Suppose [latex]i[\/latex] ranges from [latex]1[\/latex] to [latex]m[\/latex] and [latex]j[\/latex] ranges from [latex]1[\/latex] to [latex]n[\/latex], so [latex]D[\/latex] is subdivided into [latex]mn[\/latex] rectangles. This division of [latex]D[\/latex] into subrectangles gives a corresponding division of surface [latex]S[\/latex] into pieces [latex]S_{ij}[\/latex]. Choose point [latex]P_{ij}[\/latex] in each piece [latex]S_{ij}[\/latex]. Point [latex]P_{ij}[\/latex] corresponds to point [latex](u_i, v_j)[\/latex] in the parameter domain.<\/p>\r\n\r\n
Grid Curves and Tangent Vectors<\/h3>\r\n
We can form a grid with lines parallel to the [latex]u[\/latex]-axis and [latex]v[\/latex]-axis in the [latex]uv[\/latex]-plane. These grid lines correspond to a set of grid curves<\/strong> on surface [latex]S[\/latex] that is parameterized by [latex]\\mathbf{r}(u,v)[\/latex].<\/p>\r\n Without loss of generality, assume [latex]P_{ij}[\/latex] is located at the corner of two grid curves, as in Figure 9. If we think of [latex]\\mathbf{r}[\/latex] as a mapping from the [latex]uv[\/latex]-plane to [latex]\\mathbb{R}^3[\/latex], the grid curves are the image of the grid lines under [latex]\\mathbf{r}[\/latex].<\/p>\r\n Consider the grid lines through point [latex](u_i, v_j)[\/latex]. One line is given by [latex]x = u_i[\/latex], [latex]y = v[\/latex]; the other is given by [latex]x = u[\/latex], [latex]y = v_j[\/latex]. In the first grid line, the horizontal component is held constant, yielding a vertical line through [latex](u_i, v_j)[\/latex]. In the second grid line, the vertical component is held constant, yielding a horizontal line through [latex](u_i, v_j)[\/latex]. The corresponding grid curves are [latex]\\mathbf{r}(u_i, v)[\/latex] and [latex]\\mathbf{r}(u, v_j)[\/latex], and these curves intersect at point [latex]P_{ij}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1476\" align=\"aligncenter\" width=\"727\"] Now consider the vectors that are tangent to these grid curves. For grid curve [latex]{\\bf{r}}(u_i,v)[\/latex] the tangent vector at [latex]P_{ij}[\/latex] is:<\/p>\r\n [latex]{\\bf{t}}_v(P_{ij})={\\bf{r}}_v(u_i,v_j)=\\langle{x}_v(u_i,v_j),y_v(u_i,v_j),z_v(u_i,v_j)\\rangle[\/latex].<\/p>\r\n For grid curve [latex]{\\bf{r}}(u,v_j)[\/latex], the tangent vector at [latex]P_{ij}[\/latex] is:<\/p>\r\n [latex]{\\bf{t}}_u(P_{ij})={\\bf{r}}_y(u_i,v_j)=\\langle{x}_v(u_i,v_j),y_u(u_i,v_j),z_u(u_i,v_j)\\rangle[\/latex].<\/p>\r\n If vector [latex]{\\bf{N}}={\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_u(P_{ij}[\/latex]\u00a0exists and is not zero, then the tangent plane at [latex]P_{ij}[\/latex] exists (Figure 10).<\/p>\r\n If piece [latex]S_{ij}[\/latex] is small enough, then the tangent plane at point [latex]P_{ij}[\/latex] is a good approximation of piece [latex]S_{ij}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1477\" align=\"aligncenter\" width=\"636\"] The tangent plane at [latex]P_{ij}[\/latex] contains vectors [latex]\\mathbf{t}_u(P_{ij})[\/latex] and [latex]\\mathbf{t}_v(P_{ij})[\/latex]. Therefore, the parallelogram spanned by these two vectors lies in the tangent plane.<\/p>\r\n Since the original rectangle in the [latex]uv[\/latex]-plane corresponding to [latex]S_{ij}[\/latex] has width [latex]\\Delta u[\/latex] and length [latex]\\Delta v[\/latex], we approximate [latex]S_{ij}[\/latex] using the parallelogram spanned by [latex]\\Delta{u}{\\bf{t}}_u(P_{ij})[\/latex] and [latex]\\Delta{v}{\\bf{t}}_v(P_{ij})[\/latex].