{"id":5513,"date":"2022-06-02T19:02:17","date_gmt":"2022-06-02T19:02:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5513"},"modified":"2022-11-01T05:37:31","modified_gmt":"2022-11-01T05:37:31","slug":"using-the-divergence-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/using-the-divergence-theorem\/","title":{"raw":"Using the Divergence Theorem","rendered":"Using the Divergence Theorem"},"content":{"raw":"<div data-type=\"note\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Use the divergence theorem to calculate the flux of a vector field.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Apply the divergence theorem to an electrostatic field.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Using the Divergence Theorem<\/h2>\r\n<p id=\"fs-id1167793501335\">The divergence theorem translates between the flux integral of closed surface [latex]S[\/latex] and a triple integral over the solid enclosed by [latex]S[\/latex]. Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: applying the divergence theorem<\/h3>\r\nCalculate the surface integral [latex]\\displaystyle\\iint_{S}{\\bf{F}}\\cdot d{\\bf{S}}[\/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, and [latex]{\\bf{F}}=\\left\\langle\\frac{x^3}{3}+yz,\\frac{y^3}3-\\sin(xz),z-x-y\\right\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"945562173\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"945562173\"]\r\n<p id=\"fs-id1167793932468\">We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using the equation to calculate scalar surface integrals.<\/p>\r\n<p id=\"fs-id1167793629311\">By contrast, the divergence theorem allows us to calculate the single triple integral [latex]\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV[\/latex], where [latex]E[\/latex] is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}&amp;=\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV \\\\\r\n&amp;=\\displaystyle\\iiint_E(x^2+y^2+1)dV \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\displaystyle\\int_0^2(r^2+1)r \\ dzdrd\\theta \\\\\r\n&amp;=\\frac32\\displaystyle\\int_0^{2\\pi}d\\theta=3\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nUse the divergence theorem to calculate flux integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex], where [latex]S[\/latex] is the boundary of the box given by [latex]0\\leq x\\leq2[\/latex], [latex]1\\leq y\\leq4[\/latex], [latex]0\\leq z\\leq1[\/latex], and [latex]{\\bf{F}}=\\langle x^2+yz,y-z,2x+2y+2z\\rangle[\/latex] (see the following figure).\r\n\r\n[caption id=\"attachment_5430\" align=\"aligncenter\" width=\"323\"]<img class=\"size-full wp-image-5430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172609\/6.89b.jpg\" alt=\"This figure is a vector diagram in three dimensions. The box of the figure spans x from 0 to 2; y from 0 to 4; and z from 0 to 1. The vectors point up increasingly with distance from the origin; toward larger x with increasing distance from the origin; and toward smaller y values with increasing height.\" width=\"323\" height=\"302\" \/> Figure 1. The box given by [latex]0\\leq x\\leq2[\/latex], [latex]1\\leq y\\leq4[\/latex], [latex]0\\leq z\\leq1[\/latex], and [latex]{\\bf{F}}=\\langle x^2+yz,y-z,2x+2y+2z\\rangle[\/latex][\/caption][reveal-answer q=\"087582178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"087582178\"][latex]30[\/latex].[\/hidden-answer]<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: applying the divergence theorem<\/h3>\r\nLet [latex]{\\bf{v}}=\\left\\langle-\\frac{y}z,\\frac{x}z,0\\right\\rangle[\/latex] be the velocity field of a fluid. Let [latex]C[\/latex] be the solid cube given by [latex]1\\leq x \\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex], and let [latex]S[\/latex] be the boundary of this cube (see the following figure). Find the flow rate of the fluid across [latex]S[\/latex].\r\n\r\n[caption id=\"attachment_5431\" align=\"aligncenter\" width=\"478\"]<img class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172649\/6.90.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/4df0ac7e1163d69369ff76a71f516bddb63c4bcb&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;This is a figure of a diagram of the given vector field in three dimensions. The x components are \u2013y\/z, the y components are x\/z, and the z components are 0.&quot; id=&quot;13&quot;&gt;\" width=\"478\" height=\"533\" \/> Figure 2. Vector field [latex]{\\bf{v}}=\\left\\langle-\\frac{y}z,\\frac{x}z,0\\right\\rangle[\/latex].[\/caption][reveal-answer q=\"342870125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342870125\"]\r\n<p id=\"fs-id1167794032326\">The flow rate of the fluid across [latex]S[\/latex] is [latex]\\displaystyle\\iint_s{\\bf{v}}\\cdot d{\\bf{S}}[\/latex]. Before calculating this flux integral, let\u2019s discuss what the value of the integral should be. Based on\u00a0Figure 2, we see that if we place this cube in the fluid (as long as the cube doesn\u2019t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the [latex]xy[\/latex]-plane that has a center on the [latex]z[\/latex]-axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.<\/p>\r\n<p id=\"fs-id1167793928517\">To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product, and use the equation to calculate scalar surface integrals. However, using the divergence theorem makes this calculation go much more quickly:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_s{\\bf{v}}\\cdot d{\\bf{S}}&amp;=\\displaystyle\\iiint_C\\text{div }({\\bf{v}})dV \\\\\r\n&amp;=\\displaystyle\\iiint_C0 \\ dV=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793378886\">Therefore the flux is zero, as expected.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet [latex]{\\bf{v}}=\\left\\langle\\frac{x}z,\\frac{y}z,0\\right\\rangle[\/latex] be the velocity field of a fluid. Let [latex]C[\/latex] be the solid cube given by [latex]1\\leq x\\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex], and let [latex]S[\/latex] be the boundary of this cube (see the following figure). Find the flow rate of the fluid across [latex]S[\/latex].