{"id":5510,"date":"2022-06-02T19:00:41","date_gmt":"2022-06-02T19:00:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=5510"},"modified":"2022-11-01T05:34:33","modified_gmt":"2022-11-01T05:34:33","slug":"applying-stokes-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/applying-stokes-theorem\/","title":{"raw":"Applying Stokes\u2019 Theorem","rendered":"Applying Stokes\u2019 Theorem"},"content":{"raw":"
\r\n
\r\n

Learning Objectives<\/h3>\r\n
\r\n
    \r\n \t
  • Use Stokes\u2019 theorem to evaluate a line integral.<\/span><\/li>\r\n \t
  • Use Stokes\u2019 theorem to calculate a surface integral.<\/span><\/li>\r\n \t
  • Use Stokes\u2019 theorem to calculate a curl.<\/span><\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<\/div>\r\n

    Applying Stokes\u2019 Theorem<\/h2>\r\n

    Stokes\u2019 theorem translates between the flux integral of surface [latex]S[\/latex] to a line integral around the boundary of [latex]S[\/latex]. Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.<\/p>\r\n\r\n<\/div>\r\n

    \r\n
    \r\n

    Example: calculating a surface integral<\/h3>\r\nCalculate surface integral [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex], where [latex]S[\/latex] is the surface, oriented outward, in\u00a0Figure 1\u00a0and [latex]{\\bf{F}}=\\langle{z},2xy,x+y\\rangle[\/latex].\r\n\r\n[caption id=\"attachment_5419\" align=\"aligncenter\" width=\"616\"]\"<img Figure 1. A complicated surface in a vector field.[\/caption]\r\n\r\n[reveal-answer q=\"485624851\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485624851\"]\r\n

    Note that to calculate [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] without using Stokes\u2019 theorem, we would need to use the equation to calculate scalar surface integrals. Use of this equation requires a parameterization of [latex]S[\/latex]. Surface [latex]S[\/latex] is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes\u2019 theorem, noting that the boundary [latex]C[\/latex] of the surface is merely a single circle with radius 1.<\/p>\r\n

    The curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]\\langle1,1,2y\\rangle[\/latex]. By Stokes\u2019 theorem,<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex],<\/p>\r\n

    where [latex]C[\/latex] has parameterization [latex]{\\bf{r}}(t)=\\langle\\sin{t},0,1-\\cos{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. By the equation to for computing vector line integrals,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}&=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}\\langle1-\\cos{t},0,-\\sin{t}\\rangle\\cdot\\langle-\\cos{t},0\\sin{t}\\rangle{d}t \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}(-\\cos{t}+\\cos^2t-\\sin^2t)dt \\\\\r\n&=\\left[-\\sin{t}+\\frac12\\sin(2t)\\right]_0^{2\\pi} \\\\\r\n&=\\left(-\\sin{(2\\pi)}+\\frac12\\sin{(4\\pi)}\\right)-\\left(-\\sin0+\\frac12\\sin0\\right) \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    An amazing consequence of Stokes\u2019 theorem is that if [latex]S'[\/latex] is any other smooth surface with boundary [latex]C[\/latex] and the same orientation as [latex]S[\/latex], then [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=0[\/latex] because Stokes\u2019 theorem says the surface integral depends on the line integral around the boundary only.<\/p>\r\n

    In\u00a0Example \"Calculating a Surface Integral\", we calculated a surface integral simply by using information about the boundary of the surface. In general, let [latex]S_1[\/latex] and [latex]S_2[\/latex] be smooth surfaces with the same boundary [latex]C[\/latex] and the same orientation. By Stokes\u2019 theorem,<\/p>\r\n

    [latex]\\large{\\displaystyle\\iint_{S_1}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_{S_2}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}}[\/latex].<\/p>\r\n

    Therefore, if [latex]\\displaystyle\\iint_{S_1}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] is difficult to calculate but [latex]\\displaystyle\\iint_{S_2}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] is easy to calculate, Stokes\u2019 theorem allows us to calculate the easier surface integral. In\u00a0Example \"Calculating a Surface Integral\", we could have calculated [latex]\\displaystyle\\iint_{S}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] by calculating [latex]\\displaystyle\\iint_{S^\\prime}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex], where [latex]S'[\/latex] is the disk enclosed by boundary curve [latex]C[\/latex] (a much more simple surface with which to work).<\/p>\r\n

    [latex]{\\displaystyle\\iint_{S_1}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_{S_2}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}}[\/latex] shows that flux integrals of curl vector fields are\u00a0surface independent<\/span>\u00a0in the same way that line integrals of gradient fields are path independent. Recall that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a two-dimensional conservative vector field defined on a simply connected domain, [latex]f[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex], and [latex]C[\/latex] is a curve in the domain of\u00a0[latex]{\\bf{F}}[\/latex], then [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] depends only on the endpoints of [latex]C[\/latex]. Therefore if [latex]C'[\/latex] is any other curve with the same starting point and endpoint as [latex]C[\/latex] (that is, [latex]C'[\/latex] has the same orientation as [latex]C[\/latex]), then [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C^\\prime}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. In other words, the value of the integral depends on the boundary of the path only; it does not really depend on the path itself.<\/p>\r\n

