defined orientation, we can't distinguish flow \"into\" versus \"out of\" a surface.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Give the \u201cupward\u201d orientation of the graph of [latex]f(x, y)=xy[\/latex].[reveal-answer q=\"273456799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"273456799\"][latex]{\\bf{N}}(x,y)=\\left\\langle\\frac{-y}{\\sqrt{1+x^2+y^2}},\\frac{-x}{\\sqrt{1+x^2+y^2}},\\frac{1}{\\sqrt{1+x^2+y^2}}\\right\\rangle[\/latex][\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Surface Integral of a Vector Field<\/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\">Surface integrals of vector fields measure flux\u2014the amount of a vector field flowing across a surface. Just as vector line integrals measure circulation along a curve, surface integrals measure flow through a two-dimensional surface.<\/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}[\/latex] over an oriented surface [latex]S[\/latex] with unit normal vector [latex]\\mathbf{N}[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_S \\mathbf{F} \\cdot \\mathbf{N}\\,dS[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is called the <strong>flux<\/strong> of [latex]\\mathbf{F}[\/latex] across [latex]S[\/latex]. When [latex]S[\/latex] is parameterized by [latex]\\mathbf{r}(u,v)[\/latex] over domain [latex]D[\/latex]:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F}(\\mathbf{r}(u,v)) \\cdot (\\mathbf{t}_u \\times \\mathbf{t}_v)\\,dA[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Notice the magnitude terms cancel\u2014you don't need [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||[\/latex] for vector surface integrals.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Physical Interpretation<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Imagine fluid flowing through a permeable surface [latex]S[\/latex]. The flux integral measures how much fluid crosses [latex]S[\/latex] per unit time. The dot product [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] captures the component of flow perpendicular to the surface\u2014flow parallel to [latex]S[\/latex] doesn't cross it.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Applications<\/strong><\/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>Mass flux<\/strong>: [latex]\\iint_S \\rho\\mathbf{v} \\cdot d\\mathbf{S}[\/latex] measures mass flow rate (includes density [latex]\\rho[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Flow rate<\/strong>: [latex]\\iint_S \\mathbf{v} \\cdot d\\mathbf{S}[\/latex] measures volume flow rate (density omitted)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Heat flow<\/strong>: [latex]\\iint_S -k\\nabla T \\cdot d\\mathbf{S}[\/latex] measures thermal energy transfer (where [latex]k[\/latex] is thermal conductivity)<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Important Note<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Orientation matters! The flux depends on which direction you choose for [latex]\\mathbf{N}[\/latex]. Reversing orientation reverses the sign of the flux. For piecewise smooth surfaces, compute flux over each smooth piece and add the results.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S[\/latex], where [latex]{\\bf{F}}=\\langle0,-z,y\\rangle[\/latex] and [latex]S[\/latex] is the portion of the unit sphere in the first octant with outward orientation.[reveal-answer q=\"736902583\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736902583\"][latex]0[\/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=8250329&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VNMIt6NCUWE&amp;video_target=tpm-plugin-zng6rrhd-VNMIt6NCUWE\" 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\/CP6.58_transcript.html\">transcript for \u201cCP 6.58\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{v}}(x,y,z)=\\langle{x}^2+y^2,z,4y\\rangle[\/latex] m\/sec represent a velocity field of a fluid with constant density [latex]100[\/latex] kg\/m<sup>3<\/sup>. Let [latex]S[\/latex] be the half-cylinder [latex]{\\bf{r}}(u,v)=\\langle\\cos{u},\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq\\pi, \\ 0\\leq{v}\\leq2[\/latex] oriented outward. Calculate the mass flux of the fluid across [latex]S[\/latex].[reveal-answer q=\"459238062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"459238062\"]<span style=\"font-size: 1rem; text-align: initial;\">[latex]400[\/latex] kg\/sec\/m<\/span>[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">A cast-iron solid ball is given by inequality [latex]x^2+y^2+z^2\\leq1[\/latex]. The temperature at a point in a region containing the ball is [latex]T(x,y,z)=\\frac13(x^2+y^2+z^2)[\/latex]. Find the heat flow across the boundary of the solid if this boundary is oriented outward.[reveal-answer q=\"572416935\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"572416935\"][latex]-\\frac{440\\pi}3[\/latex].[\/hidden-answer]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Understand and find parametric representations of surfaces<\/li>\n<li>Calculate surface integrals of scalar functions<\/li>\n<li>Calculate surface integrals of vector fields<\/li>\n<li>Use surface integrals to solve real-world problems<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Parametric Surfaces<\/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\">Just as we parameterize curves using one parameter to trace out a path, we parameterize surfaces using two parameters to trace out a two-dimensional object in space. This is the foundation for surface integrals\u2014the higher-dimensional analog of line integrals.