\r\n\r\n[caption id=\"attachment_5432\" align=\"aligncenter\" width=\"485\"]<img class=\"size-full wp-image-5432\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172827\/6.90a.jpg\" alt=\"This is a figure of a diagram of the given vector field in three dimensions. The x components are x\/z, the y components are y\/z, and the z components are 0.\" width=\"485\" height=\"526\" \/> Figure 3. The solid cube given by [latex]1\\leq x\\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex][\/caption][reveal-answer q=\"390475920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"390475920\"][latex]9\\ln(16)[\/latex][\/hidden-answer]<\/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<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250334&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=wPx5BQt9BY8&amp;video_target=tpm-plugin-fwr205nw-wPx5BQt9BY8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.67_transcript.html\">transcript for \u201cCP 6.67\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167793275081\">Example \"Applying the Divergence Theorem\"\u00a0illustrates a remarkable consequence of the divergence theorem. Let [latex]S[\/latex] be a piecewise, smooth closed surface and let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field defined on an open region containing the surface enclosed by [latex]S[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the form [latex]{\\bf{F}}=\\langle f(y,z),g(x,z),h(x,y)\\rangle[\/latex], then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is zero. By the divergence theorem, the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]S[\/latex] is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex] where [latex]S[\/latex] is a cube and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}=\\langle\\sin(y)e^{yz},x^2z^2,\\cos(xy)e^{\\sin x}\\rangle}[\/latex].<\/p>\r\n<p id=\"fs-id1167793930007\">Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.<\/p>\r\n<p id=\"fs-id1167793944166\">We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a continuous three-dimensional vector field and [latex]P[\/latex] is a point in the domain of\u00a0[latex]{\\bf{F}}[\/latex], then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex] is a measure of the \u201coutflowing-ness\u201d of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of a fluid, then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex] is a measure of the net flow rate out of point [latex]P[\/latex] (the flow of fluid out of [latex]P[\/latex] less the flow of fluid in to [latex]P[\/latex]). To see how the divergence theorem justifies this interpretation, let [latex]B_r[\/latex] be a ball of very small radius [latex]r[\/latex] with center [latex]P[\/latex], and assume that [latex]B_r[\/latex] is in the domain of\u00a0[latex]{\\bf{F}}[\/latex]. Furthermore, assume that [latex]B_r[\/latex] has a positive, outward orientation. Since the radius of [latex]B_r[\/latex] is small and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is continuous, the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is approximately constant on [latex]B_r[\/latex]. That is, if [latex]P'[\/latex] is any point in [latex]B_r[\/latex], then [latex]\\text{div }{\\bf{F}}(P)\\approx\\text{div }{\\bf{F}}(P^\\prime)[\/latex]. Let [latex]S_r[\/latex] denote the boundary sphere of [latex]B_r[\/latex]. We can approximate the flux across [latex]S_r[\/latex] using the divergence theorem as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}&amp;=\\displaystyle\\iiint_{B_r}\\text{div }{\\bf{F}}dV \\\\\r\n&amp;\\approx\\displaystyle\\iiint_{B_r}\\text{div }{\\bf{F}}(P)dV \\\\\r\n&amp;=\\text{div }{\\bf{F}}(P)V(B_r)\r\n\\end{aligned}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial; white-space: nowrap; word-spacing: normal;\">As we shrink the radius [latex]r[\/latex] to zero via a limit, the quantity [latex]\\text{div }{\\bf{F}}(P)V(B_r)[\/latex] gets arbitrarily close to the flux. Therefore,<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{div }{\\bf{F}}(P)=\\displaystyle\\lim_{r\\to0}\\frac{1}{V(B_r)}\\displaystyle\\iint_{S_r}{\\bf{F}}\\cdot d{\\bf{S}}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793479324\">and we can consider the divergence at [latex]P[\/latex] as measuring the net rate of outward flux per unit volume at [latex]P[\/latex]. Since \u201coutflowing-ness\u201d is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<h2 data-type=\"title\">Application to Electrostatic Fields<\/h2>\r\n<p id=\"fs-id1167793420801\">The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes\u2019 theorem allowed us to translate between an integral and differential form of Faraday\u2019s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.<\/p>\r\n<p id=\"fs-id1167794187050\">One of the most common applications of the divergence theorem is to\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term291\" class=\"no-emphasis\" data-type=\"term\">electrostatic fields<\/span>. An important result in this subject is\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term292\" data-type=\"term\">Gauss\u2019 law<\/span>. This law states that if [latex]S[\/latex] is a closed surface in electrostatic field\u00a0[latex]{\\bf{E}}[\/latex], then the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge enclosed by [latex]S[\/latex] (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.<\/p>\r\n<p id=\"fs-id1167794070480\">If [latex](x, y, z)[\/latex] is a point in space, then the distance from the point to the origin is [latex]r=\\sqrt{x^2+y^2+z^2}[\/latex]. Let [latex]{\\bf{F}}_r[\/latex] denote radial vector field [latex]{\\bf{F}}_r=\\frac1{r^2}\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex]. The vector at a given position in space points in the direction of unit radial vector [latex]\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex] and is scaled by the quantity [latex]1\/r^2[\/latex]. Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector\u2019s distance from the origin. Suppose we have a stationary charge of [latex]q[\/latex] Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field\u00a0[latex]{\\bf{E}}[\/latex]\u00a0given by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r}[\/latex],<\/p>\r\n<p id=\"fs-id1167794207200\">where the approximation [latex]\\varepsilon=8.854\\times10^{-12}[\/latex] farad (F)\/m is an electric constant. (The constant [latex]\\varepsilon_0[\/latex] is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is a radial vector field similar to the gravitational field described in\u00a0Example \"A Unit Vector Field\". The difference is that this field points outward whereas the gravitational field points inward. Because<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r=\\frac{q}{4\\pi\\varepsilon_0}\\left(\\frac1{r^1}\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle\\right)}[\/latex],<\/p>\r\n<p id=\"fs-id1167793959912\">we say that electrostatic fields obey an\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term293\" data-type=\"term\">inverse-square law<\/span>. That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface [latex]S[\/latex] is zero if the charge is outside of [latex]S[\/latex], and that the flux is [latex]q\/\\varepsilon_0[\/latex] if the charge is inside of [latex]S[\/latex]. In other words, the flux across [latex]S[\/latex] is the charge inside the surface divided by constant [latex]\\varepsilon_0[\/latex]. This is a special case of Gauss\u2019 law, and here we use the divergence theorem to justify this special case.<\/p>\r\n<p id=\"fs-id1167794246555\">To show that the flux across [latex]S[\/latex] is the charge inside the surface divided by constant [latex]\\varepsilon_0[\/latex], we need two intermediate steps. First we show that the divergence of [latex]{\\bf{F}}_r[\/latex] is zero and then we show that the flux of [latex]{\\bf{F}}_r[\/latex] across any smooth surface [latex]S[\/latex] is either zero or [latex]4\\pi[\/latex]. We can then justify this special case of Gauss\u2019 law.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: the divergence of\u00a0[latex]F_r[\/latex] is Zero<\/h3>\r\nVerify that the divergence of [latex]{\\bf{F}}_r[\/latex] is zero where [latex]{\\bf{F}}_r[\/latex] is defined (away from the origin).\r\n\r\n[reveal-answer q=\"724687101\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"724687101\"]\r\n<p id=\"fs-id1167794126259\">Since [latex]r=\\sqrt{x^2+y^2+z^2}[\/latex], the quotient rule gives us<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\frac{\\partial}{\\partial{x}}\\left(\\frac{x}{r^3}\\right)&amp;=\\frac{\\partial}{\\partial{x}}\\left(\\frac{x}{(x^2+y^2+z^2)^{3\/2}}\\right) \\\\\r\n&amp;=\\frac{(x^2+y^2+z^2)^{3\/2}-x\\left[\\frac32(x^2+y^2+z^2)^{1\/2}2x\\right]}{(x^2+y^2+z^2)^3} \\\\\r\n&amp;=\\frac{r^3-3x^2r}{r^6}=\\frac{r^2-3x^2}{r^5}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793639946\">Similarly,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{y}}\\left(\\frac{y}{r^3}\\right)=\\frac{r^2-3y^2}{r^5}\\text{ and }\\frac{\\partial}{\\partial{z}}\\left(\\frac{z}{r^3}\\right)=\\frac{r^2-3z^2}{r^5}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794122035\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\text{div }{\\bf{F}}_r&amp;=\\frac{r^2-3x^2}{r^5}+\\frac{r^2-3y^2}{r^5}+\\frac{r^2-3z^2}{r^5} \\\\\r\n&amp;=\\frac{3r^2-3(x^2+y^2+z^2)}{r^5} \\\\\r\n&amp;=\\frac{3r^2-3r^2}{r^5}=0<\/p>\r\n\\end{aligned}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793711739\">Notice that since the divergence of [latex]{\\bf{F}}_r[\/latex] is zero and\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is [latex]{\\bf{F}}_r[\/latex] scaled by a constant, the divergence of electrostatic field\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is also zero (except at the origin).<\/p>\r\n\r\n<div id=\"fs-id1167794029720\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: flux across a smooth surface<\/h3>\r\n<p id=\"fs-id1167793444887\">Let [latex]S[\/latex] be a connected, piecewise smooth closed surface and let [latex]{\\bf{F}}_r=\\frac{1}{r^2}\\left\\langle\\frac{x}r,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex]. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation*}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}} = \\left\\{\r\n\\begin{array}{ll}\r\n0 &amp; \\quad \\text{if }S\\text{ does not encompass the origin} \\\\\r\n4\\pi &amp; \\quad \\text{if }S\\text{ encompasses the origin.}\r\n\\end{array}\r\n\\right.\r\n\\end{equation*}\r\n[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793294244\">In other words, this theorem says that the flux of [latex]{\\bf{F}}_r[\/latex] across any piecewise smooth closed surface [latex]S[\/latex] depends only on whether the origin is inside of [latex]S[\/latex].<\/p>\r\n\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1167793384743\">The logic of this proof follows the logic of\u00a0Example \"Using Green\u2019s Theorem on a Region with Holes\", only we use the divergence theorem rather than Green\u2019s theorem.<\/p>\r\n<p id=\"fs-id1167793931487\">First, suppose that [latex]S[\/latex] does not encompass the origin. In this case, the solid enclosed by [latex]S[\/latex] is in the domain of [latex]{\\bf{F}}_r[\/latex], and since the divergence of [latex]{\\bf{F}}_r[\/latex] is zero, we can immediately apply the divergence theorem and find that [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex] is zero.<\/p>\r\n<p id=\"fs-id1167794138267\">Now suppose that [latex]S[\/latex] does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let [latex]S_a[\/latex] be a sphere of radius [latex]a[\/latex] inside of [latex]S[\/latex] centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta=\\langle a^2\\cos\\theta\\sin^2\\phi,a^2\\sin\\theta\\sin^2\\phi,a^2\\sin\\phi\\cos\\phi\\rangle}[\/latex]<\/p>\r\n<p id=\"fs-id1167793949480\">(see\u00a0Example \"Calculating Surface Area\"). Therefore, on the surface of the sphere, the dot product [latex]{\\bf{F}}_r\\cdot{\\bf{N}}[\/latex] (in spherical coordinates) is<\/p>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{aligned}\r\n{\\bf{F}}_r\\cdot{\\bf{N}}&amp;=\\left\\langle\\frac{\\sin\\phi\\cos\\theta}{a^2},\\frac{\\sin\\phi\\sin\\theta}{a^2},\\frac{\\cos\\phi}{a^2}\\right\\rangle\\cdot\\langle a^2\\cos\\theta\\sin^2\\phi,a^2\\sin\\theta\\sin^2\\phi,a^2\\sin\\phi\\cos\\phi\\rangle \\\\\r\n&amp;=\\sin\\phi(\\langle\\sin\\phi\\cos\\theta,\\sin\\phi\\sin\\theta,\\cos\\phi\\rangle\\cdot\\langle\\sin\\phi\\cos\\theta,\\sin\\phi\\sin\\theta,\\cos\\phi\\rangle) \\\\\r\n&amp;=\\sin\\phi\r\n\\end{aligned}[\/latex].