{"id":788,"date":"2025-10-09T15:39:23","date_gmt":"2025-10-09T15:39:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=788"},"modified":"2026-05-26T15:10:06","modified_gmt":"2026-05-26T15:10:06","slug":"surface-integrals-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/surface-integrals-learn-it-1\/","title":{"raw":"Surface Integrals: Learn It 1","rendered":"Surface Integrals: Learn It 1"},"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\nWe have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.\r\n<p class=\"whitespace-normal break-words\">Just as with line integrals, there are two kinds of surface integrals: surface integrals of scalar-valued functions and surface integrals of vector fields. Before we can integrate over a surface, we need to understand how to describe surfaces mathematically using parameterizations.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Parameterizing Curves<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To calculate a line integral over curve [latex]C[\/latex], we first parameterize [latex]C[\/latex]. A parametric curve uses one parameter [latex]t[\/latex] to trace out a one-dimensional path.<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">To calculate a surface integral over surface [latex]S[\/latex], we need to parameterize [latex]S[\/latex]. A <strong>parametric surface<\/strong> uses two parameters because a surface is two-dimensional.<\/p>\r\n<p class=\"whitespace-normal break-words\">A parametric surface is given by:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex]<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice that this parameterization involves two parameters, [latex]u[\/latex] and [latex]v[\/latex], because a surface is two-dimensional, and therefore two variables are needed to trace out the surface.<\/section>\r\n<p class=\"whitespace-normal break-words\">The parameters [latex]u[\/latex] and [latex]v[\/latex] vary over a region called the <strong>parameter domain<\/strong> (or parameter space)\u2014the set of points in the [latex]uv[\/latex]-plane that can be substituted into [latex]\\mathbf{r}[\/latex]. Each choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain gives a point on the surface, just as each choice of parameter [latex]t[\/latex] gives a point on a parameterized curve. The entire surface is created by making all possible choices of [latex]u[\/latex] and [latex]v[\/latex] over the parameter domain.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>parameter domain<\/h3>\r\nGiven a parameterization of surface [latex]{\\bf{r}}(u,v)=\\langle{x}(u,v),y(u,v),z(u,v)\\rangle[\/latex], the\u00a0<strong><span id=\"505d9884-54a9-4157-83ff-87887335091b_term272\" data-type=\"term\">parameter domain<\/span><\/strong>\u00a0of the parameterization is the set of points in the [latex]uv[\/latex]-plane that can be substituted into\u00a0[latex]{\\bf{r}}[\/latex].\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167794328407\">Describe the surface [latex]S[\/latex] parameterized by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle \\cos u, \\sin u, v \\rangle,\u00a0 -\\infty &lt; u &lt; \\infty,\u00a0 -\\infty &lt; v &lt; \\infty[\/latex]<\/p>\r\n[reveal-answer q=\"466382191\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"466382191\"]\r\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of [latex]\\mathbb{R}^2[\/latex], we can choose any value for [latex]u[\/latex] and [latex]v[\/latex] and plot the corresponding point. If [latex]u=v=0[\/latex], then [latex]{\\bf{r}}(0,0)\\langle1,0,0\\rangle[\/latex], so point [latex](1, 0, 0)[\/latex] is on [latex]S[\/latex]. Similarly, points [latex]{\\bf{r}}(\\pi,2)=\\langle-1,0,2\\rangle[\/latex] and [latex]{\\bf{r}}\\left(\\frac{\\pi}2,4\\right)=(0,1,4)[\/latex] are on [latex]S[\/latex].<\/span><\/h2>\r\n<p id=\"fs-id1167793964004\">Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize [latex]S[\/latex], we visualize two families of curves that lie on [latex]S[\/latex]. In the first family of curves we hold [latex]u[\/latex] constant; in the second family of curves we hold [latex]v[\/latex] constant. This allows us to build a \u201cskeleton\u201d of the surface, thereby getting an idea of its shape.<\/p>\r\n<p id=\"fs-id1167794296350\">First, suppose that [latex]u[\/latex] is a constant [latex]K[\/latex]. Then the curve traced out by the parameterization is [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex], which gives a vertical line that goes through point [latex](\\cos{K},\\sin{K},v)[\/latex] in the [latex]xy[\/latex]-plane.<\/p>\r\n<p id=\"fs-id1167794329528\">Now suppose that [latex]v[\/latex] is a constant [latex]K[\/latex]. Then the curve traced out by the parameterization is [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex], which gives a circle in plane [latex]z=K[\/latex] with radius 1 and center [latex](0, 0, K)[\/latex].<\/p>\r\n<p id=\"fs-id1167793568953\">If [latex]u[\/latex] is held constant, then we get vertical lines; if [latex]v[\/latex] is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder [latex]x^{2}+y^{2}=1[\/latex] (Figure 1).<\/p>\r\n\r\n[caption id=\"attachment_1463\" align=\"aligncenter\" width=\"900\"]<img class=\"size-full wp-image-1463\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1.webp\" alt=\"Three diagrams in three dimensions. The first shows vertical lines around the origin. The second shows parallel circles all with center at the origin and radius of 1. The third shows the lines and circle. Together, they form the skeleton of a cylinder.\" width=\"900\" height=\"328\" \/> Figure 1. (a) Lines [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex] for [latex]K = 0, \\frac{\\pi}2, \u03c0,[\/latex] and [latex]\\frac{3\\pi}2[\/latex]. (b) Circles [latex]\\langle\\cos{u},\\sin{u},K\\rangle[\/latex] for [latex]K = \u22122, \u22121, 1,[\/latex] and [latex]2[\/latex]. (c) The lines and circles together. As [latex]u[\/latex] and [latex]v[\/latex] vary, they describe a cylinder.