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parametric surface is given by [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex], where parameters [latex]u[\/latex] and [latex]v[\/latex] vary over a <strong>parameter domain<\/strong> (a region in the [latex]uv[\/latex]-plane). Each choice of [latex](u,v)[\/latex] produces a point on the surface, and varying [latex]u[\/latex] and [latex]v[\/latex] over their entire domain traces out the complete surface.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Visualizing Surfaces<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">To understand a parametric surface, hold one parameter constant and vary the other. This creates two families of curves on the surface\u2014a &#8220;skeleton&#8221; that reveals its shape. For example, a cylinder [latex]x^2 + y^2 = R^2[\/latex] has circles (constant [latex]v[\/latex]) and vertical lines (constant [latex]u[\/latex]).<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Common Parameterizations<\/strong><\/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>Cylinders<\/strong>: [latex]\\mathbf{r}(u,v) = \\langle R\\cos u, R\\sin u, v \\rangle[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Spheres<\/strong>: [latex]\\mathbf{r}(\\phi,\\theta) = \\langle \\rho\\cos\\theta\\sin\\phi, \\rho\\sin\\theta\\sin\\phi, \\rho\\cos\\phi \\rangle[\/latex] (using spherical coordinates)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Graphs of functions<\/strong>: [latex]\\mathbf{r}(x,y) = \\langle x, y, f(x,y) \\rangle[\/latex]<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Regular and Smooth Parameterizations<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parameterization is <strong>regular<\/strong> if [latex]\\mathbf{r}_u \\times \\mathbf{r}_v \\neq \\mathbf{0}[\/latex] everywhere in the parameter domain. This ensures the image is genuinely two-dimensional (an actual surface, not a curve or point).<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A parameterization is <strong>smooth<\/strong> if [latex]\\mathbf{r}_u \\times \\mathbf{r}_v[\/latex] is continuous and never zero\u2014meaning the surface has no sharp corners. A surface is <strong>piecewise smooth<\/strong> if it has smooth faces but sharp edges where derivatives don&#8217;t exist (like a pyramid).<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Describe the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle2\\cos{u},2\\sin{u},v\\rangle,0\\leq{u}\\leq2\\pi,-\\infty < v < \\infty[\/latex].\n\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q537294031\">Show Solution<\/button><\/p>\n<div id=\"q537294031\" class=\"hidden-answer\" style=\"display: none\">Cylinder\u00a0[latex]x^{2}+y^{2}=4[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Describe the surface parameterized by [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle,-\\infty < u,\\infty,0\\leq{v}\\leq2\\pi[\/latex].\n\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q759828461\">Show Solution<\/button><\/p>\n<div id=\"q759828461\" class=\"hidden-answer\" style=\"display: none\">Cone [latex]x^{2}+y^{2}=z^{2}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Give a parameterization for the portion of cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying in the first octant.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q893249821\">Show Solution<\/button><\/p>\n<div id=\"q893249821\" class=\"hidden-answer\" style=\"display: none\">[latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle,0 < u < \\infty,0\\leq{v} < \\frac{\\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=8250327&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=odVIoTzG2b4&amp;video_target=tpm-plugin-sgncyqxs-odVIoTzG2b4\" 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\/CP6.49_transcript.html\">transcript for \u201cCP 6.49\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Is the surface parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}^{2v},v+1,\\sin{u}\\rangle[\/latex], [latex]0\\leq{u}\\leq2, \\ 0\\leq{v}\\leq3[\/latex] smooth?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q957983471\">Show Solution<\/button><\/p>\n<div id=\"q957983471\" class=\"hidden-answer\" style=\"display: none\">Yes.<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Surface Area of a Parametric Surface<\/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\">To find the surface area of a parametric surface, we extend the idea of arc length for curves to two dimensions. Just as arc length uses the derivative of a curve, surface area uses the partial derivatives of a surface parameterization.<\/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 smooth surface [latex]S[\/latex] parameterized by [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] over parameter domain [latex]D[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{Surface Area} = \\iint_D ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">where [latex]\\mathbf{t}_u = \\mathbf{r}_u[\/latex] and [latex]\\mathbf{t}_v = \\mathbf{r}_v[\/latex] are the tangent vectors along the grid curves.