<\/p>\r\nThe flux of [latex]{\\bf{F}}_r[\/latex] across [latex]S_a[\/latex] is\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^\\pi\\sin\\phi d\\phi d\\theta=4\\pi}[\/latex].<\/p>\r\n<p id=\"fs-id1167793245829\">Now, remember that we are interested in the flux across [latex]S[\/latex], not necessarily the flux across [latex]S_a[\/latex]. To calculate the flux across [latex]S[\/latex], let [latex]E[\/latex] be the solid between surfaces [latex]S_a[\/latex] and [latex]S[\/latex]. Then, the boundary of [latex]E[\/latex] consists of [latex]S_a[\/latex] and [latex]S[\/latex]. Denote this boundary by [latex]S-S_a[\/latex] to indicate that [latex]S[\/latex] is oriented outward but now [latex]S_a[\/latex] is oriented inward. We would like to apply the divergence theorem to solid\u00a0<em data-effect=\"italics\">E.<\/em>\u00a0Notice that the divergence theorem, as stated, can\u2019t handle a solid such as [latex]E[\/latex] because [latex]E[\/latex] has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green\u2019s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S-S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS&amp;=\\displaystyle\\iint_S{\\bf{F}}_r\\cdot{\\bf{N}}dS-\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS \\\\\r\n&amp;=\\displaystyle\\iiint_E\\text{div }{\\bf{F}}_r dV \\\\\r\n&amp;=\\displaystyle\\iiint_E0 \\ dV=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167794098930\">Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{F}}_r\\cdot{\\bf{N}}dS=\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdotdS=4\\pi}[\/latex],<\/p>\r\n<p id=\"fs-id1167794181311\">and we have our desired result.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1167793502247\">Now we return to calculating the flux across a smooth surface in the context of electrostatic field [latex]{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r[\/latex] of a point charge at the origin. Let [latex]S[\/latex] be a piecewise smooth closed surface that encompasses the origin. Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}&amp;=\\displaystyle\\iint_S\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r\\cdot d{\\bf{S}} \\\\\r\n&amp;=\\frac{q}{4\\pi\\varepsilon_0}\\displaystyle\\iint_S{\\bf{F}}_r\\cdot d{\\bf{S}} \\\\\r\n&amp;=\\frac{q}{\\varepsilon_0}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793629345\">If [latex]S[\/latex] does not encompass the origin, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}=\\frac{q}{4\\pi\\varepsilon_0}\\displaystyle\\iint_S{\\bf{F}}_r\\cdot d{\\bf{S}}=0}[\/latex].<\/p>\r\n<p id=\"fs-id1167794048045\">Therefore, we have justified the claim that we set out to justify: the flux across closed surface [latex]S[\/latex] is zero if the charge is outside of [latex]S[\/latex], and the flux is [latex]q\/\\varepsilon_0[\/latex] if the charge is inside of [latex]S[\/latex].<\/p>\r\n<p id=\"fs-id1167794122678\">This analysis works only if there is a single point charge at the origin. In this case, Gauss\u2019 law says that the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge enclosed by [latex]S[\/latex]. Gauss\u2019 law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss\u2019 law states that if [latex]S[\/latex] is a piecewise smooth closed surface and [latex]Q[\/latex] is the total amount of charge inside of [latex]S[\/latex], then the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is\u00a0[latex]Q\/\\varepsilon_0[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167793379049\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using Gauss' Law<\/h3>\r\nSuppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at [latex](0, 1, 1)[\/latex], [latex](1, 1, 4)[\/latex], [latex](-1, 0, 0)[\/latex], and [latex](-2, -2, 2)[\/latex]. Let\u00a0[latex]{\\bf{E}}[\/latex]\u00a0denote the electrostatic field generated by these point charges. If [latex]S[\/latex] is the sphere of radius 2 oriented outward and centered at the origin, then find [latex]\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}[\/latex].\r\n\r\n[reveal-answer q=\"240167357\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"240167357\"]\r\n<p id=\"fs-id1167793936429\">According to Gauss\u2019 law, the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge inside of\u00a0<em data-effect=\"italics\">S<\/em>\u00a0divided by the electric constant. Since [latex]S[\/latex] has radius 2, notice that only two of the charges are inside of [latex]S[\/latex]: the charge at [latex](0, 1, 1)[\/latex] and the charge at [latex](-1, 0, 0)[\/latex]. Therefore, the total charge encompassed by [latex]S[\/latex] is 0.004 and, by Gauss\u2019 law,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}=\\frac{0.004}{8.854\\times10^{-12}}\\approx4.518\\times10^9\\text{ V-m}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWork the previous example for surface [latex]S[\/latex] that is a sphere of radius 4 centered at the origin, oriented outward.\r\n\r\n[reveal-answer q=\"644230841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"644230841\"]\r\n\r\n[latex]\\approx6.777\\times10^9[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div data-type=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Use the divergence theorem to calculate the flux of a vector field.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Apply the divergence theorem to an electrostatic field.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Using the Divergence Theorem<\/h2>\n<p id=\"fs-id1167793501335\">The divergence theorem translates between the flux integral of closed surface [latex]S[\/latex] and a triple integral over the solid enclosed by [latex]S[\/latex]. Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: applying the divergence theorem<\/h3>\n<p>Calculate the surface integral [latex]\\displaystyle\\iint_{S}{\\bf{F}}\\cdot d{\\bf{S}}[\/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, and [latex]{\\bf{F}}=\\left\\langle\\frac{x^3}{3}+yz,\\frac{y^3}3-\\sin(xz),z-x-y\\right\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q945562173\">Show Solution<\/span><\/p>\n<div id=\"q945562173\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793932468\">We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using the equation to calculate scalar surface integrals.<\/p>\n<p id=\"fs-id1167793629311\">By contrast, the divergence theorem allows us to calculate the single triple integral [latex]\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV[\/latex], where [latex]E[\/latex] is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}&=\\displaystyle\\iiint_E\\text{div }{\\bf{F}}dV \\\\  &=\\displaystyle\\iiint_E(x^2+y^2+1)dV \\\\  &=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\displaystyle\\int_0^2(r^2+1)r \\ dzdrd\\theta \\\\  &=\\frac32\\displaystyle\\int_0^{2\\pi}d\\theta=3\\pi  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Use the divergence theorem to calculate flux integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex], where [latex]S[\/latex] is the boundary of the box given by [latex]0\\leq x\\leq2[\/latex], [latex]1\\leq y\\leq4[\/latex], [latex]0\\leq z\\leq1[\/latex], and [latex]{\\bf{F}}=\\langle x^2+yz,y-z,2x+2y+2z\\rangle[\/latex] (see the following figure).<\/p>\n<div id=\"attachment_5430\" style=\"width: 333px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5430\" class=\"size-full wp-image-5430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172609\/6.89b.jpg\" alt=\"This figure is a vector diagram in three dimensions. The box of the figure spans x from 0 to 2; y from 0 to 4; and z from 0 to 1. The vectors point up increasingly with distance from the origin; toward larger x with increasing distance from the origin; and toward smaller y values with increasing height.\" width=\"323\" height=\"302\" \/><\/p>\n<p id=\"caption-attachment-5430\" class=\"wp-caption-text\">Figure 1. The box given by [latex]0\\leq x\\leq2[\/latex], [latex]1\\leq y\\leq4[\/latex], [latex]0\\leq z\\leq1[\/latex], and [latex]{\\bf{F}}=\\langle x^2+yz,y-z,2x+2y+2z\\rangle[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q087582178\">Show Solution<\/span><\/p>\n<div id=\"q087582178\" class=\"hidden-answer\" style=\"display: none\">[latex]30[\/latex].<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: applying the divergence theorem<\/h3>\n<p>Let [latex]{\\bf{v}}=\\left\\langle-\\frac{y}z,\\frac{x}z,0\\right\\rangle[\/latex] be the velocity field of a fluid. Let [latex]C[\/latex] be the solid cube given by [latex]1\\leq x \\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex], and let [latex]S[\/latex] be the boundary of this cube (see the following figure). Find the flow rate of the fluid across [latex]S[\/latex].<\/p>\n<div id=\"attachment_5431\" style=\"width: 488px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5431\" class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172649\/6.90.jpg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20220422.171947\/resources\/4df0ac7e1163d69369ff76a71f516bddb63c4bcb&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;This is a figure of a diagram of the given vector field in three dimensions. The x components are \u2013y\/z, the y components are x\/z, and the z components are 0.&quot; id=&quot;13&quot;&gt;\" width=\"478\" height=\"533\" \/><\/p>\n<p id=\"caption-attachment-5431\" class=\"wp-caption-text\">Figure 2. Vector field [latex]{\\bf{v}}=\\left\\langle-\\frac{y}z,\\frac{x}z,0\\right\\rangle[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342870125\">Show Solution<\/span><\/p>\n<div id=\"q342870125\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794032326\">The flow rate of the fluid across [latex]S[\/latex] is [latex]\\displaystyle\\iint_s{\\bf{v}}\\cdot d{\\bf{S}}[\/latex]. Before calculating this flux integral, let\u2019s discuss what the value of the integral should be. Based on\u00a0Figure 2, we see that if we place this cube in the fluid (as long as the cube doesn\u2019t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the [latex]xy[\/latex]-plane that has a center on the [latex]z[\/latex]-axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.<\/p>\n<p id=\"fs-id1167793928517\">To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product, and use the equation to calculate scalar surface integrals. However, using the divergence theorem makes this calculation go much more quickly:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_s{\\bf{v}}\\cdot d{\\bf{S}}&=\\displaystyle\\iiint_C\\text{div }({\\bf{v}})dV \\\\  &=\\displaystyle\\iiint_C0 \\ dV=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793378886\">Therefore the flux is zero, as expected.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Let [latex]{\\bf{v}}=\\left\\langle\\frac{x}z,\\frac{y}z,0\\right\\rangle[\/latex] be the velocity field of a fluid. Let [latex]C[\/latex] be the solid cube given by [latex]1\\leq x\\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex], and let [latex]S[\/latex] be the boundary of this cube (see the following figure). Find the flow rate of the fluid across [latex]S[\/latex].<\/p>\n<div id=\"attachment_5432\" style=\"width: 495px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5432\" class=\"size-full wp-image-5432\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/02172827\/6.90a.jpg\" alt=\"This is a figure of a diagram of the given vector field in three dimensions. The x components are x\/z, the y components are y\/z, and the z components are 0.