{"id":1127,"date":"2021-11-09T23:42:51","date_gmt":"2021-11-09T23:42:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=1127"},"modified":"2022-11-01T05:31:33","modified_gmt":"2022-11-01T05:31:33","slug":"surface-integral-of-a-vector-field","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/surface-integral-of-a-vector-field\/","title":{"raw":"Surface Integral of a Vector Field","rendered":"Surface Integral of a Vector Field"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Explain the meaning of an oriented surface, giving an example.<\/span><\/li>\r\n \t
  • Describe the surface integral of a vector field.<\/span><\/li>\r\n \t
  • Use surface integrals to solve applied problems.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Orientation of a Surface<\/h2>\r\n

    Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an \u201corientation\u201d of the surface of integration.<\/p>\r\n

    On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an \u201cupward\u201d or \u201cdownward\u201d orientation or, in the case of surfaces such as a sphere or cylinder, an \u201coutward\u201d or \u201cinward\u201d orientation.<\/p>\r\n

    Let [latex]S[\/latex] be a smooth surface. For any point [latex](x, y, z)[\/latex] on [latex]S[\/latex], we can identify two unit normal vectors [latex]N[\/latex] and [latex]-N[\/latex]. If it is possible to choose a unit normal vector\u00a0[latex]{\\bf{N}}[\/latex]\u00a0at every point [latex](x, y, z)[\/latex] on [latex]S[\/latex] so that\u00a0[latex]{\\bf{N}}[\/latex]\u00a0varies continuously over [latex]S[\/latex], then [latex]S[\/latex] is \u201corientable<\/em>.\u201d Such a choice of unit normal vector at each point gives the\u00a0orientation of a surface [latex]S[\/latex].\u00a0<\/span>If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the \u201cnegative\u201d side and the side of the surface at which the water flows away is the \u201cpositive\u201d side. Informally, a choice of orientation gives [latex]S[\/latex] an \u201couter\u201d side and an \u201cinner\u201d side (or an \u201cupward\u201d side and a \u201cdownward\u201d side), just as a choice of orientation of a curve gives the curve \u201cforward\u201d and \u201cbackward\u201d directions.<\/p>\r\n

    Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the\u00a0positive orientation of the closed surface<\/em>\u00a0(Figure 1). We also could choose the inward normal vector at each point to give an \u201cinward\u201d orientation, which is the negative orientation of the surface.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5406\" align=\"aligncenter\" width=\"419\"]\"<img Figure 1. An oriented sphere with positive orientation.[\/caption]\r\n\r\n<\/div>\r\n

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    A portion of the graph of any smooth function [latex]z=f(x, y)[\/latex] is also orientable. If we choose the unit normal vector that points \u201cabove\u201d the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points \u201cbelow\u201d the surface at each point. To get such an orientation, we parameterize the graph of [latex]f[\/latex] in the standard way: [latex]{\\bf{r}}(x,y)=\\langle{x},y,f(x,y)\\rangle[\/latex], where [latex]x[\/latex] and [latex]y[\/latex] vary over the domain of [latex]f[\/latex]. Then, [latex]{\\bf{t}}_x=\\langle1,0,f_x\\rangle[\/latex] and [latex]{\\bf{t}}_y=\\langle0,1,f_y\\rangle[\/latex], and therefore the cross product [latex]{\\bf{t}}_x\\times{\\bf{t}}_y[\/latex] (which is normal to the surface at any point on the surface) is [latex]\\langle-f_x,-f_y,1\\rangle[\/latex]. Since the [latex]z[\/latex] component of this vector is one, the corresponding unit normal vector points \u201cupward,\u201d and the upward side of the surface is chosen to be the \u201cpositive\u201d side.<\/p>\r\n