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding the Approach<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Divide the parameter domain into small rectangles of size [latex]\\Delta u \\times \\Delta v[\/latex]. Each rectangle maps to a curved patch on the surface. At any point, the tangent vectors [latex]\\mathbf{t}_u[\/latex] and [latex]\\mathbf{t}_v[\/latex] span a parallelogram in the tangent plane that approximates the curved patch. The area of this parallelogram is [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||\\Delta u\\Delta v[\/latex]. Summing over all patches and taking a limit gives the surface area integral.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The cross product [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] captures how the parameterization stretches and distorts the surface. Its magnitude measures the &#8220;scaling factor&#8221; between area in the parameter domain and area on the actual surface.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Common Applications<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This formula recovers familiar results: the lateral surface area of a cone is [latex]\\pi r\\sqrt{h^2 + r^2}[\/latex], and the surface area of a sphere is [latex]4\\pi r^2[\/latex]. For surfaces of revolution generated by rotating [latex]y = f(x)[\/latex] around the [latex]x[\/latex]-axis, use parameterization [latex]\\mathbf{r}(x,\\theta) = \\langle x, f(x)\\cos\\theta, f(x)\\sin\\theta \\rangle[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the surface area of the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}+v,u^2,2v\\rangle[\/latex], [latex]0\\leq{u}\\leq3[\/latex], [latex]0\\leq{v}\\leq2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q847590824\">Show Solution<\/button><\/p>\n<div id=\"q847590824\" class=\"hidden-answer\" style=\"display: none\">[latex]\\approx43.02[\/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=8250326&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=BJGW8e2nxH0&amp;video_target=tpm-plugin-h4f59i0l-BJGW8e2nxH0\" 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\/CP6.51_transcript.html\">transcript for \u201cCP 6.51\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Show that the surface area of cylinder [latex]x^2+y^2=r^2[\/latex], [latex]0\\leq{z}\\leq{h}[\/latex] is [latex]2\\pi{r}h[\/latex]. Notice that this cylinder does not include the top and bottom circles.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q948375921\">Show Solution<\/button><\/p>\n<div id=\"q948375921\" class=\"hidden-answer\" style=\"display: none\">With the standard parameterization of a cylinder, The Surface Area Equation\u00a0shows that the surface area is [latex]2\\pi{r}h[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use The Surface Area Equation\u00a0to find the area of the surface of revolution obtained by rotating curve [latex]y=\\sin{x}, \\ 0\\leq{x}\\leq\\pi[\/latex] about the [latex]x[\/latex]-axis.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q253747231\">Show Solution<\/button><\/p>\n<div id=\"q253747231\" class=\"hidden-answer\" style=\"display: none\">[latex]2\\pi(\\sqrt2+\\sin{h}^{-1}(1))[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2>Surface Integral of a Scalar-Valued Function<\/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\">Surface integrals extend the concept of line integrals from one-dimensional curves to two-dimensional surfaces. Just as line integrals sum function values along a curve, surface integrals sum function values over a surface\u2014weighted by the surface area.<\/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 function [latex]f(x,y,z)[\/latex] over a surface [latex]S[\/latex] parameterized by [latex]\\mathbf{r}(u,v)[\/latex] with parameter domain [latex]D[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S f(x,y,z)\\,dS = \\iint_D f(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The notation [latex]dS[\/latex] represents an infinitesimal piece of surface area, and [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||[\/latex] converts area in the parameter domain to area on the actual surface.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Understanding the Process<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The definition mirrors scalar line integrals: divide the surface into tiny pieces [latex]S_{ij}[\/latex], evaluate [latex]f[\/latex] at a sample point in each piece, multiply by the area [latex]\\Delta S_{ij}[\/latex], and sum. Taking the limit as pieces shrink to zero gives the integral. The formula transforms this into a double integral by approximating each [latex]\\Delta S_{ij}[\/latex] using the tangent plane, giving [latex]\\Delta S_{ij} \\approx ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\Delta u\\Delta v[\/latex].<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Comparison<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For line integrals: [latex]\\int_C f,ds = \\int_a^b f(\\mathbf{r}(t)) ||\\mathbf{r}'(t)||\\,dt[\/latex] (derivative is tangent to curve)<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For surface integrals: [latex]\\iint_S f,dS = \\iint_D f(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v||\\,dA[\/latex] (cross product is perpendicular to surface)<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Applications<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]S[\/latex] represents a thin metal sheet and [latex]\\rho(x,y,z)[\/latex] is its density, then [latex]m = \\iint_S \\rho(x,y,z)\\,dS[\/latex] gives the total mass. For piecewise smooth surfaces, break the surface into smooth pieces and add the integrals.