{"id":5507,"date":"2022-06-02T18:58:38","date_gmt":"2022-06-02T18:58:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=5507"},"modified":"2022-11-01T05:30:36","modified_gmt":"2022-11-01T05:30:36","slug":"surface-integral-of-a-scalar-valued-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/surface-integral-of-a-scalar-valued-function\/","title":{"raw":"Surface Integral of a Scalar-Valued Function","rendered":"Surface Integral of a Scalar-Valued Function"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Use a surface integral to calculate the area of a given surface.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let\u2019s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain\u00a0region<\/em>\u00a0(no longer a curve) into tiny pieces and proceed in the same fashion.<\/p>\r\n

    Let [latex]S[\/latex] be a piecewise smooth surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{x}(u,v),y(u,v)z(u,v)\\rangle[\/latex] with parameter domain [latex]D[\/latex] and let [latex]f(x, y, z)[\/latex] be a function with a domain that contains [latex]S[\/latex]. For now, assume the parameter domain [latex]D[\/latex] is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). 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]. Suppose that [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 that [latex]D[\/latex] is subdivided into [latex]mn[\/latex] rectangles. This division of [latex]D[\/latex] into subrectangles gives a corresponding division of [latex]S[\/latex] into pieces [latex]S_{ij}[\/latex]. Choose point [latex]P_{ij}[\/latex] in each piece [latex]S_{ij}[\/latex], evaluate [latex]P_{ij}[\/latex] at [latex]f[\/latex], and multiply by area [latex]S_{ij}[\/latex] to form the Riemann sum<\/p>\r\n

    [latex]\\large{\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S_{ij}}}[\/latex].<\/p>\r\n

    To define a surface integral of a scalar-valued function, we let the areas of the pieces of [latex]S[\/latex] shrink to zero by taking a limit.<\/p>\r\n\r\n<\/div>\r\n

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    definition<\/h3>\r\n\r\n
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    The\u00a0surface integral of a scalar-valued function<\/span><\/strong>\u00a0of [latex]f[\/latex] over a piecewise smooth surface [latex]S[\/latex] is<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S_{ij}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.<\/p>\r\n

    The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.<\/p>\r\n

    Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas [latex]\\Delta{S_{ij}}[\/latex] with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex]:<\/p>\r\n

    [latex]\\large{{\\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>\r\n

    From the material we have already studied, we know that<\/p>\r\n

    [latex]\\large{\\Delta{S}_{ij}\\approx||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}}[\/latex].<\/p>\r\n

    Therefore,<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_Sf(x,y,z)dS\\approx\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}}[\/latex].<\/p>\r\n

    This approximation becomes arbitrarily close to [latex]\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S}_{ij}[\/latex] as we increase the number of pieces [latex]S_{ij}[\/latex] by letting [latex]m[\/latex] and [latex]n[\/latex] go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\iint_Df({\\bf{r}}(u,v)||{\\bf{t}}_u\\times{\\bf{t}}_v||}dA[\/latex].<\/p>\r\n

    The surface integral of a scalar-valued function\u00a0allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to the surface integral equation\u00a0for line integrals:<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_Cf(x,y,z)ds=\\displaystyle\\int_a^bf({\\bf{r}}(t))||{\\bf{r}}^\\prime(t)||dt}[\/latex].<\/p>\r\n

    In this case, vector [latex]{\\bf{t}}_u\\times{\\bf{t}}_v[\/latex] is perpendicular to the surface, whereas vector [latex]{\\bf{r}}^\\prime[\/latex] is tangent to the curve.<\/p>\r\n\r\n<\/div>\r\n

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    Example: Calculating a surface integral<\/h3>\r\nCalculate surface integral [latex]\\displaystyle\\iint_S5dS[\/latex], where [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{v},u^2,v\\rangle[\/latex] for [latex]0\\leq{u}\\leq2[\/latex] and\u00a0[latex]0\\leq{v}\\leq{u}[\/latex].\r\n\r\n[reveal-answer q=\"768642501\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"768642501\"]\r\n

    Notice that this parameter domain [latex]D[\/latex] is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because the surface integral equation\u00a0does not place any restrictions on the shape of the parameter domain.<\/p>\r\n