[\/caption]Notice that if [latex]x=\\cos{u}[\/latex] and [latex]y=\\sin{u}[\/latex], then [latex]x^{2}+y^{2}=1[\/latex], so points from [latex]S[\/latex] do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex] for some [latex]k[\/latex], and therefore each point on the cylinder is contained in the parameterized surface (Figure 2).[caption id=\"attachment_1465\" align=\"aligncenter\" width=\"376\"]<img class=\"size-full wp-image-1465\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6.webp\" alt=\"An image of a vertical cylinder in three dimensions with the center of its circular base located on the z axis.\" width=\"376\" height=\"467\" \/> Figure 2. Cylinder [latex]x^{2}+y^{2}=r^{2}[\/latex] has parameterization [latex]{\\bf{r}}(u,v)=\\langle{r}\\cos{u},r\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi,-\\infty&lt;v&lt;\\infty[\/latex].[\/caption]\r\n[latex]\\\\[\/latex]\r\n<strong>Analysis<\/strong>\r\n[latex]\\\\[\/latex]\r\nNotice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to [latex]0\\leq{u}\\leq\\pi, 0 &lt; v &lt;6[\/latex], then the surface would be a half-cylinder of height 6.\r\n[\/hidden-answer]<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Parameterizing Cylinders<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">It follows from the example that we can parameterize all cylinders of the form [latex]x^2 + y^2 = R^2[\/latex]. If [latex]S[\/latex] is a cylinder given by equation [latex]x^2 + y^2 = R^2[\/latex], then a parameterization of [latex]S[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle R\\cos u, R\\sin u, v \\rangle, \\quad 0 \\leq u \\leq 2\\pi, \\quad -\\infty &lt; v &lt; \\infty[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p id=\"fs-id1167793639868\">We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793571150\">Describe the surface [latex]S[\/latex] parameterized by [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u^2\\rangle,0\\le{u}\\le\\infty,0\\le{v}&lt;2\\pi[\/latex].<\/p>\r\n[reveal-answer q=\"384529048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384529048\"]\r\n<p id=\"fs-id1167793571150\">Notice that if [latex]u[\/latex] is held constant, then the resulting curve is a circle of radius [latex]u[\/latex] in plane [latex]z=u[\/latex]. Therefore, as [latex]u[\/latex] increases, the radius of the resulting circle increases. If [latex]v[\/latex] is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx^2+y^2&amp;=(u\\cos{v})^2+(u\\sin{v})^2 \\\\\r\n&amp;=u^2\\cos^2v+u^2\\sin^2v \\\\\r\n&amp;=u^2 \\\\\r\n&amp;=z\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, the surface is elliptic paraboloid [latex]x^2+y^2=z[\/latex] (Figure 3).\r\n\r\n[caption id=\"attachment_1467\" align=\"aligncenter\" width=\"619\"]<img class=\"size-full wp-image-1467\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319.png\" alt=\"Two images in three dimensions. The first shows parallel circles on the z axis with radii increasing as z increases. Vertical parabolas opening up frame the circles, forming the skeleton of a paraboloid. The second shows the elliptic paraboloid, which is made of all the possible circles and vertical parabolas in the parameter domain.\" width=\"619\" height=\"440\" \/> Figure 3. (a) Circles arise from holding [latex]u[\/latex] constant; the vertical parabolas arise from holding [latex]v[\/latex] constant. (b) An elliptic paraboloid results from all choices of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain.[\/caption]\r\n[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Give a parameterization of the cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying on or above the plane [latex]z=-2[\/latex].[reveal-answer q=\"989238235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989238235\"]The horizontal cross-section of the cone at height [latex]z=u[\/latex] is circle [latex]x^{2}+y^{2}=u^{2}[\/latex]. Therefore, a point on the cone at height [latex]u[\/latex] has coordinates [latex](u\\cos{v},u\\sin{v},u)[\/latex] for angle [latex]v[\/latex]. Hence, a parameterization of the cone is [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle[\/latex]. Since we are not interested in the entire cone, only the portion on or above plane [latex]z=-2[\/latex], the parameter domain is given by [latex]-2\\leq{u}&lt;\\infty,0\\leq{v}&lt;2\\pi[\/latex] (Figure 4).[caption id=\"attachment_1468\" align=\"aligncenter\" width=\"343\"]<img class=\"size-full wp-image-1468\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067.webp\" alt=\"A three-dimensional diagram of the cone x^2 + y^2 = z^2, which opens up along the z axis for positive z values and opens down along the z axis for negative z values. The center is at the origin.\" width=\"343\" height=\"403\" \/> Figure 4. Cone [latex]x^{2}+y^{2}=r^{2}[\/latex] has parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle[\/latex], [latex]-\\infty&lt;u&lt;\\infty, 0\\leq{v}\\leq2\\pi[\/latex].[\/caption][\/hidden-answer]<\/section>We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions.\r\n<h3>Parameterizing Spheres<\/h3>\r\n<p class=\"whitespace-normal break-words\">To parameterize a sphere, it's easiest to use spherical coordinates. The sphere of radius [latex]\\rho[\/latex] centered at the origin is given by:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(\\phi,\\theta) = \\langle \\rho\\cos\\theta\\sin\\phi, \\rho\\sin\\theta\\sin\\phi, \\rho\\cos\\phi \\rangle, \\quad 0 \\leq \\theta \\leq 2\\pi, \\quad 0 \\leq \\phi \\leq \\pi[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The idea is that as [latex]\\phi[\/latex] sweeps downward from the positive [latex]z[\/latex]-axis, a circle of radius [latex]\\rho\\sin\\phi[\/latex] is traced out by letting [latex]\\theta[\/latex] run from [latex]0[\/latex] to [latex]2\\pi[\/latex]. To see this, let [latex]\\phi[\/latex] be fixed. Then:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{aligned} x^2 + y^2 &amp;= (\\rho\\cos\\theta\\sin\\phi)^2 + (\\rho\\sin\\theta\\sin\\phi)^2 \\\\ &amp;= \\rho^2\\sin^2\\phi(\\cos^2\\theta + \\sin^2\\theta) \\\\ &amp;= \\rho^2\\sin^2\\phi \\ &amp;= (\\rho\\sin\\phi)^2 \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives the desired circle (Figure 5).