    Let [latex]S[\/latex] be a smooth orientable surface with parameterization [latex]{\\bf{r}}(u,v)[\/latex]. For each point [latex]{\\bf{r}}(a,b)[\/latex] on the surface, vectors [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex] lie in the tangent plane at that point. Vector [latex]{\\bf{t}}_u\\times{\\bf{t}}_v[\/latex] is normal to the tangent plane at [latex]{\\bf{r}}(a,b)[\/latex] and is therefore normal to [latex]S[\/latex] at that point. Therefore, the choice of unit normal vector<\/p>\r\n

    [latex]\\large{{\\bf{N}}=\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}}[\/latex]<\/p>\r\n

    gives an orientation of surface [latex]S[\/latex].<\/p>\r\n\r\n

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    Example: choosing an orientation<\/h3>\r\nGive an orientation of cylinder [latex]x^2+y^2=r^2[\/latex], [latex]0\\leq{z}\\leq{h}[\/latex].\r\n\r\n[reveal-answer q=\"814592943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"814592943\"]\r\n

    This surface has parameterization<\/p>\r\n

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

    The tangent vectors are [latex]{\\bf{t}}_u=\\langle-r\\sin{u},r\\cos{u},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,0,1\\rangle[\/latex]. To get an orientation of the surface, we compute the unit normal vector<\/p>\r\n

    [latex]{\\bf{N}}=\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}[\/latex].<\/p>\r\n

    In this case, [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle{r}\\cos{u},r\\sin{u},0\\rangle[\/latex] and therefore<\/p>\r\n

    [latex]\\||{\\bf{t}}_u\\times{\\bf{t}}_v||=\\sqrt{r^2\\cos^2u+r^2\\sin^2u}=r[\/latex].<\/p>\r\n

    An orientation of the cylinder is<\/p>\r\n

    [latex]{\\bf{N}}(u,v)=\\frac{\\langle{r}\\cos{u},r\\sin{u},0\\rangle}4=\\langle\\cos{u},\\sin{u},0\\rangle[\/latex].<\/p>\r\n

    Notice that all vectors are parallel to the [latex]xy[\/latex]-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure 2).<\/p>\r\n\r\n\r\n[caption id=\"attachment_5407\" align=\"aligncenter\" width=\"324\"]\"<img Figure 2. If all the vectors normal to a cylinder point outward, then this is an outward orientation of the cylinder.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nGive the \u201cupward\u201d orientation of the graph of [latex]f(x, y)=xy[\/latex].\r\n\r\n[reveal-answer q=\"273456799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"273456799\"]\r\n\r\n[latex]{\\bf{N}}(x,y)=\\left\\langle\\frac{-y}{\\sqrt{1+x^2+y^2}},\\frac{-x}{\\sqrt{1+x^2+y^2}},\\frac{1}{\\sqrt{1+x^2+y^2}}\\right\\rangle[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Since every curve has a \u201cforward\u201d and \u201cbackward\u201d direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called\u00a0nonorientable<\/em>. Essentially, a surface can be oriented if the surface has an \u201cinner\u201d side and an \u201couter\u201d side, or an \u201cupward\u201d side and a \u201cdownward\u201d side. Some surfaces are twisted in such a fashion that there is no well-defined notion of an \u201cinner\u201d or \u201couter\u201d side.<\/p>\r\n

    The classic example of a nonorientable surface is the M\u00f6bius strip. To create a M\u00f6bius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure 3). Because of the half-twist in the strip, the surface has no \u201couter\u201d side or \u201cinner\u201d side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5408\" align=\"aligncenter\" width=\"967\"]\"<img Figure 3. The construction of a M\u00f6bius strip.[\/caption]\r\n\r\n

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    Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.<\/p>\r\n\r\n<\/section>

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    Surface Integral of a Vector Field<\/h2>\r\n