    To use the surface integral equation\u00a0to calculate the surface integral, we first find vector [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex]. Note that [latex]{\\bf{t}}_u=\\langle1,2u,0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,0,1\\rangle[\/latex]. Therefore,<\/p>\r\n

    [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ 1&2u&0 \\\\ 0&0&1\\end{vmatrix}=\\langle2u,-1,0\\rangle[\/latex]<\/p>\r\n

    and<\/p>\r\n

    [latex]||{\\bf{t}}_u\\times{\\bf{t}}_v||=\\sqrt{1+4u^2}[\/latex].<\/p>\r\n

    By\u00a0the surface integral equation,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_S5dS&=5\\displaystyle\\iint_D\\sqrt{1+4u^2}dA \\\\\r\n&=5\\displaystyle\\int_0^2\\int_0^u\\sqrt{1+4u^2}dvdu=5\\displaystyle\\int_0^2u\\sqrt{1+4u^2}du \\\\\r\n&=5\\left[\\frac{(1+4u^2)^{3\/2}}3\\right]_0^2=\\frac{5(17^{3\/2}-1)}{12}\\approx28.79\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    Example: calculating the surface integral of a cylinder<\/h3>\r\n

    Calculate surface integral [latex]\\displaystyle\\iint_S(x+y^2)dS[\/latex], where [latex]S[\/latex] is cylinder [latex]x^2+y^2=4[\/latex], [latex]0\\leq{z}\\leq3[\/latex] (Figure 1).<\/p>\r\n\r\n[caption id=\"attachment_5401\" align=\"aligncenter\" width=\"264\"]\"<img Figure 1. Integrating function [latex]f(x,y,z)=x+y^2[\/latex] over a cylinder.[\/caption][reveal-answer q=\"247755319\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"247755319\"]\r\n

    To calculate the surface integral, we first need a parameterization of the cylinder. Following\u00a0Example \"Parameterizing a Cylinder\", a parameterization is<\/p>\r\n

    [latex]{\\bf{r}}(u,v)=\\langle2\\cos{u},2\\sin{u},v\\rangle, \\ 0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq3[\/latex].<\/p>\r\n

    The tangent vectors are [latex]{\\bf{t}}_u=\\langle\\sin{u},\\cos{u},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,0,1\\rangle[\/latex]. Then,<\/p>\r\n

    [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -\\sin{u}&\\cos{u}&0 \\\\ 0&0&1\\end{vmatrix}=\\langle\\cos{u},\\sin{u},0\\rangle[\/latex]<\/p>\r\n

    and [latex]||{\\bf{t}}_u\\times{\\bf{t}}_v||=\\sqrt{\\cos^2u+\\sin^2u}=1[\/latex]. By the surface integral equation,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_Sf(x,y,z)dS&=\\displaystyle\\iint_Df({\\bf{r}}(u,v))||{\\bf{t}}_u\\times{\\bf{t}}_v|| \\ dA \\\\\r\n&=\\displaystyle\\int_0^3\\displaystyle\\int_0^{2\\pi}(2\\cos{u}+4\\sin^2u)dydv \\\\\r\n&=\\displaystyle\\int_0^3[2\\sin{u}+2u-\\sin(2u)]_0^{2\\pi}dv=\\displaystyle\\int_0^34\\pi{d}v=12\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nCalculate [latex]\\displaystyle\\iint_S(x^2-z)dS[\/latex], where [latex]S[\/latex] is the surface with parameterization\u00a0[latex]{\\bf{r}}(u,v)=\\langle{u},u^2+v^2,1\\rangle[\/latex], [latex]0\\leq{u}\\leq2, \\ 0\\leq{v}\\leq3[\/latex].\r\n\r\n[reveal-answer q=\"834661790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834661790\"]\r\n\r\n[latex]24[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Example: calculating the surface integral of a piece of a sphere<\/h3>\r\nCalculate surface integral [latex]\\displaystyle\\iint_Sf(x,y,z)dS[\/latex], where [latex]f(x, y, z)=z^{2}[\/latex] and [latex]S[\/latex] is the surface that consists of the piece of sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex] that lies on or above plane [latex]z=1[\/latex] and the disk that is enclosed by intersection plane [latex]z=1[\/latex] and the given sphere (Figure 2).\r\n\r\n[caption id=\"attachment_5403\" align=\"aligncenter\" width=\"407\"]\"<img Figure 2. Calculating a surface integral over surface [latex]S[\/latex].[\/caption][reveal-answer q=\"223579284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"223579284\"]\r\n
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    Solution<\/span><\/h2>\r\n
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    Notice that [latex]S[\/latex] is not smooth but is piecewise smooth; [latex]S[\/latex] can be written as the union of its base [latex]S_1[\/latex] and its spherical top [latex]S_2[\/latex], and both [latex]S_1[\/latex] and [latex]S_2[\/latex] are smooth. Therefore, to calculate [latex]\\displaystyle\\iint_Sz^2dS[\/latex], we write this integral as [latex]\\displaystyle\\iint_Sz^2d{S_1}+\\displaystyle\\iint_{S_2}z^2dS[\/latex] and we calculate integrals [latex]\\displaystyle\\iint_{S_1}z^2dS[\/latex] and [latex]\\displaystyle\\iint_{S_2}z^2dS[\/latex].<\/p>\r\n

