{"id":1123,"date":"2021-11-09T23:42:10","date_gmt":"2021-11-09T23:42:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=1123"},"modified":"2022-11-01T05:20:32","modified_gmt":"2022-11-01T05:20:32","slug":"greens-theorem-on-general-regions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/greens-theorem-on-general-regions\/","title":{"raw":"Green\u2019s Theorem on General Regions","rendered":"Green\u2019s Theorem on General Regions"},"content":{"raw":"
\r\n
\r\n

Learning Objectives<\/h3>\r\n
    \r\n \t
  • Calculate circulation and flux on more general regions.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Green\u2019s Theorem on General Regions<\/h2>\r\n

    Green\u2019s theorem, as stated, applies only to regions that are simply connected\u2014that is, Green\u2019s theorem as stated so far cannot handle regions with holes. Here, we extend Green\u2019s theorem so that it does work on regions with finitely many holes (Figure 1).<\/p>\r\n\r\n\r\n[caption id=\"attachment_5306\" align=\"aligncenter\" width=\"273\"]\"<img Figure 1. Green\u2019s theorem, as stated, does not apply to a nonsimply connected region with three holes like this one.[\/caption]\r\n

    Before discussing extensions of Green\u2019s theorem, we need to go over some terminology regarding the boundary of a region. Let [latex]D[\/latex] be a region and let [latex]C[\/latex] be a component of the boundary of [latex]D[\/latex]. We say that [latex]C[\/latex] is\u00a0positively oriented<\/em>\u00a0if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our left. Therefore, the counterclockwise orientation of the boundary of a disk is a positive orientation, for example. Curve [latex]C[\/latex] is\u00a0negatively oriented<\/em>\u00a0if, as we walk along [latex]C[\/latex] in the direction of orientation, region [latex]D[\/latex] is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation, for example.<\/p>\r\n

    Let [latex]D[\/latex] be a region with finitely many holes (so that [latex]D[\/latex] has finitely many boundary curves), and denote the boundary of [latex]D[\/latex] by [latex]\\partial{D}[\/latex] (Figure 2). To extend Green\u2019s theorem so it can handle [latex]D[\/latex], we divide region [latex]D[\/latex] into two regions, [latex]D_1[\/latex] and [latex]D_2[\/latex] (with respective boundaries [latex]\\partial{D_1}[\/latex] and [latex]\\partial{D_2}[\/latex]), in such a way that [latex]D=D_1\\cup{D}_2[\/latex] and neither [latex]D_1[\/latex] nor [latex]D_2[\/latex] has any holes (Figure 2).<\/p>\r\n\r\n[caption id=\"attachment_5308\" align=\"aligncenter\" width=\"652\"]\"<img Figure 2. (a) Region [latex]D[\/latex] with an oriented boundary has three holes. (b) Region [latex]D[\/latex] split into two simply connected regions has no holes.[\/caption]\r\n

    Assume the boundary of [latex]D[\/latex] is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region [latex]D_1[\/latex] and [latex]D_2[\/latex] is positively oriented. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a vector field defined on [latex]D[\/latex], then Green\u2019s theorem says that<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\oint_{\\partial{D_1}}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\oint_{\\partial{D_2}}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\iint_{D_1}Q_x-P_ydA+\\displaystyle\\iint_{D_2}Q_x-P_ydA \\\\\r\n&=\\displaystyle\\iint_{D}(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\n

    Therefore, Green\u2019s theorem still works on a region with holes.<\/p>\r\n

    To see how this works in practice, consider annulus [latex]D[\/latex] in\u00a0Figure 3\u00a0and suppose that [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is a vector field defined on this annulus. Region [latex]D[\/latex] has a hole, so it is not simply connected. Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure 3) so that, when we divide the region into [latex]D_1[\/latex] and [latex]D_2[\/latex], we are able to keep the region on our left as we walk along a path that traverses the boundary. Let [latex]D_1[\/latex] be the upper half of the annulus and [latex]D_2[\/latex] be the lower half. Neither of these regions has holes, so we have divided [latex]D[\/latex] into two simply connected regions.<\/p>\r\n