<\/p>\r\n\r\n[caption id=\"attachment_1469\" align=\"aligncenter\" width=\"343\"]<img class=\"size-full wp-image-1469\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68.webp\" alt=\"A three-dimensional diagram of the sphere of radius rho.\" width=\"343\" height=\"371\" \/> Figure 5. The sphere of radius [latex]\\rho[\/latex] has parameterization [latex]{\\bf{r}}(\\phi,\\theta)=\\langle\\rho\\cos\\theta\\sin\\phi,\\rho\\sin\\theta\\sin\\phi,\\rho\\cos\\phi\\rangle[\/latex],[latex] 0\\leq\\theta\\leq2\\pi, 0\\leq\\phi\\leq\\pi[\/latex].[\/caption]\r\n<h3>Parameterizing Graphs of Functions<\/h3>\r\n<p class=\"whitespace-normal break-words\">To parameterize the graph of a two-variable function, let [latex]z = f(x,y)[\/latex] be a function of two variables. The simplest parameterization of the graph of [latex]f[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,y) = \\langle x, y, f(x,y) \\rangle[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]x[\/latex] and [latex]y[\/latex] vary over the domain of [latex]f[\/latex] (Figure 6).<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">For example, the graph of [latex]f(x,y) = x^2y[\/latex] can be parameterized by [latex]\\mathbf{r}(x,y) = \\langle x, y, x^2y \\rangle[\/latex], where the parameters [latex]x[\/latex] and [latex]y[\/latex] vary over the domain of [latex]f[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">If you only care about a piece of the graph of [latex]f[\/latex]\u2014say, the piece over rectangle [latex][1,3] \\times [2,5][\/latex]\u2014then you can restrict the parameter domain:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,y) = \\langle x, y, x^2y \\rangle, \\quad 1 \\leq x \\leq 3, \\quad 2 \\leq y \\leq 5[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Similarly, if [latex]S[\/latex] is a surface given by [latex]x = g(y,z)[\/latex] or [latex]y = h(x,z)[\/latex], then a parameterization of [latex]S[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(y,z) = \\langle g(y,z), y, z \\rangle[\/latex] or [latex]\\mathbf{r}(x,z) = \\langle x, h(x,z), z \\rangle[\/latex]<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">For example, the graph of paraboloid [latex]2y = x^2 + z^2[\/latex] can be parameterized by:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,z) = \\left\\langle x, \\frac{x^2 + z^2}{2}, z \\right\\rangle, \\quad 0 \\leq x \\leq \\infty, \\quad 0 \\leq z \\leq \\infty[\/latex]<\/p>\r\nNotice that we don't need to vary over the entire domain of [latex]y[\/latex] because [latex]x[\/latex] and [latex]z[\/latex] are squared.\r\n\r\n<\/section>[caption id=\"attachment_1470\" align=\"aligncenter\" width=\"423\"]<img class=\"size-full wp-image-1470\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd.webp\" alt=\"A three-dimensional diagram of a surface z = f(x,y) above its mapping in the two-dimensional x,y plane. The point (x,y) in the plane corresponds to the point z = f(x,y) on the surface.\" width=\"423\" height=\"499\" \/> Figure 6. The simplest parameterization of the graph of a function is [latex]{\\bf{r}}(x,y)=\\langle{x},y,f(x,y)\\rangle[\/latex].[\/caption]\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Smooth and Regular Parameterizations<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let's generalize the notions of smoothness and regularity from curves to parametric surfaces.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Regular Parameterizations of Curves<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Curve parameterization [latex]\\mathbf{r}(t)[\/latex], [latex]a \\leq t \\leq b[\/latex], is <strong>regular<\/strong> if [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for all [latex]t[\/latex] in [latex][a,b][\/latex]. This condition ensures the image of [latex]\\mathbf{r}[\/latex] is actually a curve, not just a point.<\/p>\r\n<p class=\"whitespace-normal break-words\">For example, [latex]\\mathbf{r}(t) = \\langle 1, 2 \\rangle[\/latex], [latex]0 \\leq t \\leq 5[\/latex], is not regular because its image is just the point \u00a0[latex](1, 2)[\/latex]\u00a0and [latex]\\mathbf{r}'(t) = \\mathbf{0}[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">We need a similar notion of regularity for surfaces to ensure a surface parameterization traces out an actual surface. Consider the parameterization:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle 0, \\cos v, 1 \\rangle, \\quad 0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\pi[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Although this appears to parameterize a surface, its image is actually just a line (Figure 7). Notice that [latex]\\mathbf{r}_u = \\langle 0, 0, 0 \\rangle[\/latex] and [latex]\\mathbf{r}_v = \\langle 0, -\\sin v, 0 \\rangle[\/latex], so the cross product [latex]\\mathbf{r}_u \\times \\mathbf{r}_v = \\mathbf{0}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The analog of the condition [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for curves is that [latex]\\mathbf{r}_u \\times \\mathbf{r}_v \\neq \\mathbf{0}[\/latex] for surfaces.<\/p>\r\n\r\n[caption id=\"attachment_1471\" align=\"aligncenter\" width=\"311\"]<img class=\"size-full wp-image-1471\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046.webp\" alt=\"A three-dimensional diagram of a line on the x,z plane where the z component is 1, the x component is 1, and the y component exists between -1 and 1.\" width=\"311\" height=\"358\" \/> Figure 7. The image of parameterization [latex]{{\\bf{r}}(u,v)=\\langle0,\\cos{v},1\\rangle, \\ 0\\leq{u}\\leq1, \\ 0\\leq{v}\\leq\\pi}[\/latex] is a line.[\/caption]<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>regular parameterization<\/h3>\r\nParameterization [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] is a\u00a0<strong><span id=\"505d9884-54a9-4157-83ff-87887335091b_term273\" data-type=\"term\">regular parameterization<\/span><\/strong>\u00a0if [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for point [latex](u, v)[\/latex] in the parameter domain.