    Analogously, suppose that [latex]S[\/latex] and [latex]S'[\/latex] are surfaces with the same boundary and same orientation, and suppose that\u00a0[latex]{\\bf{G}}[\/latex]\u00a0is a three-dimensional vector field that can be written as the curl of another vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0(so that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is like a \u201cpotential field\u201d of\u00a0[latex]{\\bf{G}}[\/latex]). By\u00a0[latex]{\\displaystyle\\iint_{S_1}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_{S_2}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}}[\/latex],<\/p>\r\n

    [latex]\\large{\\displaystyle\\int_S{\\bf{G}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_{S}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_{S^\\prime}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_{S^\\prime}\\text{curl }{\\bf{G}}\\cdot{d}{\\bf{S}}}[\/latex].<\/p>\r\n

    Therefore, the flux integral of\u00a0[latex]{\\bf{G}}[\/latex]\u00a0does not depend on the surface, only on the boundary of the surface. Flux integrals of vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of vector fields that can be written as the gradient of a scalar function are path independent.<\/p>\r\n\r\n

    \r\n
    \r\n

    try it<\/h3>\r\nUse Stokes\u2019 theorem to calculate surface integral [latex]\\displaystyle\\iint_{S}\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex], where [latex]{\\bf{F}}=\\langle{z},x,y\\rangle[\/latex] and [latex]S[\/latex] is the surface as shown in the following figure. The boundary curve, [latex]C[\/latex], is oriented clockwise when looking along the positive [latex]y[\/latex]-axis.\r\n\r\n\"<img\r\n\r\n[reveal-answer q=\"653124473\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653124473\"]\r\n\r\n[latex]-\\pi[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example: calculating a line integral<\/h3>\r\nCalculate the line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]{\\bf{F}}=\\langle{x}y,x^2+y^2+z^2,yz\\rangle[\/latex] and [latex]C[\/latex] is the boundary of the parallelogram with vertices [latex](0, 0, 1)[\/latex], [latex](0, 1, 0)[\/latex], [latex](2, 0, -1)[\/latex], and\u00a0[latex](2, 1, -2)[\/latex].\r\n\r\n[reveal-answer q=\"853672198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"853672198\"]\r\n

    To calculate the line integral directly, we need to parameterize each side of the parallelogram separately, calculate four separate line integrals, and add the result. This is not overly complicated, but it is time-consuming.<\/p>\r\n

    By contrast, let\u2019s calculate the line integral using Stokes\u2019 theorem. Let [latex]S[\/latex] denote the surface of the parallelogram. Note that [latex]S[\/latex] is the portion of the graph of [latex]z=1-x-y[\/latex] for [latex](x, y)[\/latex] varying over the rectangular region with vertices [latex](0, 0)[\/latex], [latex](0, 1)[\/latex], [latex](2, 0)[\/latex], and [latex](2, 1)[\/latex] in the [latex]xy[\/latex]-plane. Therefore, a parameterization of [latex]S[\/latex] is [latex]\\langle{x},y,1-x-y\\rangle[\/latex], [latex]0\\leq{x}\\leq2, \\ 0\\leq{y}\\leq1[\/latex]. The curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]-\\langle{z},0,x\\rangle[\/latex], and Stokes\u2019 theorem and the equation to calculate scalar surface integrals\u00a0give<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}} \\\\\r\n&=\\displaystyle\\int_0^2\\displaystyle\\int_0^1\\text{curl }{\\bf{F}}(x,y)\\cdot({\\bf{t}}_x\\times{\\bf{t}}_y)dydx \\\\\r\n&=\\displaystyle\\int_0^2\\displaystyle\\int_0^1\\langle-(1-x-y),0,x\\rangle\\cdot(\\langle1,0,-1\\rangle\\times\\langle0,1,-1\\rangle)dydx \\\\\r\n&=\\displaystyle\\int_0^2\\displaystyle\\int_0^1\\langle{x}+y-1,0,x\\rangle\\cdot\\langle1,1,1\\rangle{d}ydx \\\\\r\n&=\\displaystyle\\int_0^2\\displaystyle\\int_0^12x+y-1dydx \\\\\r\n&=3\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    try it<\/h3>\r\nUse Stokes\u2019 theorem to calculate line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]{\\bf{F}}=\\langle{z},x,y\\rangle[\/latex] and [latex]C[\/latex] is oriented clockwise and is the boundary of a triangle with vertices [latex](0, 0, 1)[\/latex],\u00a0[latex](3, 0, -2)[\/latex], and\u00a0[latex](0, 1, 2)[\/latex].\r\n\r\n[reveal-answer q=\"736428721\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736428721\"]\r\n\r\n[latex]\\frac32[\/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