    We label each piece of these new boundaries as [latex]P_i[\/latex] for some [latex]i[\/latex], as in\u00a0Figure 3. If we begin at [latex]P[\/latex] and travel along the oriented boundary, the first segment is [latex]P_1[\/latex], then [latex]P_2[\/latex], [latex]P_3[\/latex], and [latex]P_4[\/latex]. Now we have traversed [latex]D_1[\/latex] and returned to [latex]P[\/latex] Next, we start at [latex]P[\/latex] again and traverse [latex]D_2[\/latex]. Since the first piece of the boundary is the same as [latex]P_4[\/latex] in [latex]D_1[\/latex], but oriented in the opposite direction, the first piece of [latex]D_2[\/latex] is [latex]-P_4[\/latex]. Next, we have [latex]P_5[\/latex], then [latex]-P_2[\/latex], and finally [latex]P_6[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_5309\" align=\"aligncenter\" width=\"794\"]\"<img Figure 3. Breaking the annulus into two separate regions gives us two simply connected regions. The line integrals over the common boundaries cancel out.[\/caption]\r\n

    Figure 3\u00a0shows a path that traverses the boundary of [latex]D[\/latex]. Notice that this path traverses the boundary of region [latex]D_1[\/latex], returns to the starting point, and then traverses the boundary of region [latex]D_2[\/latex]. Furthermore, as we walk along the path, the region is always on our left. Notice that this traversal of the [latex]P_i[\/latex] paths covers the entire boundary of region [latex]D[\/latex]. If we had only traversed one portion of the boundary of [latex]D[\/latex], then we cannot apply Green\u2019s theorem to [latex]D[\/latex].<\/p>\r\n

    The boundary of the upper half of the annulus, therefore, is [latex]P_1\\cup{P_2}\\cup{P_3}\\cup{P_4}[\/latex] and the boundary of the lower half of the annulus is [latex]-P_4\\cup{P_5}\\cup-{P_2}\\cup{P_6}[\/latex]. Then, Green\u2019s theorem implies<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\int_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{-P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_4}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}-\\displaystyle\\int{P_2}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int+{P_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_3}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_5}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int{P_6}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int_{\\partial{D}_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{\\partial{D}_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\iint_{D_1}(Q_x-P_y)dA+\\displaystyle\\iint_{D_2}(Q_x-P_y)dA \\\\\r\n&=\\displaystyle\\iint_{D}(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, we arrive at the equation found in Green\u2019s theorem\u2014namely,\r\n

    <\/div>\r\n

    [latex]\\large{\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_D(Q_z-P_y)dA}[\/latex].<\/p>\r\n

    The same logic implies that the flux form of Green\u2019s theorem can also be extended to a region with finitely many holes:<\/p>\r\n

    [latex]\\large{\\displaystyle\\oint_{C}{\\bf{F}}\\cdot{\\bf{D}}ds=\\displaystyle\\iint_D(P_x+Q_y)dA}[\/latex].<\/p>\r\n\r\n

    \r\n
    \r\n

    Example: using green's theorem on a region with holes<\/h3>\r\n

    Calculate integral<\/p>\r\n

    [latex]\\large{\\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy}[\/latex],<\/p>\r\n

    where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]1\\leq{\\bf{r}}\\leq2,0\\leq \\theta\\leq2\\pi[\/latex].<\/p>\r\n[reveal-answer q=\"437584636\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437584636\"]\r\n

    Although [latex]D[\/latex] is not simply connected, we can use the extended form of Green\u2019s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\oint_{\\partial{D}}\\left(\\sin{x}-\\frac{x^3}3\\right)dx+\\left(\\frac{y^3}3+\\sin{y}\\right)dy&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&=\\displaystyle\\iint_D(x^2+y^2)dA \\\\\r\n&=\\displaystyle\\int_0^\r\n{2\\pi}\\displaystyle\\int_1^2r^3drd\\theta=\\displaystyle\\int_0^{2\\pi}\\frac{15}4d\\theta \\\\\r\n&=\\frac{15\\pi}2\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Example: using the extended form of green's theorem<\/h3>\r\nLet [latex]{\\bf{F}}=\\langle{P},Q\\rangle=\\langle\\frac{y}{x^2+y^2},-\\frac{y}{x^2+y^2}[\/latex] and let [latex]C[\/latex] be any simple closed curve in a plane oriented counterclockwise. What are the possible values of [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]?\r\n\r\n[reveal-answer q=\"329492936\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"329492936\"]\r\n