\r\n\r\n<\/section>If parameterization [latex]\\mathbf{r}[\/latex] is regular, then the image of [latex]\\mathbf{r}[\/latex] is a two-dimensional object, as a surface should be. Throughout this module, we assume all parameterizations [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] are regular.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Smooth Parameterizations of Curves<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Curve parameterization [latex]\\mathbf{r}(t)[\/latex], [latex]a \\leq t \\leq b[\/latex], is <strong>smooth<\/strong> if [latex]\\mathbf{r}'(t)[\/latex] is continuous and [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for all [latex]t[\/latex] in [latex][a,b][\/latex]. Informally, a curve parameterization is smooth if the resulting curve has no sharp corners.<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>smooth\u00a0parameterization<\/h3>\r\nA surface parameterization [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] is\u00a0<strong>smooth\u00a0<\/strong>if vector [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain.\r\n\r\n<\/section>A surface may also be <strong>piecewise smooth<\/strong> if it has smooth faces but also has locations where the directional derivatives do not exist.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Which of the figures in\u00a0Figure 8\u00a0is smooth?\r\n\r\n[caption id=\"attachment_1474\" align=\"aligncenter\" width=\"882\"]<img class=\"size-full wp-image-1474\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3.webp\" alt=\"Two three-dimensional figures. The first surface is smooth. It looks like a tire with a large hole in the middle. The second is piecewise smooth. It is a pyramid with a rectangular base and four sides.\" width=\"882\" height=\"410\" \/> Figure 8. (a) This surface is smooth. (b) This surface is piecewise smooth.[\/caption]\r\n\r\n[reveal-answer q=\"248348349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"248348349\"]\r\n<p id=\"fs-id1167793373625\">The surface in\u00a0Figure 8(a) can be parameterized by<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}(u,v)=\\langle(2+\\cos{v})\\cos{u},(2+\\cos{v})\\sin{u},\\sin{v}\\rangle \\ 0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq2\\pi[\/latex]<\/p>\r\n<p id=\"fs-id1167793900590\">(we can use technology to verify). Notice that vectors<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}_u=\\langle-(2+\\cos{v})\\sin{u},(2+\\cos{v})\\cos{u},0\\rangle\\text{ and }{\\bf{r}}_v=\\langle-\\sin{v}\\cos{u},-\\sin{v}\\sin{u},\\cos{v}\\rangle[\/latex]<\/p>\r\n<p id=\"fs-id1167794070993\">exist for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain, and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{r}}_v&amp;=\\begin{vmatrix}\r\n{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\\r\n-(2+\\cos{v})\\sin{u}&amp;(2+\\cos{v})\\cos{u}&amp;0 \\\\\r\n-\\sin{v}\\cos{u}&amp;-\\sin{v}\\sin{u}&amp;\\cos{v}\r\n\\end{vmatrix} \\\\\r\n&amp;=[(2+\\cos{v})\\cos{u}\\cos{v}]{\\bf{i}}+[(2+\\cos{v})\\sin{u}\\cos{v}]{\\bf{j}}+[(2+\\cos{v})\\sin{v}\\sin^2u+(2+\\cos{v})\\sin{v}\\cos^2u]{\\bf{k}} \\\\\r\n&amp;=[(2+\\cos{v})\\cos{u}\\cos{v}]{\\bf{i}}+[(2+\\cos{v})\\sin{u}\\cos{v}]{\\bf{j}}+[(2+\\cos{v})\\sin{v}]{\\bf{k}}\r\n\\end{aligned}[\/latex]<\/p>\r\n.\r\n<p id=\"fs-id1167794126254\">The\u00a0[latex]\\bf{k}[\/latex]\u00a0component of this vector is zero only if [latex]v=0[\/latex] or [latex]v=\\pi[\/latex]. If [latex]v=0[\/latex] or [latex]v=\\pi[\/latex], then the only choices for [latex]u[\/latex] that make the [latex]\\bf{j}[\/latex] component zero are [latex]u=0[\/latex] or [latex]u=\\pi[\/latex]. But, these choices of [latex]u[\/latex] do not make the [latex]\\bf{i}[\/latex] component zero. Therefore, [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.<\/p>\r\n<p id=\"fs-id1167794126003\">In the pyramid in\u00a0Figure 8(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of five smooth faces, and thus this surface is piecewise smooth.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]325849[\/ohm_question]<\/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<p>We have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.<\/p>\n<p class=\"whitespace-normal break-words\">Just as with line integrals, there are two kinds of surface integrals: surface integrals of scalar-valued functions and surface integrals of vector fields. Before we can integrate over a surface, we need to understand how to describe surfaces mathematically using parameterizations.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Parameterizing Curves<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To calculate a line integral over curve [latex]C[\/latex], we first parameterize [latex]C[\/latex]. A parametric curve uses one parameter [latex]t[\/latex] to trace out a one-dimensional path.<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">To calculate a surface integral over surface [latex]S[\/latex], we need to parameterize [latex]S[\/latex]. A <strong>parametric surface<\/strong> uses two parameters because a surface is two-dimensional.<\/p>\n<p class=\"whitespace-normal break-words\">A parametric surface is given by:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex]<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice that this parameterization involves two parameters, [latex]u[\/latex] and [latex]v[\/latex], because a surface is two-dimensional, and therefore two variables are needed to trace out the surface.<\/section>\n<p class=\"whitespace-normal break-words\">The parameters [latex]u[\/latex] and [latex]v[\/latex] vary over a region called the <strong>parameter domain<\/strong> (or parameter space)\u2014the set of points in the [latex]uv[\/latex]-plane that can be substituted into [latex]\\mathbf{r}[\/latex]. Each choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain gives a point on the surface, just as each choice of parameter [latex]t[\/latex] gives a point on a parameterized curve. The entire surface is created by making all possible choices of [latex]u[\/latex] and [latex]v[\/latex] over the parameter domain.