{"id":791,"date":"2025-10-09T15:39:31","date_gmt":"2025-10-09T15:39:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=791"},"modified":"2025-11-06T19:29:30","modified_gmt":"2025-11-06T19:29:30","slug":"stokes-theorem-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/stokes-theorem-learn-it-1\/","title":{"raw":"Stokes\u2019 Theorem: Learn It 1","rendered":"Stokes\u2019 Theorem: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use Stokes' theorem to relate line integrals to surface integrals<\/li>\r\n \t<li>Use Stokes\u2019 theorem to calculate a surface integral<\/li>\r\n \t<li>Use Stokes\u2019 theorem to calculate a curl<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Stokes\u2019 Theorem<\/h2>\r\n<p class=\"whitespace-normal break-words\">Stokes' theorem is a higher-dimensional generalization of Green's theorem. Like the Fundamental Theorem for Line Integrals and Green's theorem, it generalizes the Fundamental Theorem of Calculus to higher dimensions by relating an integral over a geometric object to an integral over its boundary.<\/p>\r\n<p class=\"whitespace-normal break-words\">Specifically, Stokes' theorem relates a surface integral over surface [latex]S[\/latex] to a line integral around the boundary of [latex]S[\/latex]. This connection allows us to translate between line integrals and surface integrals, and it links the concepts of curl and circulation. The theorem has important applications in fluid mechanics and electromagnetism, including the derivation of Faraday's law.<\/p>\r\n<p class=\"whitespace-normal break-words\">Stokes' theorem says we can calculate the flux of curl [latex]\\mathbf{F}[\/latex] across surface [latex]S[\/latex] by knowing information only about the values of [latex]\\mathbf{F}[\/latex] along the boundary of [latex]S[\/latex]. Conversely, we can calculate the line integral of vector field [latex]\\mathbf{F}[\/latex] along the boundary of surface [latex]S[\/latex] by translating to a double integral of the curl of [latex]\\mathbf{F}[\/latex] over [latex]S[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]S[\/latex] be an oriented smooth surface with unit normal vector [latex]\\mathbf{N}[\/latex]. Furthermore, suppose the boundary of [latex]S[\/latex] is a simple closed curve [latex]C[\/latex]. The orientation of [latex]S[\/latex] induces the positive orientation of [latex]C[\/latex] if, as you walk in the positive direction around [latex]C[\/latex] with your head pointing in the direction of [latex]\\mathbf{N}[\/latex], the surface is always on your left.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Stokes' theorem<\/h3>\r\n<p id=\"fs-id1167793502349\">Let [latex]S[\/latex] be a piecewise smooth oriented surface with a boundary that is a simple closed curve [latex]C[\/latex] with positive orientation (Figure 1). If [latex]{\\bf{F}}[\/latex] is a vector field with component functions that have continuous partial derivatives on an open region containing [latex]S[\/latex], then:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\r\n\r\n<\/section>\r\n\r\n[caption id=\"attachment_1508\" align=\"aligncenter\" width=\"413\"]<img class=\"size-full wp-image-1508\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279.webp\" alt=\"A diagram of a surface S in three dimensions. The orientation of the curve C around its boundary is positive. Various normals are drawn coming off of the surface.\" width=\"413\" height=\"449\" \/> Figure 1. Stokes\u2019 theorem relates the flux integral over the surface to a line integral around the boundary of the surface. Note that the orientation of the curve is positive.