    With the idea of orientable surfaces in place, we are now ready to define a\u00a0surface integral of a vector field<\/span>.<\/strong> The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a two-dimensional vector field and [latex]C[\/latex] is a plane curve, then the definition of the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] involved chopping [latex]C[\/latex] into small pieces, choosing a point inside each piece, and calculating [latex]{\\bf{F}}\\cdot{\\bf{N}}[\/latex] at the point (where\u00a0[latex]{\\bf{N}}[\/latex]\u00a0is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface [latex]S[\/latex] into small pieces, choose a point in the small (two-dimensional) piece, and calculate [latex]{\\bf{F}}\\cdot{\\bf{N}}[\/latex] at the point.<\/p>\r\n

    To place this definition in a real-world setting, let [latex]S[\/latex] be an oriented surface with unit normal vector\u00a0[latex]{\\bf{N}}[\/latex]. Let\u00a0[latex]{\\bf{v}}[\/latex]\u00a0be a velocity field of a fluid flowing through [latex]S[\/latex], and suppose the fluid has density [latex]\\rho(x,y,z)[\/latex]. Imagine the fluid flows through [latex]S[\/latex], but [latex]S[\/latex] is completely permeable so that it does not impede the fluid flow (Figure 4). The\u00a0mass flux<\/span><\/strong>\u00a0of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across [latex]S[\/latex]?<\/p>\r\n\r\n

    [caption id=\"attachment_5409\" align=\"aligncenter\" width=\"231\"]\"<img Figure 4. Fluid flows across a completely permeable surface [latex]S[\/latex].[\/caption]<\/div>\r\n<\/section>\r\n
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    The rate of flow, measured in mass per unit time per unit area, is [latex]\\rho{\\bf{N}}[\/latex]. To calculate the mass flux across [latex]S[\/latex], chop [latex]S[\/latex] into small pieces [latex]S_{ij}[\/latex]. If [latex]S_{ij}[\/latex] is small enough, then it can be approximated by a tangent plane at some point [latex]P[\/latex] in [latex]S_{ij}[\/latex]. Therefore, the unit normal vector at [latex]P[\/latex] can be used to approximate [latex]{\\bf{N}}(x,y,z)[\/latex] across the entire piece [latex]S_{ij}[\/latex], because the normal vector to a plane does not change as we move across the plane. The component of the vector [latex]\\rho{\\bf{v}}[\/latex] at [latex]P[\/latex] in the direction of\u00a0[latex]{\\bf{N}}[\/latex]\u00a0is [latex]\\rho{\\bf{v}}\\cdot{\\bf{N}}[\/latex] at [latex]P[\/latex]. Since [latex]S_{ij}[\/latex] is small, the dot product [latex]\\rho{\\bf{v}}\\cdot{\\bf{N}}[\/latex] changes very little as we vary across [latex]S_{ij}[\/latex], and therefore [latex]\\rho{\\bf{v}}\\cdot{\\bf{N}}[\/latex] can be taken as approximately constant across [latex]S_{ij}[\/latex]. To approximate the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] (and not just locally at point [latex]P[\/latex]), we need to multiply [latex](\\rho{\\bf{v}}\\cdot{\\bf{N}})(P)[\/latex] by the area of [latex]S_{ij}[\/latex]. Therefore, the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of\u00a0[latex]{\\bf{N}}[\/latex]\u00a0can be approximated by [latex](\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex], where\u00a0[latex]{\\bf{N}}[\/latex], [latex]\\rho[\/latex], and\u00a0[latex]{\\bf{v}}[\/latex]\u00a0are all evaluated at [latex]P[\/latex] (Figure 5). This is analogous to the flux of two-dimensional vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across plane curve [latex]C[\/latex], in which we approximated flux across a small piece of [latex]C[\/latex] with the expression [latex]({\\bf{F}}\\cdot{\\bf{N}})\\nabla{s}[\/latex]. To approximate the mass flux across [latex]S[\/latex], form the sum [latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex]. As pieces [latex]S_{ij}[\/latex] get smaller, the sum [latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex] gets arbitrarily close to the mass flux. Therefore, the mass flux is<\/p>\r\n

    [latex]\\displaystyle\\iint_s\\rho{\\bf{v}}\\cdot{\\bf{N}}dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].<\/p>\r\n