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>parameter domain<\/h3>\n<p>Given a parameterization of surface [latex]{\\bf{r}}(u,v)=\\langle{x}(u,v),y(u,v),z(u,v)\\rangle[\/latex], the\u00a0<strong><span id=\"505d9884-54a9-4157-83ff-87887335091b_term272\" data-type=\"term\">parameter domain<\/span><\/strong>\u00a0of the parameterization is the set of points in the [latex]uv[\/latex]-plane that can be substituted into\u00a0[latex]{\\bf{r}}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167794328407\">Describe the surface [latex]S[\/latex] parameterized by<\/p>\n<p style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle \\cos u, \\sin u, v \\rangle,\u00a0 -\\infty < u < \\infty,\u00a0 -\\infty < v < \\infty[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q466382191\">Show Solution<\/button><\/p>\n<div id=\"q466382191\" class=\"hidden-answer\" style=\"display: none\">\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of [latex]\\mathbb{R}^2[\/latex], we can choose any value for [latex]u[\/latex] and [latex]v[\/latex] and plot the corresponding point. If [latex]u=v=0[\/latex], then [latex]{\\bf{r}}(0,0)\\langle1,0,0\\rangle[\/latex], so point [latex](1, 0, 0)[\/latex] is on [latex]S[\/latex]. Similarly, points [latex]{\\bf{r}}(\\pi,2)=\\langle-1,0,2\\rangle[\/latex] and [latex]{\\bf{r}}\\left(\\frac{\\pi}2,4\\right)=(0,1,4)[\/latex] are on [latex]S[\/latex].<\/span><\/h2>\n<p id=\"fs-id1167793964004\">Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize [latex]S[\/latex], we visualize two families of curves that lie on [latex]S[\/latex]. In the first family of curves we hold [latex]u[\/latex] constant; in the second family of curves we hold [latex]v[\/latex] constant. This allows us to build a \u201cskeleton\u201d of the surface, thereby getting an idea of its shape.<\/p>\n<p id=\"fs-id1167794296350\">First, suppose that [latex]u[\/latex] is a constant [latex]K[\/latex]. Then the curve traced out by the parameterization is [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex], which gives a vertical line that goes through point [latex](\\cos{K},\\sin{K},v)[\/latex] in the [latex]xy[\/latex]-plane.<\/p>\n<p id=\"fs-id1167794329528\">Now suppose that [latex]v[\/latex] is a constant [latex]K[\/latex]. Then the curve traced out by the parameterization is [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex], which gives a circle in plane [latex]z=K[\/latex] with radius 1 and center [latex](0, 0, K)[\/latex].<\/p>\n<p id=\"fs-id1167793568953\">If [latex]u[\/latex] is held constant, then we get vertical lines; if [latex]v[\/latex] is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder [latex]x^{2}+y^{2}=1[\/latex] (Figure 1).<\/p>\n<figure id=\"attachment_1463\" aria-describedby=\"caption-attachment-1463\" style=\"width: 900px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1463\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1.webp\" alt=\"Three diagrams in three dimensions. The first shows vertical lines around the origin. The second shows parallel circles all with center at the origin and radius of 1. The third shows the lines and circle. Together, they form the skeleton of a cylinder.\" width=\"900\" height=\"328\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1.webp 900w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1-300x109.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1-768x280.webp 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1-65x24.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1-225x82.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153123\/8222eebba73b30b683ca0265ea289ae54c5f3dd1-350x128.webp 350w\" sizes=\"auto, (max-width: 900px) 100vw, 900px\" \/><figcaption id=\"caption-attachment-1463\" class=\"wp-caption-text\">Figure 1. (a) Lines [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex] for [latex]K = 0, \\frac{\\pi}2, \u03c0,[\/latex] and [latex]\\frac{3\\pi}2[\/latex]. (b) Circles [latex]\\langle\\cos{u},\\sin{u},K\\rangle[\/latex] for [latex]K = \u22122, \u22121, 1,[\/latex] and [latex]2[\/latex]. (c) The lines and circles together. As [latex]u[\/latex] and [latex]v[\/latex] vary, they describe a cylinder.<\/figcaption><\/figure>\n<p>Notice that if [latex]x=\\cos{u}[\/latex] and [latex]y=\\sin{u}[\/latex], then [latex]x^{2}+y^{2}=1[\/latex], so points from [latex]S[\/latex] do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle [latex]\\langle\\cos{K},\\sin{K},v\\rangle[\/latex] for some [latex]k[\/latex], and therefore each point on the cylinder is contained in the parameterized surface (Figure 2).<\/p>\n<figure id=\"attachment_1465\" aria-describedby=\"caption-attachment-1465\" style=\"width: 376px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1465\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6.webp\" alt=\"An image of a vertical cylinder in three dimensions with the center of its circular base located on the z axis.\" width=\"376\" height=\"467\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6.webp 376w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6-242x300.webp 242w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6-65x81.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6-225x279.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153147\/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6-350x435.webp 350w\" sizes=\"auto, (max-width: 376px) 100vw, 376px\" \/><figcaption id=\"caption-attachment-1465\" class=\"wp-caption-text\">Figure 2. Cylinder [latex]x^{2}+y^{2}=r^{2}[\/latex] has parameterization [latex]{\\bf{r}}(u,v)=\\langle{r}\\cos{u},r\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi,-\\infty&lt;v&lt;\\infty[\/latex].<\/figcaption><\/figure>\n<p>[latex]\\\\[\/latex]<br \/>\n<strong>Analysis<\/strong><br \/>\n[latex]\\\\[\/latex]<br \/>\nNotice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to [latex]0\\leq{u}\\leq\\pi, 0 < v <6[\/latex], then the surface would be a half-cylinder of height 6.