    First, we calculate [latex]\\displaystyle\\iint_{S_1}z^2dS[\/latex]. To calculate this integral we need a parameterization of [latex]S_1[\/latex]. This surface is a disk in plane [latex]z=1[\/latex] centered at [latex](0, 0, 1)[\/latex]. To parameterize this disk, we need to know its radius. Since the disk is formed where plane [latex]z=1[\/latex] intersects sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex], we can substitute [latex]z=1[\/latex] into equation [latex]x^{2}+y^{2}+z^{2}=4[\/latex]:<\/p>\r\n

    [latex]x^2+y^2+1=4\\Rightarrow{x}^2+y^2=3[\/latex].<\/p>\r\n

    Therefore, the radius of the disk is [latex]\\sqrt3[\/latex] and a parameterization of [latex]S_1[\/latex] is [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},1\\rangle[\/latex], [latex]0\\leq{u}\\leq\\sqrt3[\/latex],[latex]0\\leq{v}\\leq2\\pi[\/latex]. The tangent vectors are [latex]{\\bf{t}}_u=\\langle\\cos{v},\\sin{v},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle-u\\sin{v},u\\cos{v},0\\rangle[\/latex], and thus<\/p>\r\n

    [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ \\cos{v}&\\sin{v}&0 \\\\-u\\sin{v}&u\\cos{v}&0\\end{vmatrix}=\\langle0,0,u\\cos^2v+u\\sin^2v\\rangle=\\langle0,0,u\\rangle[\/latex].<\/p>\r\n

    The magnitude of this vector is [latex]u[\/latex]. Therefore,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_1}z^2dS&=\\displaystyle\\int_0^{\\sqrt3}\\displaystyle\\int_0^{2\\pi}f({\\bf{r}}(u,v))||{\\bf{t}}_u\\times{\\bf{t}}_v|| \\ dvdu \\\\\r\n&=\\displaystyle\\int_0^{\\sqrt3}\\displaystyle\\int_0^{2\\pi}u \\ dvdu \\\\\r\n&=2\\pi\\displaystyle\\int_0^{\\sqrt3} \\ udu \\\\\r\n&=3\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

    Now we calculate [latex]\\displaystyle\\iint_{S_2} \\ dS[\/latex]. To calculate this integral, we need a parameterization of [latex]S_2[\/latex]. The parameterization of full sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex] is<\/p>\r\n

    [latex]{\\bf{r}}(\\phi,\\theta)=\\langle2\\cos\\theta\\sin\\phi,2\\sin\\theta\\sin\\phi,2\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi[\/latex].<\/p>\r\n