<\/p>\r\n In other words, we scale the tangent vectors by the constants [latex]\\Delta u[\/latex] and [latex]\\Delta v[\/latex] to match the scale of the original division of rectangles in the parameter domain.<\/p>\r\n Therefore, the area of the parallelogram used to approximate the area of [latex]S_{ij}[\/latex] is:<\/p>\r\n [latex]\\Delta{S_{ij}}\\approx||(\\Delta{u}{\\bf{t}}_u(P_{ij}))\\times(\\Delta{v}{\\bf{t}}_v(P_{ij}))=||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex].<\/p>\r\n Varying point [latex]P_{ij}[\/latex] over all pieces [latex]S_{ij}[\/latex] and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 11).<\/p>\r\n\r\n[caption id=\"attachment_1478\" align=\"aligncenter\" width=\"255\"] Let [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] with parameter domain [latex]D[\/latex] be a smooth parameterization of surface [latex]S[\/latex]. Furthermore, assume that [latex]S[\/latex] is traced out only once as [latex](u, v)[\/latex] varies over [latex]D[\/latex]. The\u00a0surface area<\/span><\/strong> of [latex]S[\/latex] is:<\/p>\r\n\r\n [latex]\\large{\\displaystyle\\iint_D||{\\bf{t}}_u\\times{\\bf{t}}_v||dA}[\/latex],<\/p>\r\nwhere\u00a0[latex]{\\bf{t}}_u=\\left\\langle\\frac{\\partial{x}}{\\partial{u}},\\frac{\\partial{y}}{\\partial{u}},\\frac{\\partial{z}}{\\partial{u}}\\right\\rangle\\text{ and }{\\bf{t}}_v=\\left\\langle\\frac{\\partial{x}}{\\partial{v}},\\frac{\\partial{y}}{\\partial{v}},\\frac{\\partial{z}}{\\partial{v}}\\right\\rangle[\/latex].\r\n\r\n<\/section> [latex]{\\bf{s}}(u,v)=\\langle{k}v\\cos{u},kv\\sin{u},v\\rangle, \\ 0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq{h}[\/latex].<\/p>\r\n The idea behind this parameterization is that for a fixed [latex]v[\/latex] value, the circle swept out by letting [latex]u[\/latex] vary is the circle at height [latex]v[\/latex] and radius [latex]kv[\/latex]. As [latex]v[\/latex] increases, the parameterization sweeps out a \u201cstack\u201d of circles, resulting in the desired cone.<\/p>\r\n\r\n[caption id=\"attachment_1479\" align=\"aligncenter\" width=\"262\"] With a parameterization in hand, we can calculate the surface area of the cone using the surface area of a parametric surface equation. The tangent vectors are [latex]{\\bf{t}}_u=\\langle-kv\\sin{u},kv\\cos{u},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle{k}\\cos{u},k\\sin{u},1\\rangle[\/latex]. Therefore,<\/p>\r\n [latex]\\begin{aligned}\r\n{\\bf{t}}_u\\times{\\bf{t}}_v&=\\begin{vmatrix}\r\n{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\\r\n-kv\\sin{u}&kv\\cos{u}&0 \\\\\r\n{k}\\cos{u}&k\\sin{u}&1\r\n\\end{vmatrix} \\\\\r\n&=\\langle{k}v\\cos{u},kv\\sin{u},-k^2v\\sin^2y-k^2v\\cos^2u\\rangle \\\\\r\n&=\\langle{k}v\\cos{u},kv\\sin{u},-k^2v\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\n The magnitude of this vector is<\/p>\r\n [latex]\\begin{aligned}\r\n||\\langle{k}v\\cos{u},kv\\sin{u},-k^2v\\rangle||&=\\sqrt{k^2v^2\\cos^2u+k^2v^2\\sin^2u+k^4v^2} \\\\\r\n&=\\sqrt{k^2v^2+k^4v^2} \\\\\r\n&=kv\\sqrt{1+k^2}\r\n\\end{aligned}[\/latex].