\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Parameterizing Cylinders<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">It follows from the example that we can parameterize all cylinders of the form [latex]x^2 + y^2 = R^2[\/latex]. If [latex]S[\/latex] is a cylinder given by equation [latex]x^2 + y^2 = R^2[\/latex], then a parameterization of [latex]S[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle R\\cos u, R\\sin u, v \\rangle, \\quad 0 \\leq u \\leq 2\\pi, \\quad -\\infty < v < \\infty[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1167793639868\">We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793571150\">Describe the surface [latex]S[\/latex] parameterized by [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u^2\\rangle,0\\le{u}\\le\\infty,0\\le{v}<2\\pi[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q384529048\">Show Solution<\/button><\/p>\n<div id=\"q384529048\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that if [latex]u[\/latex] is held constant, then the resulting curve is a circle of radius [latex]u[\/latex] in plane [latex]z=u[\/latex]. Therefore, as [latex]u[\/latex] increases, the radius of the resulting circle increases. If [latex]v[\/latex] is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x^2+y^2&=(u\\cos{v})^2+(u\\sin{v})^2 \\\\  &=u^2\\cos^2v+u^2\\sin^2v \\\\  &=u^2 \\\\  &=z  \\end{aligned}[\/latex].<\/p>\n<p>Therefore, the surface is elliptic paraboloid [latex]x^2+y^2=z[\/latex] (Figure 3).<\/p>\n<figure id=\"attachment_1467\" aria-describedby=\"caption-attachment-1467\" style=\"width: 619px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1467\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319.png\" alt=\"Two images in three dimensions. The first shows parallel circles on the z axis with radii increasing as z increases. Vertical parabolas opening up frame the circles, forming the skeleton of a paraboloid. The second shows the elliptic paraboloid, which is made of all the possible circles and vertical parabolas in the parameter domain.\" width=\"619\" height=\"440\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319.png 619w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319-300x213.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319-65x46.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319-225x160.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153555\/Screenshot-2025-10-17-113319-350x249.png 350w\" sizes=\"auto, (max-width: 619px) 100vw, 619px\" \/><figcaption id=\"caption-attachment-1467\" class=\"wp-caption-text\">Figure 3. (a) Circles arise from holding [latex]u[\/latex] constant; the vertical parabolas arise from holding [latex]v[\/latex] constant. (b) An elliptic paraboloid results from all choices of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Give a parameterization of the cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying on or above the plane [latex]z=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q989238235\">Show Solution<\/button><\/p>\n<div id=\"q989238235\" class=\"hidden-answer\" style=\"display: none\">The horizontal cross-section of the cone at height [latex]z=u[\/latex] is circle [latex]x^{2}+y^{2}=u^{2}[\/latex]. Therefore, a point on the cone at height [latex]u[\/latex] has coordinates [latex](u\\cos{v},u\\sin{v},u)[\/latex] for angle [latex]v[\/latex]. Hence, a parameterization of the cone is [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle[\/latex]. Since we are not interested in the entire cone, only the portion on or above plane [latex]z=-2[\/latex], the parameter domain is given by [latex]-2\\leq{u}<\\infty,0\\leq{v}<2\\pi[\/latex] (Figure 4).\n\n<figure id=\"attachment_1468\" aria-describedby=\"caption-attachment-1468\" style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1468\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067.webp\" alt=\"A three-dimensional diagram of the cone x^2 + y^2 = z^2, which opens up along the z axis for positive z values and opens down along the z axis for negative z values. The center is at the origin.\" width=\"343\" height=\"403\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067.webp 343w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067-255x300.webp 255w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067-65x76.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153705\/80b41e00683076a5a361597b58344c406e2d8067-225x264.webp 225w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><figcaption id=\"caption-attachment-1468\" class=\"wp-caption-text\">Figure 4. Cone [latex]x^{2}+y^{2}=r^{2}[\/latex] has parameterization [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle[\/latex], [latex]-\\infty&lt;u&lt;\\infty, 0\\leq{v}\\leq2\\pi[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<p>We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions.<\/p>\n<h3>Parameterizing Spheres<\/h3>\n<p class=\"whitespace-normal break-words\">To parameterize a sphere, it&#8217;s easiest to use spherical coordinates. The sphere of radius [latex]\\rho[\/latex] centered at the origin is given by:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(\\phi,\\theta) = \\langle \\rho\\cos\\theta\\sin\\phi, \\rho\\sin\\theta\\sin\\phi, \\rho\\cos\\phi \\rangle, \\quad 0 \\leq \\theta \\leq 2\\pi, \\quad 0 \\leq \\phi \\leq \\pi[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The idea is that as [latex]\\phi[\/latex] sweeps downward from the positive [latex]z[\/latex]-axis, a circle of radius [latex]\\rho\\sin\\phi[\/latex] is traced out by letting [latex]\\theta[\/latex] run from [latex]0[\/latex] to [latex]2\\pi[\/latex]. To see this, let [latex]\\phi[\/latex] be fixed. Then:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{aligned} x^2 + y^2 &= (\\rho\\cos\\theta\\sin\\phi)^2 + (\\rho\\sin\\theta\\sin\\phi)^2 \\\\ &= \\rho^2\\sin^2\\phi(\\cos^2\\theta + \\sin^2\\theta) \\\\ &= \\rho^2\\sin^2\\phi \\ &= (\\rho\\sin\\phi)^2 \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives the desired circle (Figure 5).<\/p>\n<figure id=\"attachment_1469\" aria-describedby=\"caption-attachment-1469\" style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1469\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68.webp\" alt=\"A three-dimensional diagram of the sphere of radius rho.