{"id":1131,"date":"2021-11-09T23:43:35","date_gmt":"2021-11-09T23:43:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=1131"},"modified":"2022-11-01T05:36:54","modified_gmt":"2022-11-01T05:36:54","slug":"the-divergence-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/the-divergence-theorem\/","title":{"raw":"The Divergence Theorem","rendered":"The Divergence Theorem"},"content":{"raw":"
\r\n
\r\n

Learning Objectives<\/h3>\r\n
    \r\n \t
  • Explain the meaning of the divergence theorem.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n
    \r\n

    Overview of Theorems<\/h2>\r\n

    Before examining the divergence theorem, it is helpful to begin with an overview of the versions of the\u00a0Fundamental Theorem of Calculus<\/span>\u00a0we have discussed:<\/p>\r\n\r\n<\/section><\/div>\r\n

    \r\n
      \r\n \t
    1. The Fundamental Theorem of Calculus<\/strong>:\r\n<\/span>\r\n
      \r\n

      [latex]\\large{\\displaystyle\\int_a^bf^\\prime(x)dx=f(b)-f(a)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n\r\n<\/span>This theorem relates the integral of derivative [latex]f'[\/latex] over line segment [latex][a,b][\/latex] along the [latex]x[\/latex]-axis to a difference of [latex]f[\/latex] evaluated on the boundary.<\/li>\r\n \t

    2. The Fundamental Theorem for Line Integrals<\/strong>:\r\n<\/span>\r\n
      \r\n

      [latex]\\large{\\displaystyle\\int_C\\nabla f\\cdot d{\\bf{r}}=f(P_1)-f(P_0)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n\r\n<\/span>where [latex]P_0[\/latex] is the initial point of [latex]C[\/latex] and [latex]P_1[\/latex] is the terminal point of [latex]C[\/latex]. The\u00a0Fundamental Theorem for Line Integrals<\/span>\u00a0allows path [latex]C[\/latex] to be a path in a plane or in space, not just a line segment on the [latex]x[\/latex]-axis. If we think of the gradient as a derivative, then this theorem relates an integral of derivative [latex]\\nabla f[\/latex] over path [latex]C[\/latex] to a difference of [latex]f[\/latex] evaluated on the boundary of [latex]C[\/latex].<\/li>\r\n \t

    3. Green\u2019s theorem, circulation form<\/strong>:\r\n<\/span>\r\n
      \r\n

      [latex]\\large{\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\int_C{\\bf{F}}\\cdot d{\\bf{r}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n\r\n<\/span>Since [latex]Q_x-P_y=\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}[\/latex] and curl is a derivative of sorts,\u00a0Green\u2019s theorem<\/span>\u00a0relates the integral of derivative curl[latex]{\\bf{F}}[\/latex]\u00a0over planar region [latex]D[\/latex] to an integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over the boundary of [latex]D[\/latex].<\/li>\r\n \t

    4. Green\u2019s theorem, flux form<\/strong>:\r\n<\/span>\r\n
      \r\n

      [latex]\\large{\\displaystyle\\iint_D(P_x+Q_y)dA=\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds}[\/latex].<\/p>\r\n\r\n<\/div>\r\n\r\n<\/span>Since [latex]P_x+Q_y=\\text{div }{\\bf{F}}[\/latex] and divergence is a derivative of sorts, the flux form of Green\u2019s theorem relates the integral of derivative div\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over planar region [latex]D[\/latex] to an integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over the boundary of [latex]D[\/latex].<\/li>\r\n \t

    5. Stokes\u2019 theorem<\/strong>:\r\n<\/span>\r\n
      [latex]\\large{\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot ds=\\displaystyle\\int_C{\\bf{F}}\\cdot d{\\bf{r}}}[\/latex]<\/div>\r\n\r\n<\/span>If we think of the curl as a derivative of sorts, then\u00a0Stokes\u2019 theorem<\/span>\u00a0relates the integral of derivative curl[latex]{\\bf{F}}[\/latex]\u00a0over surface [latex]S[\/latex] (not necessarily planar) to an integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over the boundary of [latex]S[\/latex].<\/li>\r\n<\/ol>\r\n

