{"id":5504,"date":"2022-06-02T18:56:46","date_gmt":"2022-06-02T18:56:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=5504"},"modified":"2022-11-01T05:29:59","modified_gmt":"2022-11-01T05:29:59","slug":"parametric-surfaces","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/parametric-surfaces\/","title":{"raw":"Parametric Surfaces","rendered":"Parametric Surfaces"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Find the parametric representations of a cylinder, a cone, and a sphere.<\/span><\/li>\r\n \t
  • Describe the surface integral of a scalar-valued function over a parametric surface.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n
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    Parametric Surfaces<\/h2>\r\n

    A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.<\/p>\r\n

    However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve [latex]C[\/latex], we first need to parameterize [latex]C[\/latex]. In a similar way, to calculate a surface integral over surface [latex]S[\/latex], we need to parameterize [latex]S[\/latex]. That is, we need a working concept of a\u00a0parameterized surface<\/span><\/strong>\u00a0(or a\u00a0parametric surface<\/span><\/strong>), in the same way that we already have a concept of a parameterized curve.<\/p>\r\n

    A parameterized surface is given by a description of the form<\/p>\r\n

    [latex]\\large{{\\bf{r}}(u,v)=\\langle{x}(u,v),y(u,v),z(u,v)\\rangle}[\/latex].<\/p>\r\n

    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. The parameters [latex]u[\/latex] and [latex]v[\/latex] vary over a region called the parameter domain, or\u00a0parameter space<\/span><\/strong>\u2014the set of points in the [latex]uv[\/latex]-plane that can be substituted into\u00a0[latex]{\\bf{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 a 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

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    definition<\/h3>\r\n\r\n
    \r\n\r\nGiven a parameterization of surface [latex]{\\bf{r}}(u,v)=\\langle{x}(u,v),y(u,v),z(u,v)\\rangle[\/latex], the\u00a0parameter 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<\/div>\r\n
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    Example: parameterizing a cylinder<\/h3>\r\n

    Describe the surface [latex]S[\/latex] parameterized by<\/p>\r\n

    [latex]\\large{{\\bf{r}}(u,v)=\\langle\\cos{u},\\sin{u},v\\rangle,-\\infty<u<\\infty,-\\infty<v<\\infty}[\/latex].<\/p>\r\n[reveal-answer q=\"466382191\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"466382191\"]\r\n

    Solution<\/span><\/h2>\r\n
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    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].<\/p>\r\n

    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

    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

    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

    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_5335\" align=\"aligncenter\" width=\"900\"]\"<img 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_5338\" align=\"aligncenter\" width=\"376\"]\"<img 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<v<\\infty[\/latex].[\/caption]\r\n

    Analysis<\/h2>\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<v<6[\/latex], then the surface would be a half-cylinder of height 6.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    try it<\/h3>\r\nDescribe 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].\r\n\r\n[reveal-answer q=\"537294031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"537294031\"]\r\n\r\nCylinder\u00a0[latex]x^{2}+y^{2}=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    It follows from\u00a0Example \"Parameterizing a Cylinder\"\u00a0that 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

    [latex]\\large{{\\bf{r}}(u,v)=\\langle{R}\\cos{u},R\\sin{u},v\\rangle,0\\leq{u}\\leq2\\pi,-\\infty<v<\\infty}[\/latex].<\/p>\r\n

    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

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    Example: describing a surface<\/h3>\r\n
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    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>\r\n\r\n<\/div>\r\n<\/div>\r\n[reveal-answer q=\"384529048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384529048\"]\r\n

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    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

    [latex]\\begin{aligned}\r\nx^2+y^2&=(u\\cos{v})^2+(u\\sin{v})^2 \\\\\r\n&=u^2\\cos^2v+u^2\\sin^2v \\\\\r\n&=u^2 \\\\\r\n&=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_5348\" align=\"aligncenter\" width=\"622\"]\"<img 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]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nDescribe 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].\r\n\r\n[reveal-answer q=\"759828461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759828461\"]\r\n\r\nCone [latex]x^{2}+y^{2}=z^{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Example: finding a parameterization<\/h3>\r\nGive a parameterization of the cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying on or above the plane [latex]z=-2[\/latex].\r\n\r\n[reveal-answer q=\"989238235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989238235\"]\r\n\r\nThe 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).\r\n\r\n[caption id=\"attachment_5349\" align=\"aligncenter\" width=\"343\"]\"<img 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<u<\\infty, 0\\leq{v}\\leq2\\pi[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n
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    try it<\/h3>\r\nGive a parameterization for the portion of cone [latex]x^{2}+y^{2}=z^{2}[\/latex] lying in the first octant.\r\n\r\n[reveal-answer q=\"893249821\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"893249821\"]\r\n\r\n[latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},u\\rangle,0<u<\\infty,0\\leq{v}<\\frac{\\pi}2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n