\" width=\"343\" height=\"371\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68.webp 343w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68-277x300.webp 277w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68-65x70.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153804\/7b37adfac687f3ede4c12b1624213f7c408c0a68-225x243.webp 225w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><figcaption id=\"caption-attachment-1469\" class=\"wp-caption-text\">Figure 5. The sphere of radius [latex]\\rho[\/latex] has parameterization [latex]{\\bf{r}}(\\phi,\\theta)=\\langle\\rho\\cos\\theta\\sin\\phi,\\rho\\sin\\theta\\sin\\phi,\\rho\\cos\\phi\\rangle[\/latex],[latex] 0\\leq\\theta\\leq2\\pi, 0\\leq\\phi\\leq\\pi[\/latex].<\/figcaption><\/figure>\n<h3>Parameterizing Graphs of Functions<\/h3>\n<p class=\"whitespace-normal break-words\">To parameterize the graph of a two-variable function, let [latex]z = f(x,y)[\/latex] be a function of two variables. The simplest parameterization of the graph of [latex]f[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,y) = \\langle x, y, f(x,y) \\rangle[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]x[\/latex] and [latex]y[\/latex] vary over the domain of [latex]f[\/latex] (Figure 6).<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">For example, the graph of [latex]f(x,y) = x^2y[\/latex] can be parameterized by [latex]\\mathbf{r}(x,y) = \\langle x, y, x^2y \\rangle[\/latex], where the parameters [latex]x[\/latex] and [latex]y[\/latex] vary over the domain of [latex]f[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">If you only care about a piece of the graph of [latex]f[\/latex]\u2014say, the piece over rectangle [latex][1,3] \\times [2,5][\/latex]\u2014then you can restrict the parameter domain:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,y) = \\langle x, y, x^2y \\rangle, \\quad 1 \\leq x \\leq 3, \\quad 2 \\leq y \\leq 5[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Similarly, if [latex]S[\/latex] is a surface given by [latex]x = g(y,z)[\/latex] or [latex]y = h(x,z)[\/latex], then a parameterization of [latex]S[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(y,z) = \\langle g(y,z), y, z \\rangle[\/latex] or [latex]\\mathbf{r}(x,z) = \\langle x, h(x,z), z \\rangle[\/latex]<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">For example, the graph of paraboloid [latex]2y = x^2 + z^2[\/latex] can be parameterized by:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{r}(x,z) = \\left\\langle x, \\frac{x^2 + z^2}{2}, z \\right\\rangle, \\quad 0 \\leq x \\leq \\infty, \\quad 0 \\leq z \\leq \\infty[\/latex]<\/p>\n<p>Notice that we don&#8217;t need to vary over the entire domain of [latex]y[\/latex] because [latex]x[\/latex] and [latex]z[\/latex] are squared.<\/p>\n<\/section>\n<figure id=\"attachment_1470\" aria-describedby=\"caption-attachment-1470\" style=\"width: 423px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1470\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd.webp\" alt=\"A three-dimensional diagram of a surface z = f(x,y) above its mapping in the two-dimensional x,y plane. The point (x,y) in the plane corresponds to the point z = f(x,y) on the surface.\" width=\"423\" height=\"499\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd.webp 423w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd-254x300.webp 254w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd-65x77.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd-225x265.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153831\/0dc3336b5ac8a3f543f7960e6dffd31b165532bd-350x413.webp 350w\" sizes=\"auto, (max-width: 423px) 100vw, 423px\" \/><figcaption id=\"caption-attachment-1470\" class=\"wp-caption-text\">Figure 6. The simplest parameterization of the graph of a function is [latex]{\\bf{r}}(x,y)=\\langle{x},y,f(x,y)\\rangle[\/latex].<\/figcaption><\/figure>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Smooth and Regular Parameterizations<\/h3>\n<p class=\"whitespace-normal break-words\">Let&#8217;s generalize the notions of smoothness and regularity from curves to parametric surfaces.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Regular Parameterizations of Curves<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Curve parameterization [latex]\\mathbf{r}(t)[\/latex], [latex]a \\leq t \\leq b[\/latex], is <strong>regular<\/strong> if [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for all [latex]t[\/latex] in [latex][a,b][\/latex]. This condition ensures the image of [latex]\\mathbf{r}[\/latex] is actually a curve, not just a point.<\/p>\n<p class=\"whitespace-normal break-words\">For example, [latex]\\mathbf{r}(t) = \\langle 1, 2 \\rangle[\/latex], [latex]0 \\leq t \\leq 5[\/latex], is not regular because its image is just the point \u00a0[latex](1, 2)[\/latex]\u00a0and [latex]\\mathbf{r}'(t) = \\mathbf{0}[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">We need a similar notion of regularity for surfaces to ensure a surface parameterization traces out an actual surface. Consider the parameterization:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathbf{r}(u,v) = \\langle 0, \\cos v, 1 \\rangle, \\quad 0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\pi[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Although this appears to parameterize a surface, its image is actually just a line (Figure 7). Notice that [latex]\\mathbf{r}_u = \\langle 0, 0, 0 \\rangle[\/latex] and [latex]\\mathbf{r}_v = \\langle 0, -\\sin v, 0 \\rangle[\/latex], so the cross product [latex]\\mathbf{r}_u \\times \\mathbf{r}_v = \\mathbf{0}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The analog of the condition [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for curves is that [latex]\\mathbf{r}_u \\times \\mathbf{r}_v \\neq \\mathbf{0}[\/latex] for surfaces.