<\/p>\r\n By the surface area of a parametric surface equation, the surface area of the cone is<\/p>\r\n [latex]\\begin{aligned}\r\n\\displaystyle\\iint_D||{\\bf{t}}_u\\times{\\bf{t}}_v||dA&=\\displaystyle\\int_0^h\\displaystyle\\int_0^{2\\pi}kv\\sqrt{1+k^2}dudv \\\\\r\n&=2\\pi{k}\\sqrt{1+k^2}\\displaystyle\\int_0^hvdv \\\\\r\n&=2\\pi{k}\\sqrt{1+k^2}\\left[\\frac{v^2}2\\right]_0^h \\\\\r\n&=\\pi{k}h^2\\sqrt{1+k^2}\r\n\\end{aligned}[\/latex].<\/p>\r\n Since [latex]k=\\tan\\alpha=r\/h[\/latex],<\/p>\r\n [latex]\\begin{aligned}\r\n\\pi{k}h^2\\sqrt{1+k^2}&=\\pi\\frac{r}hh^2\\sqrt{1+\\frac{r^2}{h^2}} \\\\\r\n&=\\pi{r}h\\sqrt{1+\\frac{r^2}{h^2}} \\\\\r\n&=\\pi{r}\\sqrt{h^2+h^2\\left(\\frac{r^2}{h^2}\\right)} \\\\\r\n&=\\pi{r}\\sqrt{h^2+r^2}\r\n\\end{aligned}[\/latex].<\/p>\r\n Therefore, the lateral surface area of the cone is [latex]\\pi{r}\\sqrt{h^2+r^2}[\/latex].<\/p>\r\n Analysis<\/span><\/strong><\/p>\r\n The surface area of a right circular cone with radius [latex]r[\/latex] and height [latex]h[\/latex] is usually given as [latex]\\pi{r}^2+\\pi{r}\\sqrt{h^2+r^2}[\/latex]. The reason for this is that the circular base is included as part of the cone, and therefore the area of the base [latex]\\pi{r}^2[\/latex] is added to the lateral surface area [latex]\\pi{r}\\sqrt{h^2+r^2}[\/latex] that we found.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section> Show that the surface area of the sphere [latex]x^{2}+y^{2}+z^{2}=r^{2}[\/latex] is [latex]4\\pi{r}^2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n[reveal-answer q=\"452745923\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"452745923\"]\r\n The sphere has parameterization<\/p>\r\n [latex]\\langle{r}\\cos\\theta\\sin\\phi,r\\sin\\theta\\sin\\phi,r\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi[\/latex].<\/p>\r\n The tangent vectors are<\/p>\r\n [latex]{\\bf{t}}_\\theta=\\langle-r\\sin\\theta\\sin\\phi,r\\cos\\theta\\sin\\phi,0\\rangle\\text{ and }{\\bf{t}}_\\phi=\\langle{r}\\cos\\theta\\cos\\phi,r\\sin\\theta\\cos\\phi,-r\\sin\\phi\\rangle[\/latex].<\/p>\r\n Therefore,<\/p>\r\n [latex]\\begin{aligned}\r\n{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta&=\\langle{r}^2\\cos\\theta\\sin^2\\phi,r^2\\sin\\theta\\sin^2\\phi,r^2\\sin^2\\theta\\sin\\phi\\cos\\phi+r^2\\cos^2\\theta\\sin\\phi\\cos\\phi\\rangle \\\\\r\n&=\\langle{r}^2\\cos\\theta\\sin^2\\phi,r^2\\sin\\theta\\sin^2\\phi,r^2\\sin\\phi\\cos\\phi\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\n Now,<\/p>\r\n [latex]\\begin{aligned}\r\n||{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta||&=\\sqrt{r^4\\sin^4\\phi\\cos^2\\theta+r^4\\sin^4\\phi\\sin^2\\theta+r^4\\sin^2\\phi\\cos^2\\phi} \\\\\r\n&=\\sqrt{r^4\\sin^4\\phi+r^4\\sin^2\\phi\\cos^2\\phi} \\\\\r\n&=r^2\\sqrt{\\sin^2\\phi} \\\\\r\n&=r^2\\sin\\phi\r\n\\end{aligned}[\/latex].<\/p>\r\n Notice that [latex]\\phi\\geq0[\/latex] on the parameter domain because [latex]0\\leq\\phi\\leq\\pi[\/latex], and this justifies equation [latex]\\sqrt{\\sin^2\\phi}=\\sin\\phi[\/latex]. The surface area of the sphere is<\/p>\r\n [latex]\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi}r^2\\sin\\phi{d}\\phi{d}\\theta=r^2\\displaystyle\\int_0^{2\\pi}2d\\theta=4\\pi{r}^2[\/latex].<\/p>\r\n We have derived the familiar formula for the surface area of a sphere using surface integrals.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. This lets us calculate the surface area of a surface of revolution using the same techniques.