      Stating the Divergence Theorem<\/h2>\r\n

      The divergence theorem follows the general pattern of these other theorems. If we think of divergence as a derivative of sorts, then the\u00a0divergence theorem<\/span>\u00a0relates a triple integral of derivative div[latex]{\\bf{F}}[\/latex]\u00a0over a solid to a flux integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over a closed surface [latex]S[\/latex] to a triple integral of the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over the solid enclosed by [latex]S[\/latex].<\/p>\r\n\r\n

      \r\n

      theorem: The divergence theorem<\/h3>\r\n\r\n
      \r\n

      Let [latex]S[\/latex] be a piecewise, smooth closed surface that encloses solid [latex]E[\/latex] in space. Assume that [latex]S[\/latex] is oriented outward, and let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field with continuous partial derivatives on an open region containing [latex]E[\/latex] (Figure 1). Then<\/p>\r\n

      [latex]\\large{\\displaystyle\\iiint_e\\text{div }{\\bf{F}}dV=\\displaystyle\\int_C{\\bf{F}}\\cdot d{\\bf{S}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"attachment_5424\" align=\"aligncenter\" width=\"458\"]\"<img Figure 1. The divergence theorem relates a flux integral across a closed surface [latex]S[\/latex] to a triple integral over solid\u00a0[latex]E[\/latex] enclosed by the surface.[\/caption]\r\n

      Recall that the flux form of Green\u2019s theorem states that [latex]\\displaystyle\\iint_D\\text{div }{\\bf{F}}dA=\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds[\/latex]. Therefore, the divergence theorem is a version of Green\u2019s theorem in one higher dimension.<\/p>\r\n

      The proof of the divergence theorem is beyond the scope of this text. However, we look at an informal proof that gives a general feel for why the theorem is true, but does not prove the theorem with full rigor. This explanation follows the informal explanation given for why Stokes\u2019 theorem is true.<\/p>\r\n\r\n

      \r\n

      Proof<\/h3>\r\n

      Let [latex]B[\/latex] be a small box with sides parallel to the coordinate planes inside [latex]E[\/latex] (Figure 2). Let the center of [latex]B[\/latex] have coordinates [latex](x, y, z)[\/latex] and suppose the edge lengths are [latex]\\Delta x[\/latex], [latex]\\Delta y[\/latex], and [latex]\\Delta z[\/latex] (Figure 2(b)). The normal vector out of the top of the box is\u00a0[latex]{\\bf{k}}[\/latex]\u00a0and the normal vector out of the bottom of the box is [latex]-{\\bf{k}}[\/latex]. The dot product of [latex]{\\bf{F}}=\\langle P,Q,R\\rangle[\/latex] with\u00a0[latex]{\\bf{k}}[\/latex]\u00a0is [latex]R[\/latex] and the dot product with [latex]-{\\bf{k}}[\/latex] is [latex]-R[\/latex]. The area of the top of the box (and the bottom of the box) [latex]\\Delta S[\/latex] is\u00a0[latex]\\Delta x\\Delta y[\/latex].<\/p>\r\n\r\n<\/section>[caption id=\"attachment_5426\" align=\"aligncenter\" width=\"820\"]\"<img Figure 2. (a) A small box [latex]B[\/latex] inside surface\u00a0[latex]E[\/latex] has sides parallel to the coordinate planes. (b) Box\u00a0[latex]B[\/latex] has side lengths [latex]{\\Delta x}[\/latex],[latex]{\\Delta y}[\/latex], and\u00a0[latex]{\\Delta z}[\/latex] (c) If we look at the side view of [latex]B[\/latex], we see that, since [latex](x, y, z)[\/latex] is the center of the box, to get to the top of the box we must travel a vertical distance of\u00a0[latex]{\\Delta z}\/2[\/latex] up from [latex](x, y, z)[\/latex]. Similarly, to get to the bottom of the box we must travel a distance\u00a0[latex]{\\Delta z}\/2[\/latex] down from [latex](x, y, z)[\/latex].[\/caption]<\/div>\r\n

      <\/div>\r\n
      \r\n

      The flux out of the top of the box can be approximated by [latex]R\\left(x,y,z+\\frac{\\Delta z}{2}\\right)\\Delta x\\Delta y[\/latex] (Figure 2(c)) and the flux out of the bottom of the box is [latex]-R\\left(x,y,z-\\frac{\\Delta z}{2}\\right)\\Delta x\\Delta y[\/latex]. If we denote the difference between these values as [latex]\\Delta R[\/latex], then the net flux in the vertical direction can be approximated by [latex]\\Delta R\\Delta x\\Delta y[\/latex]. However,<\/p>\r\n