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]\\displaystyle\\iint_S(x^2-z)dS[\/latex], where [latex]S[\/latex] is the surface with parameterization\u00a0[latex]{\\bf{r}}(u,v)=\\langle{u},u^2+v^2,1\\rangle[\/latex], [latex]0\\leq{u}\\leq2, \\ 0\\leq{v}\\leq3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q834661790\">Show Solution<\/button><\/p>\n<div id=\"q834661790\" class=\"hidden-answer\" style=\"display: none\">[latex]24[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate line integral [latex]\\displaystyle\\iint_S(x-y)dS[\/latex], where [latex]S[\/latex] is cylinder [latex]x^2+y^2=1[\/latex],[latex]0\\leq{z}\\leq2[\/latex], including the circular top and bottom.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q843749219\">Show Solution<\/button><\/p>\n<div id=\"q843749219\" class=\"hidden-answer\" style=\"display: none\">[latex]0[\/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=8250328&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=RpZfvx5TarU&amp;video_target=tpm-plugin-15l3ck1c-RpZfvx5TarU\" 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\/CP6.55_transcript.html\">transcript for \u201cCP 6.55\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A piece of metal has a shape that is modeled by paraboloid [latex]z=x^2+y^2[\/latex], [latex]0\\leq{z}\\leq4[\/latex], and the density of the metal is given by [latex]\\rho(x,y,z)=z+1[\/latex]. Find the mass of the piece of metal.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q682307653\">Show Solution<\/button><\/p>\n<div id=\"q682307653\" class=\"hidden-answer\" style=\"display: none\">[latex]38.401\\pi\\approx120.640[\/latex].<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Orientation of a Surface<\/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\">Just as vector line integrals require an oriented curve (a direction of travel), surface integrals of vector fields require an oriented surface. An orientation specifies which side of the surface is &#8220;positive&#8221;\u2014think of it as choosing an &#8220;outward&#8221; vs. &#8220;inward&#8221; or &#8220;upward&#8221; vs. &#8220;downward&#8221; side.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>What Makes a Surface Orientable?<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A smooth surface [latex]S[\/latex] is <strong>orientable<\/strong> if you can choose a unit normal vector [latex]\\mathbf{N}[\/latex] at every point on [latex]S[\/latex] such that [latex]\\mathbf{N}[\/latex] varies continuously across the entire surface. This continuous field of normal vectors defines the orientation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a closed surface like a sphere, the outward-pointing normal gives the <strong>positive orientation<\/strong>, while the inward-pointing normal gives the <strong>negative orientation<\/strong>. For graphs of functions [latex]z = f(x,y)[\/latex], the upward-pointing normal (positive [latex]z[\/latex]-component) typically defines the positive orientation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding an Orientation from a Parameterization<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]S[\/latex] is parameterized by [latex]\\mathbf{r}(u,v)[\/latex], the cross product [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] is perpendicular to the surface at each point. The unit normal vector:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{N} = \\frac{\\mathbf{t}_u \\times \\mathbf{t}_v}{||\\mathbf{t}_u \\times \\mathbf{t}_v||}[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">gives an orientation of the surface. The direction depends on the order of the cross product\u2014reversing [latex]u[\/latex] and [latex]v[\/latex] flips the orientation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Nonorientable Surfaces<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Not all surfaces can be oriented. The famous <strong>M\u00f6bius strip<\/strong>\u2014formed by giving a rectangular strip a half-twist before connecting the ends\u2014has no consistent &#8220;inside&#8221; or &#8220;outside.&#8221; If you place a normal vector on the strip and move it continuously around the surface, it points in the opposite direction when it returns to its starting point. Such surfaces have only one side and cannot support vector surface integrals.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Orientation is essential for defining surface integrals of vector fields (flux integrals). Without a well-defined orientation, we can&#8217;t distinguish flow &#8220;into&#8221; versus &#8220;out of&#8221; a surface.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Give the \u201cupward\u201d orientation of the graph of [latex]f(x, y)=xy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q273456799\">Show Solution<\/button><\/p>\n<div id=\"q273456799\" class=\"hidden-answer\" style=\"display: none\">[latex]{\\bf{N}}(x,y)=\\left\\langle\\frac{-y}{\\sqrt{1+x^2+y^2}},\\frac{-x}{\\sqrt{1+x^2+y^2}},\\frac{1}{\\sqrt{1+x^2+y^2}}\\right\\rangle[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Surface Integral of a Vector Field<\/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\">Surface integrals of vector fields measure flux\u2014the amount of a vector field flowing across a surface. Just as vector line integrals measure circulation along a curve, surface integrals measure flow through a two-dimensional surface.