\" width=\"485\" height=\"526\" \/><\/p>\n<p id=\"caption-attachment-5432\" class=\"wp-caption-text\">Figure 3. The solid cube given by [latex]1\\leq x\\leq4[\/latex], [latex]2\\leq y\\leq5[\/latex], [latex]1\\leq z\\leq4[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390475920\">Show Solution<\/span><\/p>\n<div id=\"q390475920\" class=\"hidden-answer\" style=\"display: none\">[latex]9\\ln(16)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250334&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=wPx5BQt9BY8&amp;video_target=tpm-plugin-fwr205nw-wPx5BQt9BY8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.67_transcript.html\">transcript for \u201cCP 6.67\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167793275081\">Example &#8220;Applying the Divergence Theorem&#8221;\u00a0illustrates a remarkable consequence of the divergence theorem. Let [latex]S[\/latex] be a piecewise, smooth closed surface and let\u00a0[latex]{\\bf{F}}[\/latex]\u00a0be a vector field defined on an open region containing the surface enclosed by [latex]S[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0has the form [latex]{\\bf{F}}=\\langle f(y,z),g(x,z),h(x,y)\\rangle[\/latex], then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is zero. By the divergence theorem, the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]S[\/latex] is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex] where [latex]S[\/latex] is a cube and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{F}}=\\langle\\sin(y)e^{yz},x^2z^2,\\cos(xy)e^{\\sin x}\\rangle}[\/latex].<\/p>\n<p id=\"fs-id1167793930007\">Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.<\/p>\n<p id=\"fs-id1167793944166\">We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a continuous three-dimensional vector field and [latex]P[\/latex] is a point in the domain of\u00a0[latex]{\\bf{F}}[\/latex], then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex] is a measure of the \u201coutflowing-ness\u201d of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0represents the velocity field of a fluid, then the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0at [latex]P[\/latex] is a measure of the net flow rate out of point [latex]P[\/latex] (the flow of fluid out of [latex]P[\/latex] less the flow of fluid in to [latex]P[\/latex]). To see how the divergence theorem justifies this interpretation, let [latex]B_r[\/latex] be a ball of very small radius [latex]r[\/latex] with center [latex]P[\/latex], and assume that [latex]B_r[\/latex] is in the domain of\u00a0[latex]{\\bf{F}}[\/latex]. Furthermore, assume that [latex]B_r[\/latex] has a positive, outward orientation. Since the radius of [latex]B_r[\/latex] is small and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is continuous, the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is approximately constant on [latex]B_r[\/latex]. That is, if [latex]P'[\/latex] is any point in [latex]B_r[\/latex], then [latex]\\text{div }{\\bf{F}}(P)\\approx\\text{div }{\\bf{F}}(P^\\prime)[\/latex]. Let [latex]S_r[\/latex] denote the boundary sphere of [latex]B_r[\/latex]. We can approximate the flux across [latex]S_r[\/latex] using the divergence theorem as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}&=\\displaystyle\\iiint_{B_r}\\text{div }{\\bf{F}}dV \\\\  &\\approx\\displaystyle\\iiint_{B_r}\\text{div }{\\bf{F}}(P)dV \\\\  &=\\text{div }{\\bf{F}}(P)V(B_r)  \\end{aligned}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial; white-space: nowrap; word-spacing: normal;\">As we shrink the radius [latex]r[\/latex] to zero via a limit, the quantity [latex]\\text{div }{\\bf{F}}(P)V(B_r)[\/latex] gets arbitrarily close to the flux. Therefore,<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{div }{\\bf{F}}(P)=\\displaystyle\\lim_{r\\to0}\\frac{1}{V(B_r)}\\displaystyle\\iint_{S_r}{\\bf{F}}\\cdot d{\\bf{S}}}[\/latex]<\/p>\n<p id=\"fs-id1167793479324\">and we can consider the divergence at [latex]P[\/latex] as measuring the net rate of outward flux per unit volume at [latex]P[\/latex]. Since \u201coutflowing-ness\u201d is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.<\/p>\n<\/div>\n<div data-type=\"note\">\n<h2 data-type=\"title\">Application to Electrostatic Fields<\/h2>\n<p id=\"fs-id1167793420801\">The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes\u2019 theorem allowed us to translate between an integral and differential form of Faraday\u2019s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.<\/p>\n<p id=\"fs-id1167794187050\">One of the most common applications of the divergence theorem is to\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term291\" class=\"no-emphasis\" data-type=\"term\">electrostatic fields<\/span>. An important result in this subject is\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term292\" data-type=\"term\">Gauss\u2019 law<\/span>. This law states that if [latex]S[\/latex] is a closed surface in electrostatic field\u00a0[latex]{\\bf{E}}[\/latex], then the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge enclosed by [latex]S[\/latex] (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.<\/p>\n<p id=\"fs-id1167794070480\">If [latex](x, y, z)[\/latex] is a point in space, then the distance from the point to the origin is [latex]r=\\sqrt{x^2+y^2+z^2}[\/latex]. Let [latex]{\\bf{F}}_r[\/latex] denote radial vector field [latex]{\\bf{F}}_r=\\frac1{r^2}\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex]. The vector at a given position in space points in the direction of unit radial vector [latex]\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex] and is scaled by the quantity [latex]1\/r^2[\/latex]. Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector\u2019s distance from the origin. Suppose we have a stationary charge of [latex]q[\/latex] Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field\u00a0[latex]{\\bf{E}}[\/latex]\u00a0given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r}[\/latex],<\/p>\n<p id=\"fs-id1167794207200\">where the approximation [latex]\\varepsilon=8.854\\times10^{-12}[\/latex] farad (F)\/m is an electric constant. (The constant [latex]\\varepsilon_0[\/latex] is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is a radial vector field similar to the gravitational field described in\u00a0Example &#8220;A Unit Vector Field&#8221;. The difference is that this field points outward whereas the gravitational field points inward. Because<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r=\\frac{q}{4\\pi\\varepsilon_0}\\left(\\frac1{r^1}\\left\\langle\\frac{x}y,\\frac{y}r,\\frac{z}r\\right\\rangle\\right)}[\/latex],<\/p>\n<p id=\"fs-id1167793959912\">we say that electrostatic fields obey an\u00a0<span id=\"d77de5bc-07e0-40d3-a789-c2f01549dd9b_term293\" data-type=\"term\">inverse-square law<\/span>. That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface [latex]S[\/latex] is zero if the charge is outside of [latex]S[\/latex], and that the flux is [latex]q\/\\varepsilon_0[\/latex] if the charge is inside of [latex]S[\/latex]. In other words, the flux across [latex]S[\/latex] is the charge inside the surface divided by constant [latex]\\varepsilon_0[\/latex]. This is a special case of Gauss\u2019 law, and here we use the divergence theorem to justify this special case.<\/p>\n<p id=\"fs-id1167794246555\">To show that the flux across [latex]S[\/latex] is the charge inside the surface divided by constant [latex]\\varepsilon_0[\/latex], we need two intermediate steps. First we show that the divergence of [latex]{\\bf{F}}_r[\/latex] is zero and then we show that the flux of [latex]{\\bf{F}}_r[\/latex] across any smooth surface [latex]S[\/latex] is either zero or [latex]4\\pi[\/latex]. We can then justify this special case of Gauss\u2019 law.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: the divergence of\u00a0[latex]F_r[\/latex] is Zero<\/h3>\n<p>Verify that the divergence of [latex]{\\bf{F}}_r[\/latex] is zero where [latex]{\\bf{F}}_r[\/latex] is defined (away from the origin).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q724687101\">Show Solution<\/span><\/p>\n<div id=\"q724687101\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794126259\">Since [latex]r=\\sqrt{x^2+y^2+z^2}[\/latex], the quotient rule gives us<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\frac{\\partial}{\\partial{x}}\\left(\\frac{x}{r^3}\\right)&=\\frac{\\partial}{\\partial{x}}\\left(\\frac{x}{(x^2+y^2+z^2)^{3\/2}}\\right) \\\\  &=\\frac{(x^2+y^2+z^2)^{3\/2}-x\\left[\\frac32(x^2+y^2+z^2)^{1\/2}2x\\right]}{(x^2+y^2+z^2)^3} \\\\  &=\\frac{r^3-3x^2r}{r^6}=\\frac{r^2-3x^2}{r^5}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793639946\">Similarly,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial}{\\partial{y}}\\left(\\frac{y}{r^3}\\right)=\\frac{r^2-3y^2}{r^5}\\text{ and }\\frac{\\partial}{\\partial{z}}\\left(\\frac{z}{r^3}\\right)=\\frac{r^2-3z^2}{r^5}}[\/latex].<\/p>\n<p id=\"fs-id1167794122035\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\text{div }{\\bf{F}}_r&=\\frac{r^2-3x^2}{r^5}+\\frac{r^2-3y^2}{r^5}+\\frac{r^2-3z^2}{r^5} \\\\  &=\\frac{3r^2-3(x^2+y^2+z^2)}{r^5} \\\\  &=\\frac{3r^2-3r^2}{r^5}=0<\/p>\n<p>  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793711739\">Notice that since the divergence of [latex]{\\bf{F}}_r[\/latex] is zero and\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is [latex]{\\bf{F}}_r[\/latex] scaled by a constant, the divergence of electrostatic field\u00a0[latex]{\\bf{E}}[\/latex]\u00a0is also zero (except at the origin).<\/p>\n<div id=\"fs-id1167794029720\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: flux across a smooth surface<\/h3>\n<p id=\"fs-id1167793444887\">Let [latex]S[\/latex] be a connected, piecewise smooth closed surface and let [latex]{\\bf{F}}_r=\\frac{1}{r^2}\\left\\langle\\frac{x}r,\\frac{y}r,\\frac{z}r\\right\\rangle[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation*}  \\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}} = \\left\\{  \\begin{array}{ll}  0 & \\quad \\text{if }S\\text{ does not encompass the origin} \\\\  4\\pi & \\quad \\text{if }S\\text{ encompasses the origin.}  \\end{array}  \\right.  \\end{equation*}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167793294244\">In other words, this theorem says that the flux of [latex]{\\bf{F}}_r[\/latex] across any piecewise smooth closed surface [latex]S[\/latex] depends only on whether the origin is inside of [latex]S[\/latex].<\/p>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1167793384743\">The logic of this proof follows the logic of\u00a0Example &#8220;Using Green\u2019s Theorem on a Region with Holes&#8221;, only we use the divergence theorem rather than Green\u2019s theorem.<\/p>\n<p id=\"fs-id1167793931487\">First, suppose that [latex]S[\/latex] does not encompass the origin. In this case, the solid enclosed by [latex]S[\/latex] is in the domain of [latex]{\\bf{F}}_r[\/latex], and since the divergence of [latex]{\\bf{F}}_r[\/latex] is zero, we can immediately apply the divergence theorem and find that [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot d{\\bf{S}}[\/latex] is zero.<\/p>\n<p id=\"fs-id1167794138267\">Now suppose that [latex]S[\/latex] does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let [latex]S_a[\/latex] be a sphere of radius [latex]a[\/latex] inside of [latex]S[\/latex] centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta=\\langle a^2\\cos\\theta\\sin^2\\phi,a^2\\sin\\theta\\sin^2\\phi,a^2\\sin\\phi\\cos\\phi\\rangle}[\/latex]<\/p>\n<p id=\"fs-id1167793949480\">(see\u00a0Example &#8220;Calculating Surface Area&#8221;). Therefore, on the surface of the sphere, the dot product [latex]{\\bf{F}}_r\\cdot{\\bf{N}}[\/latex] (in spherical coordinates) is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{F}}_r\\cdot{\\bf{N}}&=\\left\\langle\\frac{\\sin\\phi\\cos\\theta}{a^2},\\frac{\\sin\\phi\\sin\\theta}{a^2},\\frac{\\cos\\phi}{a^2}\\right\\rangle\\cdot\\langle a^2\\cos\\theta\\sin^2\\phi,a^2\\sin\\theta\\sin^2\\phi,a^2\\sin\\phi\\cos\\phi\\rangle \\\\  &=\\sin\\phi(\\langle\\sin\\phi\\cos\\theta,\\sin\\phi\\sin\\theta,\\cos\\phi\\rangle\\cdot\\langle\\sin\\phi\\cos\\theta,\\sin\\phi\\sin\\theta,\\cos\\phi\\rangle) \\\\  &=\\sin\\phi  \\end{aligned}[\/latex].