[\/caption]\r\n\r\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\r\n<p class=\"whitespace-normal break-words\"><strong>Stokes' Theorem and Green's Theorem<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Suppose surface [latex]S[\/latex] is a flat region in the [latex]xy[\/latex]-plane with upward orientation. Then the unit normal vector is [latex]\\mathbf{k}[\/latex] and surface integral [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] is actually the double integral [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}{d}A[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">In this special case, Stokes' theorem gives:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}{d}A[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">However, this is the circulation form of Green's theorem, which shows us that Green's theorem is a special case of Stokes' theorem. Green's theorem can only handle surfaces in a plane, but Stokes' theorem can handle surfaces in a plane or in space.<\/p>\r\n\r\n<\/section>The complete proof of Stokes\u2019 theorem is beyond the scope of this text. We look at an intuitive explanation for the truth of the theorem and then see proof of the theorem in the special case that surface [latex]S[\/latex] is a portion of a graph of a function, and [latex]S[\/latex], the boundary of [latex]S[\/latex], and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are all fairly tame.\r\n\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793696474\">First, we look at an informal proof of the theorem. This proof is not rigorous, but it is meant to give a general feeling for why the theorem is true. Let [latex]S[\/latex] be a surface and let [latex]D[\/latex] be a small piece of the surface so that [latex]D[\/latex] does not share any points with the boundary of [latex]S[\/latex]. We choose [latex]D[\/latex] to be small enough so that it can be approximated by an oriented square [latex]E[\/latex]. Let [latex]D[\/latex] inherit its orientation from [latex]S[\/latex], and give [latex]E[\/latex] the same orientation. This square has four sides; denote them [latex]E_l[\/latex], [latex]E_r[\/latex], [latex]E_u[\/latex], and [latex]E_d[\/latex] for the left, right, up, and down sides, respectively. On the square, we can use the flux form of Green\u2019s theorem:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{E_l+E_d+E_r+E_u}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_E\\text{curl }{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_E\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793958564\">To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface (Figure 2). By Green\u2019s theorem, the flux across each approximating square is a line integral over its boundary. Let [latex]F[\/latex] be an approximating square with an orientation inherited from [latex]S[\/latex] and with a right side [latex]E_l[\/latex] (so [latex]F[\/latex] is to the left of [latex]E[\/latex]). Let [latex]F_r[\/latex] denote the right side of [latex]F[\/latex]; then, [latex]E_l=-F_r[\/latex]. In other words, the right side of [latex]F[\/latex] is the same curve as the left side of [latex]E[\/latex], just oriented in the opposite direction. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{E_l}{\\bf{F}}\\cdot{d}{\\bf{r}}=-\\displaystyle\\int_{F_r}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793250345\">As we add up all the fluxes over all the squares approximating surface [latex]S[\/latex], line integrals [latex]\\displaystyle\\int_{E_l}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] and [latex]\\displaystyle\\int_{F_r}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]cancel each other out. The same goes for the line integrals over the other three sides of [latex]E[\/latex]. These three line integrals cancel out with the line integral of the lower side of the square above [latex]E[\/latex], the line integral over the left side of the square to the right of [latex]E[\/latex], and the line integral over the upper side of the square below [latex]E[\/latex] (Figure 3). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of [latex]S[\/latex]. Therefore, the sum of all the fluxes (which, by Green\u2019s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of [latex]S[\/latex]. In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1510\" align=\"aligncenter\" width=\"403\"]<img class=\"size-full wp-image-1510\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d.webp\" alt=\"A diagram of a surface S sectioned into small pieces in a grid \u2013 they are small enough to be approximated by a square E. The orientation of each square is the same as S, shown with counterclockwise arrows in each square.. The N and F vectors are drawn coming off of one square.\" width=\"403\" height=\"298\" \/> Figure 2. Chop the surface into small pieces. The pieces should be small enough that they can be approximated by a square.[\/caption]\r\n\r\n[caption id=\"attachment_1511\" align=\"aligncenter\" width=\"680\"]<img class=\"size-full wp-image-1511\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362.webp\" alt=\"Two diagrams labeled A and B. A contains two squares, F and E. Both have arrows along the sides in a counterclockwise manner. The right side of F is labeled F_r. The left, right, upper, and lower sides of E are labeled E_l, E_r, E_u, and E_d, respectively. B contains five squares. F and E are drawn as they are in diagram A. Above, to the right, and below E there are three other squares: G, H, and I, respectively. All have arrows along their sides in a counterclockwise manner. The bottom side of G is labeled G_d, the left side of H is labeled H_l, and the upper side of I is labeled I_u.\" width=\"680\" height=\"404\" \/> Figure 3. (a) The line integral along [latex]{E_l}[\/latex] cancels out the line integral along [latex]{F_r}[\/latex] because [latex]{E_l}=-{F_r}[\/latex]. (b) The line integral along any of the sides of [latex]E[\/latex] cancels out with the line integral along a side of an adjacent approximating square.[\/caption]Let\u2019s now look at a rigorous proof of the theorem in the special case that [latex]S[\/latex] is the graph of function [latex]z=f(x, y)[\/latex], where [latex]x[\/latex] and [latex]y[\/latex] vary over a bounded, simply connected region [latex]D[\/latex] of finite area (Figure 4). Furthermore, assume that [latex]f[\/latex] has continuous second-order partial derivatives. Let [latex]C[\/latex] denote the boundary of [latex]S[\/latex] and let [latex]C'[\/latex] denote the boundary of [latex]D[\/latex]. Then, [latex]D[\/latex] is the \u201cshadow\u201d of [latex]S[\/latex] in the plane and [latex]C'[\/latex] is the \u201cshadow\u201d of [latex]C[\/latex]. Suppose that [latex]S[\/latex] is oriented upward. The counterclockwise orientation of [latex]C[\/latex] is positive, as is the counterclockwise orientation of [latex]C^\\prime[\/latex]. Let [latex]{\\bf{F}}(x,y,z)=\\langle{P},Q,R\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives.[caption id=\"attachment_1512\" align=\"aligncenter\" width=\"261\"]<img class=\"size-full wp-image-1512\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae.webp\" alt=\"A diagram of a surface S in three dimensions, where z is a function of x and y notated as z=f(x,y). The normal N is drawn pointing up and away from the surface. D is the shadow, or projection of S in the (x,y)-plane. The curve around S is labeled C, and C\u2019 is the projection of C in the (x,y)-plane. Arrows are drawn on C, the boundary of S, in a counterclockwise manner.\" width=\"261\" height=\"260\" \/> Figure 4. [latex]D[\/latex] is the \u201cshadow,\u201d or projection, of\u00a0[latex]S[\/latex] in the plane and [latex]C'[\/latex] is the projection of [latex]C[\/latex].