      [latex]\\large{\\Delta R\\Delta x\\Delta y=\\left(\\frac{\\Delta R}{\\Delta z}\\right)\\Delta x\\Delta y\\Delta z\\approx\\left(\\frac{\\partial R}{\\partial z}\\right)\\Delta V}[\/latex].<\/p>\r\n

      Therefore, the net flux in the vertical direction can be approximated by [latex]\\left(\\frac{\\partial R}{\\partial z}\\right)\\Delta V[\/latex]. Similarly, the net flux in the [latex]x[\/latex]-direction can be approximated by [latex]\\left(\\frac{\\partial P}{\\partial x}\\right)\\Delta V[\/latex] and the net flux in the [latex]y[\/latex]-direction can be approximated by [latex]\\left(\\frac{\\partial Q}{\\partial y}\\right)\\Delta V[\/latex]. Adding the fluxes in all three directions gives an approximation of the total flux out of the box:<\/p>\r\n

      Total flux [latex]\\approx\\left(\\frac{\\partial P}{\\partial x}+\\frac{\\partial Q}{\\partial y}+\\frac{\\partial R}{\\partial z}\\right)\\Delta V=\\text{div }{\\bf{F}}\\Delta V[\/latex].<\/p>\r\n

      This approximation becomes arbitrarily close to the value of the total flux as the volume of the box shrinks to zero.<\/p>\r\n

      The sum of [latex]\\text{div }{\\bf{F}}\\Delta V[\/latex] over all the small boxes approximating [latex]E[\/latex] is approximately [latex]\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV[\/latex]. On the other hand, the sum of [latex]\\text{div }{\\bf{F}}\\Delta V[\/latex] over all the small boxes approximating [latex]E[\/latex] is the sum of the fluxes over all these boxes. Just as in the informal proof of Stokes\u2019 theorem, adding these fluxes over all the boxes results in the cancelation of a lot of the terms. If an approximating box shares a face with another approximating box, then the flux over one face is the negative of the flux over the shared face of the adjacent box. These two integrals cancel out. When adding up all the fluxes, the only flux integrals that survive are the integrals over the faces approximating the boundary of [latex]E[\/latex]. As the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over [latex]S[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      Example: verifying the divergence theorem<\/h3>\r\nVerify the divergence theorem for vector field [latex]{\\bf{F}}=\\langle x-y,x+z, z-y\\rangle[\/latex] and surface [latex]S[\/latex] that consists of cone [latex]x^{2}+y^{2}=z^{2}[\/latex], [latex]0\\leq z\\leq1[\/latex], and the circular top of the cone (see the following figure). Assume this surface is positively oriented.\r\n\r\n[caption id=\"attachment_5429\" align=\"aligncenter\" width=\"431\"]\"This Figure 3. The vector field [latex]{\\bf{F}}=\\langle x-y,x+z, z-y\\rangle[\/latex] and surface [latex]S[\/latex] that consists of cone [latex]x^{2}+y^{2}=z^{2}[\/latex], [latex]0\\leq z\\leq1[\/latex], and the circular top of the cone[\/caption][reveal-answer q=\"727712381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"727712381\"]\r\n

      Let [latex]E[\/latex] be the solid cone enclosed by [latex]S[\/latex]. To verify the theorem for this example, we show that [latex]\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV=\\displaystyle\\iint_s{\\bf{F}}\\cdot d{\\bf{S}}[\/latex] by calculating each integral separately.<\/p>\r\n

      To compute the triple integral, note that [latex]\\text{div }{\\bf{F}}=P_x+Q_y+R_z=2[\/latex], and therefore the triple integral is<\/p>\r\n

      [latex]\\large{\\begin{aligned}\r\n\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV&=2\\displaystyle\\iiint_EdV \\\\\r\n&=2\\text{(volume of E)}\r\n\\end{aligned}}[\/latex].<\/p>\r\n

      The volume of a right circular cone is given by [latex]\\pi{r}^2\\frac{h}3[\/latex]. In this case, [latex]h=r=1[\/latex]. Therefore,<\/p>\r\n