<\/p>\r\n Let [latex]y = f(x) \\geq 0[\/latex] be a positive single-variable function on the domain [latex]a \\leq x \\leq b[\/latex], and let [latex]S[\/latex] be the surface obtained by rotating [latex]f[\/latex] about the [latex]x[\/latex]-axis (Figure 13). Let [latex]\\theta[\/latex] be the angle of rotation. Then [latex]S[\/latex] can be parameterized with parameters [latex]x[\/latex] and [latex]\\theta[\/latex] by:<\/p>\r\n [latex]\\mathbf{r}(x,\\theta) = \\langle x, f(x)\\cos\\theta, f(x)\\sin\\theta \\rangle, \\quad a \\leq x \\leq b, \\quad 0 \\leq \\theta \\leq 2\\pi[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n This surface has parameterization<\/p>\r\n [latex]{\\bf{r}}(x,\\theta)\\langle{x},x^2\\cos\\theta,x^2\\sin\\theta\\rangle, \\ 0\\leq{x}\\leq{b}, \\ 0\\leq{x}\\leq2\\pi[\/latex].<\/p>\r\n The tangent vectors are<\/p>\r\n[latex]{\\bf{t}}_x=\\langle1,2x\\cos\\theta,2x\\sin\\theta\\rangle\\text{ and }{\\bf{t}}_\\theta=\\langle0,-x^2\\sin\\theta,-x^2\\cos\\theta\\rangle[\/latex].\r\n\r\nTherefore,\r\n [latex]\\begin{aligned}\r\n{\\bf{t}}_x\\times{\\bf{t}}_\\theta&=\\langle2x^3\\cos^2\\theta+2x^3\\sin^2\\theta,-x^2\\cos\\theta,-x^2\\sin\\theta\\rangle \\\\\r\n&=\\langle2x^3,-x^2\\cos\\theta,-x^2\\sin\\theta\\rangle\r\n\\end{aligned}[\/latex]<\/p>\r\n and<\/p>\r\n [latex]\\begin{aligned}\r\n{\\bf{t}}_x\\times{\\bf{t}}_\\theta&=\\sqrt{4x^6+x^4\\cos^2\\theta+x^4\\sin^2\\theta} \\\\\r\n&=\\sqrt{4x^6+x^4} \\\\\r\n&=x^2\\sqrt{4x^2+1}\r\n\\end{aligned}[\/latex].<\/p>\r\n The area of the surface of revolution is<\/p>\r\n [latex]\\begin{aligned}\r\n\\displaystyle\\int_0^b\\int_0^\\pi{x}^2\\sqrt{4x^2+1}d\\theta{d}x&=2\\pi\\int_0^bx^2\\sqrt{4x^2+1}dx \\\\\r\n&=2\\pi\\left[\\frac1{64}(2\\sqrt{4x^2+1}(8x^3+x)\\sin{h}^{-1}(2x))\\right]_0^b \\\\\r\n&=2\\pi\\left[\\frac1{64}(2\\sqrt{4b^2+1}(8b^3+b)\\sin{h}^{-1}(2b))\\right]\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":" Now that we can parameterize surfaces, we need to define the surface area of a parametric surface. This definition provides the foundation for surface integrals.<\/p>\n Let [latex]S[\/latex] be a surface with parameterization [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] over some parameter domain [latex]D[\/latex]. We assume the surface parameterization is continuously differentiable<\/strong>\u2014meaning each component function has continuous partial derivatives. For simplicity, assume [latex]D[\/latex] is a rectangle (this material extends to nonrectangular domains). Divide rectangle [latex]D[\/latex] into subrectangles [latex]D_{ij}[\/latex] with horizontal width [latex]\\Delta u[\/latex] and vertical length [latex]\\Delta v[\/latex].<\/p>\n Suppose [latex]i[\/latex] ranges from [latex]1[\/latex] to [latex]m[\/latex] and [latex]j[\/latex] ranges from [latex]1[\/latex] to [latex]n[\/latex], so [latex]D[\/latex] is subdivided into [latex]mn[\/latex] rectangles. This division of [latex]D[\/latex] into subrectangles gives a corresponding division of surface [latex]S[\/latex] into pieces [latex]S_{ij}[\/latex]. Choose point [latex]P_{ij}[\/latex] in each piece [latex]S_{ij}[\/latex]. Point [latex]P_{ij}[\/latex] corresponds to point [latex](u_i, v_j)[\/latex] in the parameter domain.<\/p>\n We can form a grid with lines parallel to the [latex]u[\/latex]-axis and [latex]v[\/latex]-axis in the [latex]uv[\/latex]-plane. These grid lines correspond to a set of grid curves<\/strong> on surface [latex]S[\/latex] that is parameterized by [latex]\\mathbf{r}(u,v)[\/latex].