[\/caption]\r\n<p id=\"fs-id1167794051238\">We take the standard parameterization of [latex]S:x=x,y=y,z=g(x,y)[\/latex]. The tangent vectors are [latex]{\\bf{t}}_x=\\langle1,0,g_x\\rangle[\/latex],<span style=\"font-size: 1rem; text-align: initial;\">and [latex]{\\bf{t}}_y=\\langle0,1,g_y\\rangle[\/latex], and therefore, [latex]{\\bf{t}}_x\\times{\\bf{t}}_y=\\langle-g_x,-g_y,1\\rangle[\/latex]. By the equation to calculate scalar surface integrals<\/span><span style=\"font-size: 1rem; text-align: initial;\">,<\/span><\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_D[-(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}[\/latex],<\/p>\r\n<p id=\"fs-id1167793509977\">where the partial derivatives are all evaluated at [latex](x,y,g(x,y))[\/latex], making the integrand depend on [latex]x[\/latex] and [latex]y[\/latex] only. Suppose [latex]\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] is a parameterization of [latex]C'[\/latex]. Then, a parameterization of [latex]C[\/latex] is [latex]\\langle{x}(t),y(t),g(x(t),y(t))\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Armed with these parameterizations, the Chain rule, and Green\u2019s theorem, and keeping in mind that [latex]P[\/latex],\u00a0[latex]Q[\/latex], and [latex]R[\/latex] are all functions of [latex]x[\/latex] and [latex]y[\/latex], we can evaluate line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_a^b(Px^\\prime(t)+Qy^\\prime(t)+Rz^\\prime(t))dt \\\\\r\n&amp;=\\displaystyle\\int_a^b\\left[Px^\\prime(t)+Qy^\\prime(t)+R\\left(\\frac{\\partial{z}}{\\partial{x}}\\frac{dx}{dt}+\\frac{\\partial{z}}{\\partial{y}}\\frac{dy}{dt}\\right)\\right]dt \\\\\r\n&amp;=\\displaystyle\\int_a^b\\left[\\left((P+R\\frac{\\partial{z}}{\\partial{x}}\\right)x^\\prime(t)+\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)y^\\prime(t)\\right]dt \\\\\r\n&amp;=\\displaystyle\\int_{C^\\prime}\\left(P+R\\frac{\\partial{z}}{\\partial{x}}\\right)dx+\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)dy \\\\\r\n&amp;=\\displaystyle\\iint_D\\left[\\frac{\\partial}{\\partial{x}}\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)-\\frac{\\partial}{\\partial{y}}\\left(P+R\\frac{\\partial{z}}{\\partial{x}}\\right)\\right]dA \\\\\r\n&amp;=\\displaystyle\\iint_D\\left(\\frac{\\partial{Q}}{\\partial{x}}+\\frac{\\partial{Q}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{x}}+\\frac{\\partial{R}}{\\partial{x}}\\frac{\\partial{z}}{\\partial{y}}+\\frac{\\partial{R}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{x}}\\frac{\\partial{z}}{\\partial{y}}+R\\frac{\\partial^2{z}}{\\partial{x}\\partial{y}}\\right)-\\left(\\frac{\\partial{P}}{\\partial{y}}+\\frac{\\partial{P}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{y}}+\\frac{\\partial{R}}{\\partial{y}}\\frac{\\partial{z}}{\\partial{x}}+\\frac{\\partial{R}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{y}}\\frac{\\partial{z}}{\\partial{x}}+R\\frac{\\partial^2{z}}{\\partial{y}\\partial{x}}\\right)dA\r\n\\end{aligned}}[\/latex].<\/p>\r\nBy Clairaut\u2019s theorem, [latex]\\frac{\\partial^2{z}}{\\partial{x}\\partial{y}}=\\frac{\\partial^2{z}}{\\partial{y}\\partial{x}}[\/latex]. Therefore, four of the terms disappear from this double integral, and we are left with\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_D[-R(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}[\/latex],<\/p>\r\n<p id=\"fs-id1167793372539\">which equals [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">We have shown that Stokes' theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field [latex]\\mathbf{F}[\/latex] is conservative.<\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]\\mathbf{F}[\/latex] is conservative, the curl of [latex]\\mathbf{F}[\/latex] is zero, so [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=0[\/latex]. Since the boundary of [latex]S[\/latex] is a closed curve,[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is also zero.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Verify that Stokes\u2019 theorem is true for vector field [latex]{\\bf{F}}(x,y,z)=\\langle{y},2z,x^2\\rangle[\/latex] and surface [latex]S[\/latex], where [latex]S[\/latex] is the paraboloid [latex]z=4-x^{2}-y^{2}[\/latex]. Assume the surface is outward oriented and [latex]z\\geq0[\/latex]\r\n\r\n[caption id=\"attachment_1514\" align=\"aligncenter\" width=\"450\"]<img class=\"size-full wp-image-1514\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1.webp\" alt=\"A diagram in three dimensions of a hemisphere in a vector field. The arrows of the vector field follow the shape of the hemisphere, which is located in quadrants 2 and 3 of the (x, y) plane and stretches up and down into the z-plane. The center of the hemisphere is at the origin. The normal N is drawn stretching up and away from the hemisphere.