      [latex]\\large{\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV=2}\\text{(volume of E)}=\\frac{2\\pi}{3}[\/latex].<\/p>\r\n

      To compute the flux integral, first note that [latex]S[\/latex] is piecewise smooth; [latex]S[\/latex] can be written as a union of smooth surfaces. Therefore, we break the flux integral into two pieces: one flux integral across the circular top of the cone and one flux integral across the remaining portion of the cone. Call the circular top [latex]S_1[\/latex] and the portion under the top [latex]S_2[\/latex]. We start by calculating the flux across the circular top of the cone. Notice that [latex]S_1[\/latex] has parameterization<\/p>\r\n

      [latex]\\large{{\\bf{r}}(u,v)=\\langle u,\\cos v,u\\sin v,1\\rangle, \\ 0\\leq u\\leq1, \\ 0\\leq v\\leq2\\pi}[\/latex].<\/p>\r\n

      Then, the tangent vectors are [latex]{\\bf{t}}_u\\langle\\cos v,\\sin v, 0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle-u\\cos v,u\\sin v,0\\rangle[\/latex]. Therefore, the flux across [latex]S_1[\/latex] is<\/p>\r\n

      [latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_1}{\\bf{F}}\\cdot d{\\bf{S}}&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}{\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v)dA \\\\\r\n&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}\\langle u\\cos v-u\\sin v,u\\cos v+1,1-u\\sin v\\rangle\\cdot\\langle0,0,u\\rangle \\ dvdu \\\\\r\n&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}u-u^2\\sin v \\ dvdu=\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

      We now calculate the flux over [latex]S_2[\/latex]. A parameterization of this surface is<\/p>\r\n

      [latex]\\large{{\\bf{r}}(u,v)=\\langle u,\\cos v,u\\sin v,1\\rangle, \\ 0\\leq u\\leq1, \\ 0\\leq v\\leq2\\pi}[\/latex].<\/p>\r\n

      The tangent vectors are [latex]{\\bf{t}}_u\\langle\\cos v,\\sin v, 1\\rangle[\/latex] and [latex]{\\bf{t}}_v\\langle-u\\sin v,u\\cos v, 0\\rangle[\/latex], so the cross product is<\/p>\r\n

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

      Notice that the negative signs on the [latex]x[\/latex] and [latex]y[\/latex] components induce the negative (or inward) orientation of the cone. Since the surface is positively oriented, we use vector [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle u\\cos v,u\\sin v,-u\\rangle[\/latex] in the flux integral. The flux across [latex]S_2[\/latex] is then<\/p>\r\n

      [latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_2}{\\bf{F}}\\cdot d{\\bf{S}}&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}{\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v)dA \\\\\r\n&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}\\langle u\\cos v-u\\sin v,u\\cos v+u,u-u\\sin v\\rangle\\cdot\\langle u\\cos v,u\\sin v,-u\\rangle \\\\\r\n&=\\displaystyle\\int_0^1\\displaystyle\\int_0^{2\\pi}u^2\\cos^2v+2u^2\\sin v-u^2 \\ dvdu=-\\frac{\\pi}3\r\n\\end{aligned}[\/latex].<\/p>\r\nThe total flux across [latex]S[\/latex] is\r\n

      [latex]\\large{\\displaystyle\\iint_{S_2}{\\bf{F}}\\cdot d{\\bf{S}}=\\displaystyle\\iint_{S_1}{\\bf{F}}\\cdot d{\\bf{S}}+\\displaystyle\\iint_{S_2}{\\bf{F}}\\cdot d{\\bf{S}}=\\frac{2\\pi}3=\\displaystyle\\iiint_e\\text{div }{\\bf{F}}dV}[\/latex],<\/p>\r\n

      and we have verified the divergence theorem for this example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      \r\n

      try it<\/h3>\r\nVerify the divergence theorem for vector field [latex]{\\bf{F}}(x,y,z)=\\langle x+y+z,y,2x-y\\rangle[\/latex] and surface [latex]S[\/latex] given by the cylinder [latex]x^{2}+y^{2}=1[\/latex], [latex]0\\leq z\\leq3[\/latex] plus the circular top and bottom of the cylinder. Assume that [latex]S[\/latex] is positively oriented.\r\n\r\n[reveal-answer q=\"934283621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934283621\"]\r\n\r\nBoth integrals equal\u00a0[latex]6\\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