<\/p>\n<figure id=\"attachment_1471\" aria-describedby=\"caption-attachment-1471\" style=\"width: 311px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1471\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046.webp\" alt=\"A three-dimensional diagram of a line on the x,z plane where the z component is 1, the x component is 1, and the y component exists between -1 and 1.\" width=\"311\" height=\"358\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046.webp 311w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046-261x300.webp 261w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046-65x75.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17153859\/745227e56c04a5df74ee49af2b05fce09d079046-225x259.webp 225w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><figcaption id=\"caption-attachment-1471\" class=\"wp-caption-text\">Figure 7. The image of parameterization [latex]{{\\bf{r}}(u,v)=\\langle0,\\cos{v},1\\rangle, \\ 0\\leq{u}\\leq1, \\ 0\\leq{v}\\leq\\pi}[\/latex] is a line.<\/figcaption><\/figure>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>regular parameterization<\/h3>\n<p>Parameterization [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] is a\u00a0<strong><span id=\"505d9884-54a9-4157-83ff-87887335091b_term273\" data-type=\"term\">regular parameterization<\/span><\/strong>\u00a0if [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for point [latex](u, v)[\/latex] in the parameter domain.<\/p>\n<\/section>\n<p>If parameterization [latex]\\mathbf{r}[\/latex] is regular, then the image of [latex]\\mathbf{r}[\/latex] is a two-dimensional object, as a surface should be. Throughout this module, we assume all parameterizations [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] are regular.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Smooth Parameterizations of Curves<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Curve parameterization [latex]\\mathbf{r}(t)[\/latex], [latex]a \\leq t \\leq b[\/latex], is <strong>smooth<\/strong> if [latex]\\mathbf{r}'(t)[\/latex] is continuous and [latex]\\mathbf{r}'(t) \\neq \\mathbf{0}[\/latex] for all [latex]t[\/latex] in [latex][a,b][\/latex]. Informally, a curve parameterization is smooth if the resulting curve has no sharp corners.<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>smooth\u00a0parameterization<\/h3>\n<p>A surface parameterization [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] is\u00a0<strong>smooth\u00a0<\/strong>if vector [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain.<\/p>\n<\/section>\n<p>A surface may also be <strong>piecewise smooth<\/strong> if it has smooth faces but also has locations where the directional derivatives do not exist.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Which of the figures in\u00a0Figure 8\u00a0is smooth?<\/p>\n<figure id=\"attachment_1474\" aria-describedby=\"caption-attachment-1474\" style=\"width: 882px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1474\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3.webp\" alt=\"Two three-dimensional figures. The first surface is smooth. It looks like a tire with a large hole in the middle. The second is piecewise smooth. It is a pyramid with a rectangular base and four sides.\" width=\"882\" height=\"410\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3.webp 882w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3-300x139.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3-768x357.webp 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3-65x30.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3-225x105.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154252\/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3-350x163.webp 350w\" sizes=\"auto, (max-width: 882px) 100vw, 882px\" \/><figcaption id=\"caption-attachment-1474\" class=\"wp-caption-text\">Figure 8. (a) This surface is smooth. (b) This surface is piecewise smooth.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q248348349\">Show Solution<\/button><\/p>\n<div id=\"q248348349\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793373625\">The surface in\u00a0Figure 8(a) can be parameterized by<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}(u,v)=\\langle(2+\\cos{v})\\cos{u},(2+\\cos{v})\\sin{u},\\sin{v}\\rangle \\ 0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq2\\pi[\/latex]<\/p>\n<p id=\"fs-id1167793900590\">(we can use technology to verify). Notice that vectors<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}_u=\\langle-(2+\\cos{v})\\sin{u},(2+\\cos{v})\\cos{u},0\\rangle\\text{ and }{\\bf{r}}_v=\\langle-\\sin{v}\\cos{u},-\\sin{v}\\sin{u},\\cos{v}\\rangle[\/latex]<\/p>\n<p id=\"fs-id1167794070993\">exist for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain, and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{r}}_v&=\\begin{vmatrix}  {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\  -(2+\\cos{v})\\sin{u}&(2+\\cos{v})\\cos{u}&0 \\\\  -\\sin{v}\\cos{u}&-\\sin{v}\\sin{u}&\\cos{v}  \\end{vmatrix} \\\\  &=[(2+\\cos{v})\\cos{u}\\cos{v}]{\\bf{i}}+[(2+\\cos{v})\\sin{u}\\cos{v}]{\\bf{j}}+[(2+\\cos{v})\\sin{v}\\sin^2u+(2+\\cos{v})\\sin{v}\\cos^2u]{\\bf{k}} \\\\  &=[(2+\\cos{v})\\cos{u}\\cos{v}]{\\bf{i}}+[(2+\\cos{v})\\sin{u}\\cos{v}]{\\bf{j}}+[(2+\\cos{v})\\sin{v}]{\\bf{k}}  \\end{aligned}[\/latex]<\/p>\n<p>.<\/p>\n<p id=\"fs-id1167794126254\">The\u00a0[latex]\\bf{k}[\/latex]\u00a0component of this vector is zero only if [latex]v=0[\/latex] or [latex]v=\\pi[\/latex]. If [latex]v=0[\/latex] or [latex]v=\\pi[\/latex], then the only choices for [latex]u[\/latex] that make the [latex]\\bf{j}[\/latex] component zero are [latex]u=0[\/latex] or [latex]u=\\pi[\/latex]. But, these choices of [latex]u[\/latex] do not make the [latex]\\bf{i}[\/latex] component zero. Therefore, [latex]{\\bf{r}}_u\\times{\\bf{r}}_v[\/latex] is not zero for any choice of [latex]u[\/latex] and [latex]v[\/latex] in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.<\/p>\n<p id=\"fs-id1167794126003\">In the pyramid in\u00a0Figure 8(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. 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