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

    We use the extended form of Green\u2019s theorem to show that [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is either [latex]0[\/latex] or [latex]-2\\pi[\/latex]\u2013that is, no matter how crazy curve [latex]C[\/latex] is, the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0along [latex]C[\/latex] can have only one of two possible values. We consider two cases: the case when [latex]C[\/latex] encompasses the origin and the case when [latex]C[\/latex] does not encompass the origin.<\/p>\r\n\r\n

    Case 1: C Does Not Encompass the Origin<\/h2>\r\n
    \r\n

    In this case, the region enclosed by [latex]C[\/latex] is simply connected because the only hole in the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is at the origin. We showed in our discussion of cross-partials that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial condition. If we restrict the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0just to [latex]C[\/latex] and the region it encloses, then\u00a0[latex]{\\bf{F}}[\/latex]\u00a0with this restricted domain is now defined on a simply connected domain. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partial property on its restricted domain, the field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative on this simply connected region and hence the circulation [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is zero.<\/p>\r\n\r\n

    Case 2: C Does Encompass the Origin<\/h2>\r\n<\/section>
    \r\n

    In this case, the region enclosed by [latex]C[\/latex] is not simply connected because this region contains a hole at the origin. Let [latex]C_1[\/latex] be a circle of radius\u00a0a<\/em>\u00a0centered at the origin so that [latex]C_1[\/latex] is entirely inside the region enclosed by [latex]C[\/latex] (Figure 4). Give [latex]C_1[\/latex] a clockwise orientation.<\/p>\r\n\r\n[caption id=\"attachment_5310\" align=\"aligncenter\" width=\"454\"]\"<img Figure 4. Choose circle [latex]C_1[\/latex] centered at the origin that is contained entirely inside [latex]C[\/latex].[\/caption]\r\n

    Let [latex]D[\/latex] be the region between [latex]C_1[\/latex] and [latex]C[\/latex], and [latex]C[\/latex] is orientated counterclockwise. By the extended version of Green\u2019s theorem,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\iint_DQ_x-P_ydA \\\\\r\n&=\\displaystyle\\iint_D-\\frac{y^2-x^2}{(x^2+y^2)^2}+\\frac{y^2-x^2}{(x^2+y^2)^2}dA \\\\\r\n&=0\r\n\\end{aligned}[\/latex],<\/p>\r\n

    and therefore<\/p>\r\n

    [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=-\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\r\n

    Since [latex]C_1[\/latex] is a specific curve, we can evaluate [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. Let<\/p>\r\n

    [latex]x=a\\cos{t},y=a\\sin{t},0\\leq{t}\\leq2\\pi[\/latex]<\/p>\r\n

    be a parameterization of [latex]C_1[\/latex]. Then,<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^{2\\pi}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}\\left\\langle-\\frac{\\sin{(t)}}a,-\\frac{\\cos{(t)}}1\\right\\rangle\\cdot\\langle-a\\sin{(t)},-\\cos{(t)}\\rangle dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}\\sin^2(t)+\\cos^2(t)dt=\\displaystyle\\int_0^{2\\pi}dt=2\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

    Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}s=-2\\pi[\/latex].<\/p>\r\n\r\n<\/section><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    try it<\/h3>\r\nCalculate integral [latex]\\displaystyle\\oint_{\\partial{D}}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], where [latex]D[\/latex] is the annulus given by the polar inequalities [latex]2\\leq{r}\\leq5,0\\leq\\theta\\leq2\\pi[\/latex], and [latex]{\\bf{F}}(x,y)=\\langle{x}^3,5x+e^y\\sin{y}\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"743192057\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"743192057\"]\r\n\r\n[latex]105\\pi[\/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