<\/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}[\/latex] over an oriented surface [latex]S[\/latex] with unit normal vector [latex]\\mathbf{N}[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_S \\mathbf{F} \\cdot \\mathbf{N}\\,dS[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is called the <strong>flux<\/strong> of [latex]\\mathbf{F}[\/latex] across [latex]S[\/latex]. When [latex]S[\/latex] is parameterized by [latex]\\mathbf{r}(u,v)[\/latex] over domain [latex]D[\/latex]:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F}(\\mathbf{r}(u,v)) \\cdot (\\mathbf{t}_u \\times \\mathbf{t}_v)\\,dA[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Notice the magnitude terms cancel\u2014you don&#8217;t need [latex]||\\mathbf{t}_u \\times \\mathbf{t}_v||[\/latex] for vector surface integrals.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Physical Interpretation<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Imagine fluid flowing through a permeable surface [latex]S[\/latex]. The flux integral measures how much fluid crosses [latex]S[\/latex] per unit time. The dot product [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] captures the component of flow perpendicular to the surface\u2014flow parallel to [latex]S[\/latex] doesn&#8217;t cross it.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Applications<\/strong><\/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>Mass flux<\/strong>: [latex]\\iint_S \\rho\\mathbf{v} \\cdot d\\mathbf{S}[\/latex] measures mass flow rate (includes density [latex]\\rho[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Flow rate<\/strong>: [latex]\\iint_S \\mathbf{v} \\cdot d\\mathbf{S}[\/latex] measures volume flow rate (density omitted)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Heat flow<\/strong>: [latex]\\iint_S -k\\nabla T \\cdot d\\mathbf{S}[\/latex] measures thermal energy transfer (where [latex]k[\/latex] is thermal conductivity)<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Important Note<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Orientation matters! The flux depends on which direction you choose for [latex]\\mathbf{N}[\/latex]. Reversing orientation reverses the sign of the flux. For piecewise smooth surfaces, compute flux over each smooth piece and add the results.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S[\/latex], where [latex]{\\bf{F}}=\\langle0,-z,y\\rangle[\/latex] and [latex]S[\/latex] is the portion of the unit sphere in the first octant with outward orientation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q736902583\">Show Solution<\/button><\/p>\n<div id=\"q736902583\" class=\"hidden-answer\" style=\"display: none\">[latex]0[\/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=8250329&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VNMIt6NCUWE&amp;video_target=tpm-plugin-zng6rrhd-VNMIt6NCUWE\" 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\/CP6.58_transcript.html\">transcript for \u201cCP 6.58\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Let [latex]{\\bf{v}}(x,y,z)=\\langle{x}^2+y^2,z,4y\\rangle[\/latex] m\/sec represent a velocity field of a fluid with constant density [latex]100[\/latex] kg\/m<sup>3<\/sup>. Let [latex]S[\/latex] be the half-cylinder [latex]{\\bf{r}}(u,v)=\\langle\\cos{u},\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq\\pi, \\ 0\\leq{v}\\leq2[\/latex] oriented outward. Calculate the mass flux of the fluid across [latex]S[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q459238062\">Show Solution<\/button><\/p>\n<div id=\"q459238062\" class=\"hidden-answer\" style=\"display: none\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]400[\/latex] kg\/sec\/m<\/span><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A cast-iron solid ball is given by inequality [latex]x^2+y^2+z^2\\leq1[\/latex]. The temperature at a point in a region containing the ball is [latex]T(x,y,z)=\\frac13(x^2+y^2+z^2)[\/latex]. Find the heat flow across the boundary of the solid if this boundary is oriented outward.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q572416935\">Show Solution<\/button><\/p>\n<div id=\"q572416935\" class=\"hidden-answer\" style=\"display: none\">[latex]-\\frac{440\\pi}3[\/latex].<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":16,"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-790","chapter","type-chapter","status-publish","hentry"],"part":438,"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\/790","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":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions"}],"predecessor-version":[{"id":2246,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions\/2246"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/438"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/790\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=790"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=790"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=790"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=790"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}