    Since we are only taking the piece of the sphere on or above plane [latex]z=1[\/latex], we have to restrict the domain of [latex]\\phi[\/latex]. To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in\u00a0Figure 3\u00a0(the [latex]\\sqrt3[\/latex] comes from the fact that the base of [latex]S[\/latex] is a disk with radius [latex]\\sqrt3[\/latex]). Therefore, the tangent of [latex]\\phi[\/latex] is [latex]\\sqrt3[\/latex], which implies that [latex]\\phi[\/latex] is [latex]\\pi\/6[\/latex]. We now have a parameterization of [latex]S_2[\/latex]:<\/p>\r\n

    [latex]{\\bf{r}}(\\phi,\\theta)=\\langle2\\cos\\theta\\sin\\phi,2\\sin\\theta\\sin\\phi,2\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi\/3[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5404\" align=\"aligncenter\" width=\"422\"]\"<img Figure 3. The maximum value of [latex]\\phi[\/latex] has a tangent value of [latex]{\\sqrt3}[\/latex][\/caption]\r\n

    The tangent vectors are<\/p>\r\n[latex]{\\bf{t}}_\\phi=\\langle2\\cos\\theta\\cos\\phi,2\\sin\\theta\\cos\\phi,-2\\sin\\phi\\rangle\\text{ and }{\\bf{t}}_\\theta=\\langle-2\\sin\\theta\\sin\\phi,u\\cos\\theta\\sin\\phi,0\\rangle[\/latex],\r\n

    and thus<\/p>\r\n

    [latex]\\begin{aligned}\r\n{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta&=\\begin{vmatrix}\r\n{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\\r\n2\\cos\\theta\\cos\\phi&2\\sin\\theta\\cos\\phi&-2\\sin\\phi \\\\\r\n-2\\sin\\theta\\sin\\phi&2\\cos\\theta\\sin\\phi&0\r\n\\end{vmatrix} \\\\\r\n&=\\langle4\\cos\\theta\\sin^2\\phi,4\\sin\\theta\\sin^2\\phi,4\\cos^2\\theta\\sin\\phi+4\\sin^2\\theta\\cos\\phi\\sin\\phi\\rangle \\\\\r\n&=\\langle4\\cos\\theta\\sin^2\\phi,4\\sin\\theta\\sin^2\\phi,4\\cos\\phi\\sin\\phi\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\n

    The magnitude of this vector is<\/p>\r\n

    [latex]\\begin{aligned}\r\n||{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta||&=\\sqrt{16\\cos^2\\theta\\sin^4\\phi+16\\sin^2\\theta\\sin^4\\phi+16\\cos^2\\phi\\sin^2\\phi} \\\\\r\n&=4\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}\r\n\\end{aligned}[\/latex].<\/p>\r\n

    Therefore,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_2}z^2dS&=\\displaystyle\\int_0^{\\pi\/3}\\displaystyle\\int_0^{2\\pi}f({\\bf{r}}(\\phi,\\theta))||{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta|| \\ d\\theta{d}\\phi \\\\\r\n&=\\displaystyle\\int_0^{\\pi\/3}\\displaystyle\\int_0^{2\\pi}16\\cos^2\\phi\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}d\\theta{d}\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}d\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sin\\phi\\sqrt{\\sin^2\\phi+\\cos^2\\phi}d\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sin\\phi{d}\\phi \\\\\r\n&=32\\pi\\left[-\\frac{\\cos^3\\phi}3\\right]_0^{\\pi\/6}=32\\pi\\left[\\frac13-\\frac{\\sqrt3}8\\right]=\\frac{28\\pi}3\r\n\\end{aligned}[\/latex].<\/p>\r\nSince\r\n

    [latex]\\displaystyle\\iint_Sz^2 \\ dS=\\displaystyle\\iint_{S_1}z^2 \\ dS+\\displaystyle\\iint_{S_2}z^2 \\ dS=3\\pi\\frac{28\\pi}3=\\frac{37\\pi}3[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

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    Analysis<\/span><\/h2>\r\n

    In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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    try it<\/h3>\r\nCalculate line integral [latex]\\displaystyle\\iint_S(x-y)dS[\/latex], where [latex]S[\/latex] is cylinder [latex]x^2+y^2=1[\/latex],[latex]0\\leq{z}\\leq2[\/latex], including the circular top and bottom.\r\n\r\n[reveal-answer q=\"843749219\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843749219\"]\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