<\/p>\n<p>The flux of [latex]{\\bf{F}}_r[\/latex] across [latex]S_a[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS=\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^\\pi\\sin\\phi d\\phi d\\theta=4\\pi}[\/latex].<\/p>\n<p id=\"fs-id1167793245829\">Now, remember that we are interested in the flux across [latex]S[\/latex], not necessarily the flux across [latex]S_a[\/latex]. To calculate the flux across [latex]S[\/latex], let [latex]E[\/latex] be the solid between surfaces [latex]S_a[\/latex] and [latex]S[\/latex]. Then, the boundary of [latex]E[\/latex] consists of [latex]S_a[\/latex] and [latex]S[\/latex]. Denote this boundary by [latex]S-S_a[\/latex] to indicate that [latex]S[\/latex] is oriented outward but now [latex]S_a[\/latex] is oriented inward. We would like to apply the divergence theorem to solid\u00a0<em data-effect=\"italics\">E.<\/em>\u00a0Notice that the divergence theorem, as stated, can\u2019t handle a solid such as [latex]E[\/latex] because [latex]E[\/latex] has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green\u2019s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_{S-S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS&=\\displaystyle\\iint_S{\\bf{F}}_r\\cdot{\\bf{N}}dS-\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdot{\\bf{N}}dS \\\\  &=\\displaystyle\\iiint_E\\text{div }{\\bf{F}}_r dV \\\\  &=\\displaystyle\\iiint_E0 \\ dV=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167794098930\">Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{F}}_r\\cdot{\\bf{N}}dS=\\displaystyle\\iint_{S_a}{\\bf{F}}_r\\cdotdS=4\\pi}[\/latex],<\/p>\n<p id=\"fs-id1167794181311\">and we have our desired result.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167793502247\">Now we return to calculating the flux across a smooth surface in the context of electrostatic field [latex]{\\bf{E}}=\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r[\/latex] of a point charge at the origin. Let [latex]S[\/latex] be a piecewise smooth closed surface that encompasses the origin. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}&=\\displaystyle\\iint_S\\frac{q}{4\\pi\\varepsilon_0}{\\bf{F}}_r\\cdot d{\\bf{S}} \\\\  &=\\frac{q}{4\\pi\\varepsilon_0}\\displaystyle\\iint_S{\\bf{F}}_r\\cdot d{\\bf{S}} \\\\  &=\\frac{q}{\\varepsilon_0}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793629345\">If [latex]S[\/latex] does not encompass the origin, then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}=\\frac{q}{4\\pi\\varepsilon_0}\\displaystyle\\iint_S{\\bf{F}}_r\\cdot d{\\bf{S}}=0}[\/latex].<\/p>\n<p id=\"fs-id1167794048045\">Therefore, we have justified the claim that we set out to justify: the flux across closed surface [latex]S[\/latex] is zero if the charge is outside of [latex]S[\/latex], and the flux is [latex]q\/\\varepsilon_0[\/latex] if the charge is inside of [latex]S[\/latex].<\/p>\n<p id=\"fs-id1167794122678\">This analysis works only if there is a single point charge at the origin. In this case, Gauss\u2019 law says that the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge enclosed by [latex]S[\/latex]. Gauss\u2019 law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss\u2019 law states that if [latex]S[\/latex] is a piecewise smooth closed surface and [latex]Q[\/latex] is the total amount of charge inside of [latex]S[\/latex], then the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is\u00a0[latex]Q\/\\varepsilon_0[\/latex].<\/p>\n<div id=\"fs-id1167793379049\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: using Gauss&#8217; Law<\/h3>\n<p>Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at [latex](0, 1, 1)[\/latex], [latex](1, 1, 4)[\/latex], [latex](-1, 0, 0)[\/latex], and [latex](-2, -2, 2)[\/latex]. Let\u00a0[latex]{\\bf{E}}[\/latex]\u00a0denote the electrostatic field generated by these point charges. If [latex]S[\/latex] is the sphere of radius 2 oriented outward and centered at the origin, then find [latex]\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q240167357\">Show Solution<\/span><\/p>\n<div id=\"q240167357\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793936429\">According to Gauss\u2019 law, the flux of\u00a0[latex]{\\bf{E}}[\/latex]\u00a0across [latex]S[\/latex] is the total charge inside of\u00a0<em data-effect=\"italics\">S<\/em>\u00a0divided by the electric constant. Since [latex]S[\/latex] has radius 2, notice that only two of the charges are inside of [latex]S[\/latex]: the charge at [latex](0, 1, 1)[\/latex] and the charge at [latex](-1, 0, 0)[\/latex]. Therefore, the total charge encompassed by [latex]S[\/latex] is 0.004 and, by Gauss\u2019 law,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S{\\bf{E}}\\cdot d{\\bf{S}}=\\frac{0.004}{8.854\\times10^{-12}}\\approx4.518\\times10^9\\text{ V-m}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Work the previous example for surface [latex]S[\/latex] that is a sphere of radius 4 centered at the origin, oriented outward.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q644230841\">Show Solution<\/span><\/p>\n<div id=\"q644230841\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\approx6.777\\times10^9[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5513\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 6.67. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":36,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 6.67\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5513","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5513","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5513\/revisions"}],"predecessor-version":[{"id":6165,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5513\/revisions\/6165"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/24"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5513\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5513"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5513"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5513"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}