<\/p>\n Without loss of generality, assume [latex]P_{ij}[\/latex] is located at the corner of two grid curves, as in Figure 9. If we think of [latex]\\mathbf{r}[\/latex] as a mapping from the [latex]uv[\/latex]-plane to [latex]\\mathbb{R}^3[\/latex], the grid curves are the image of the grid lines under [latex]\\mathbf{r}[\/latex].<\/p>\n Consider the grid lines through point [latex](u_i, v_j)[\/latex]. One line is given by [latex]x = u_i[\/latex], [latex]y = v[\/latex]; the other is given by [latex]x = u[\/latex], [latex]y = v_j[\/latex]. In the first grid line, the horizontal component is held constant, yielding a vertical line through [latex](u_i, v_j)[\/latex]. In the second grid line, the vertical component is held constant, yielding a horizontal line through [latex](u_i, v_j)[\/latex]. The corresponding grid curves are [latex]\\mathbf{r}(u_i, v)[\/latex] and [latex]\\mathbf{r}(u, v_j)[\/latex], and these curves intersect at point [latex]P_{ij}[\/latex].<\/p>\n Now consider the vectors that are tangent to these grid curves. For grid curve [latex]{\\bf{r}}(u_i,v)[\/latex] the tangent vector at [latex]P_{ij}[\/latex] is:<\/p>\n [latex]{\\bf{t}}_v(P_{ij})={\\bf{r}}_v(u_i,v_j)=\\langle{x}_v(u_i,v_j),y_v(u_i,v_j),z_v(u_i,v_j)\\rangle[\/latex].<\/p>\n For grid curve [latex]{\\bf{r}}(u,v_j)[\/latex], the tangent vector at [latex]P_{ij}[\/latex] is:<\/p>\n [latex]{\\bf{t}}_u(P_{ij})={\\bf{r}}_y(u_i,v_j)=\\langle{x}_v(u_i,v_j),y_u(u_i,v_j),z_u(u_i,v_j)\\rangle[\/latex].<\/p>\n If vector [latex]{\\bf{N}}={\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_u(P_{ij}[\/latex]\u00a0exists and is not zero, then the tangent plane at [latex]P_{ij}[\/latex] exists (Figure 10).<\/p>\n If piece [latex]S_{ij}[\/latex] is small enough, then the tangent plane at point [latex]P_{ij}[\/latex] is a good approximation of piece [latex]S_{ij}[\/latex].<\/p>\n The tangent plane at [latex]P_{ij}[\/latex] contains vectors [latex]\\mathbf{t}_u(P_{ij})[\/latex] and [latex]\\mathbf{t}_v(P_{ij})[\/latex]. Therefore, the parallelogram spanned by these two vectors lies in the tangent plane.<\/p>\n Since the original rectangle in the [latex]uv[\/latex]-plane corresponding to [latex]S_{ij}[\/latex] has width [latex]\\Delta u[\/latex] and length [latex]\\Delta v[\/latex], we approximate [latex]S_{ij}[\/latex] using the parallelogram spanned by [latex]\\Delta{u}{\\bf{t}}_u(P_{ij})[\/latex] and [latex]\\Delta{v}{\\bf{t}}_v(P_{ij})[\/latex].<\/p>\n In other words, we scale the tangent vectors by the constants [latex]\\Delta u[\/latex] and [latex]\\Delta v[\/latex] to match the scale of the original division of rectangles in the parameter domain.<\/p>\n Therefore, the area of the parallelogram used to approximate the area of [latex]S_{ij}[\/latex] is:<\/p>\n [latex]\\Delta{S_{ij}}\\approx||(\\Delta{u}{\\bf{t}}_u(P_{ij}))\\times(\\Delta{v}{\\bf{t}}_v(P_{ij}))=||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex].<\/p>\n Varying point [latex]P_{ij}[\/latex] over all pieces [latex]S_{ij}[\/latex] and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 11).