    This is a surface integral of a vector field. Letting the vector field [latex]\\rho{\\bf{v}}[\/latex] be an arbitrary vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0leads to the following definition.<\/p>\r\n\r\n[caption id=\"attachment_5410\" align=\"aligncenter\" width=\"497\"]\"<img Figure 5. The mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of\u00a0[latex]{\\bf{N}}[\/latex] can be approximated by [latex](\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].[\/caption]\r\n

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    Definition<\/h3>\r\n\r\n
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    Let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a continuous vector field with a domain that contains oriented surface [latex]S[\/latex] with unit normal vector\u00a0[latex]{\\bf{N}}[\/latex]. The\u00a0surface integral<\/span>\u00a0of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over [latex]S[\/latex] is<\/p>\r\n

    [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    Notice the parallel between this definition and the definition of vector line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}{d}s[\/latex]. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex] is called the\u00a0flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across <\/em>[latex]S[\/latex], just as integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}{d}s[\/latex] is the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across curve [latex]C[\/latex]. A surface integral over a vector field is also called a\u00a0flux integral<\/span>.<\/p>\r\n

    Just as with vector line integrals, surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex] is easier to compute after surface [latex]S[\/latex] has been parameterized. Let [latex]{\\bf{r}}(u,v)[\/latex] be a parameterization of [latex]S[\/latex] with parameter domain [latex]D[\/latex]. Then, the unit normal vector is given by [latex]{\\bf{N}}=\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}[\/latex] and, from the surface integral equation, we have<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S&=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S \\\\\r\n&=\\displaystyle\\iint_S{\\bf{F}}\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}dS \\\\\r\n&=\\displaystyle\\iint_D\\left({\\bf{F}}({\\bf{r}}(u,v))\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}\\right)||{\\bf{t}}_u\\times{\\bf{t}}_v||dA \\\\\r\n&=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, to compute a surface integral over a vector field we can use the equation\r\n

    [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex].<\/p>\r\n\r\n

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    Example: calculating a surface integral<\/h3>\r\nCalculate the surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex], where [latex]{\\bf{F}}=\\langle-y,x,0\\rangle[\/latex] and [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{u},v^2-u,u+v\\rangle[\/latex], [latex]0\\leq{u}\\leq3, \\ 0\\leq{v}\\leq4[\/latex].\r\n\r\n[reveal-answer q=\"812467498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812467498\"]\r\n

    The tangent vectors are [latex]{\\bf{t}}_u=\\langle1,-1,1\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,2v,1\\rangle[\/latex]. Therefore,<\/p>\r\n

    [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle-1-2v,-1,2v\\rangle[\/latex].<\/p>\r\n

    By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex],<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3{\\bf{F}}({\\bf{r}}(u,v))\\cdot{\\bf{t}}_u\\times{\\bf{t}}_v)dudv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3\\langle{u}-v^2,u,0\\rangle\\cdot\\langle-1-2v,-1,2v\\rangle{d}udv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3[(u-v^2)(-1-2v)-u]dudv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3(2v^3+v^2-2uv-2u]dudv \\\\\r\n&=\\displaystyle\\int_0^4[2v^3u+v^2u-vu^2-u^2]_0^3dv \\\\\r\n&=\\displaystyle\\int_0^4(6v^3+3v^2-9v-9)dv \\\\\r\n&=\\left[\\frac{3v^2}2+v^3-\\frac{9v^2}2-9v\\right]_0^4 \\\\\r\n&=340\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]S[\/latex] is [latex]340[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nCalculate surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S[\/latex], where [latex]{\\bf{F}}=\\langle0,-z,y\\rangle[\/latex] and [latex]S[\/latex] is the portion of the unit sphere in the first octant with outward orientation.\r\n\r\n[reveal-answer q=\"736902583\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736902583\"]\r\n\r\n[latex]0[\/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