<\/p>\n Let [latex]{\\bf{r}}(u,v)=\\langle(x(u,v),y(u,v),z(u,v)\\rangle[\/latex] with parameter domain [latex]D[\/latex] be a smooth parameterization of surface [latex]S[\/latex]. Furthermore, assume that [latex]S[\/latex] is traced out only once as [latex](u, v)[\/latex] varies over [latex]D[\/latex]. The\u00a0surface area<\/span><\/strong> of [latex]S[\/latex] is:<\/p>\n [latex]\\large{\\displaystyle\\iint_D||{\\bf{t}}_u\\times{\\bf{t}}_v||dA}[\/latex],<\/p>\n where\u00a0[latex]{\\bf{t}}_u=\\left\\langle\\frac{\\partial{x}}{\\partial{u}},\\frac{\\partial{y}}{\\partial{u}},\\frac{\\partial{z}}{\\partial{u}}\\right\\rangle\\text{ and }{\\bf{t}}_v=\\left\\langle\\frac{\\partial{x}}{\\partial{v}},\\frac{\\partial{y}}{\\partial{v}},\\frac{\\partial{z}}{\\partial{v}}\\right\\rangle[\/latex].<\/p>\n<\/section>\n![\"Two](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154425\/554bdb200e243c7a7fe7695f959bcc99d6c61c86.webp\")
Figure 9. Grid lines on a parameter domain correspond to grid curves on a surface.[\/caption]\r\n![\"Two](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154449\/7a8f20e54fbe300cbc4ec7c1c79f355a7c0f670a.webp\")
Figure 10. If the cross product of vectors [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex] exists, then there is a tangent plane.[\/caption]\r\nApproximating Surface Area<\/h3>\r\n
![\"A](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154515\/5446eff034279beb6c8d1c545bd42983566a72fb.webp\")
Figure 11. The parallelogram spanned by [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex] approximates the piece of surface [latex]S_{ij}[\/latex].[\/caption]surface area of a parametric surface<\/h3>\r\n
Before calculating the surface area of this cone using the surface area of a parametric surface equation, we need a parameterization. We assume this cone is in [latex]\\mathbb{R}^3[\/latex] with its vertex at the origin (Figure 12). To obtain a parameterization, let [latex]\\alpha[\/latex] be the angle that is swept out by starting at the positive [latex]z[\/latex]-axis and ending at the cone, and let [latex]k=\\tan\\alpha[\/latex]. For a height value [latex]v[\/latex] with [latex]0\\leq{v}\\leq{h}[\/latex], the radius of the circle formed by intersecting the cone with plane [latex]z=v[\/latex] is [latex]kv[\/latex]. Therefore, a parameterization of this cone is<\/span><\/h2>\r\n
![\"A](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154604\/b15442ba83e41b09bb6363dbd68df733254ba936.webp\")
Figure 12. The right circular cone with radius [latex]r = kh[\/latex] and height [latex]h[\/latex] has parameterization [latex]{\\bf{s}}(u,v)=\\langle{k}v\\cos{u},kv\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, 0\\leq{v}\\leq{h}[\/latex].[\/caption]\r\nSurfaces of Revolution<\/h3>\r\n
![\"Two](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154706\/62e003469f226d24fb2fc7ce6d152985d58f6522.webp\")
Figure 13. We can parameterize a surface of revolution by [latex]{\\bf{r}}(x,\\theta)=\\langle{x},f(x)\\cos\\theta,f(x)\\sin\\theta\\rangle[\/latex], [latex]a\\leq{x}\\leq{b}, 0\\leq{x}\\leq2\\pi[\/latex].[\/caption]<\/div>\r\n![\"A](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17154958\/7d61bcda55b02e38101709cea6667b3baa333904.webp\")
Figure 14. A surface integral can be used to calculate the surface area of this solid of revolution.[\/caption]\r\n\r\n[reveal-answer q=\"538742370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538742370\"]\r\nSurface Area of a Parametric Surface<\/h2>\n
Grid Curves and Tangent Vectors<\/h3>\n
Approximating Surface Area<\/h3>\n
surface area of a parametric surface<\/h3>\n