\" width=\"450\" height=\"453\" \/> Figure 5.\u00a0Verifying Stokes\u2019 theorem for a hemisphere in a vector field.[\/caption]\r\n\r\n[reveal-answer q=\"384759728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384759728\"]\r\n<p id=\"fs-id1167793371929\">As a surface integral, you have [latex]g(x, y)=4-x^{2}-y^{2}[\/latex], [latex]g_x=-2y[\/latex], and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\ \\frac{\\partial}{\\partial{x}}&amp;\\frac{\\partial}{\\partial{y}}&amp;\\frac{\\partial}{\\partial{z}} \\\\ y&amp;2z&amp;x^2\\end{vmatrix}=\\langle-2,-2x,-1\\rangle}[\/latex].<\/p>\r\n<p id=\"fs-id1167793918257\">By the equation to calculate scalar surface integrals,<\/p>\r\n\r\n<div id=\"fs-id1167793400648\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}&amp;=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}({\\bf{r}}(\\phi,\\theta))\\cdot{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta)dA \\\\\r\n&amp;=\\displaystyle\\iint_D\\langle-2,-2x,-1\\rangle\\cdot\\langle2x,2y,1\\rangle{d}A \\\\\r\n&amp;=\\displaystyle\\int_{-2}^2\\displaystyle\\int_{\\sqrt{4-x^2}}^{\\sqrt{4-x^2}}(-4x-4xy-1)dydx \\\\\r\n&amp;=\\displaystyle\\int_{-2}^2\\left(8x\\sqrt{4-x^2}-2\\sqrt{4-x^2}\\right){d}x \\\\\r\n&amp;=4\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793831988\">As a line integral, you can parameterize [latex]C[\/latex] by [latex]{\\bf{r}}(t)=\\langle2\\cos{t},2\\sin{t},0\\rangle, \\ 0\\leq{t}\\leq2\\pi[\/latex]. By the equation to calculate scalar surface integrals,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&amp;=\\displaystyle\\int_0^{2\\pi}\\langle2\\sin{t},0,4\\cos^2t\\rangle\\cdot\\langle-2\\sin{t},2\\cos{t},0\\rangle{d}t \\\\\r\n&amp;=\\displaystyle\\int_0^{2\\pi}-4\\sin^2tdt=-4\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"eip-idm922876032\">Therefore, we have verified Stokes' theorem for this example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use Stokes&#8217; theorem to relate line integrals to surface integrals<\/li>\n<li>Use Stokes\u2019 theorem to calculate a surface integral<\/li>\n<li>Use Stokes\u2019 theorem to calculate a curl<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Stokes\u2019 Theorem<\/h2>\n<p class=\"whitespace-normal break-words\">Stokes&#8217; theorem is a higher-dimensional generalization of Green&#8217;s theorem. Like the Fundamental Theorem for Line Integrals and Green&#8217;s theorem, it generalizes the Fundamental Theorem of Calculus to higher dimensions by relating an integral over a geometric object to an integral over its boundary.<\/p>\n<p class=\"whitespace-normal break-words\">Specifically, Stokes&#8217; theorem relates a surface integral over surface [latex]S[\/latex] to a line integral around the boundary of [latex]S[\/latex]. This connection allows us to translate between line integrals and surface integrals, and it links the concepts of curl and circulation. The theorem has important applications in fluid mechanics and electromagnetism, including the derivation of Faraday&#8217;s law.<\/p>\n<p class=\"whitespace-normal break-words\">Stokes&#8217; theorem says we can calculate the flux of curl [latex]\\mathbf{F}[\/latex] across surface [latex]S[\/latex] by knowing information only about the values of [latex]\\mathbf{F}[\/latex] along the boundary of [latex]S[\/latex]. Conversely, we can calculate the line integral of vector field [latex]\\mathbf{F}[\/latex] along the boundary of surface [latex]S[\/latex] by translating to a double integral of the curl of [latex]\\mathbf{F}[\/latex] over [latex]S[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]S[\/latex] be an oriented smooth surface with unit normal vector [latex]\\mathbf{N}[\/latex]. Furthermore, suppose the boundary of [latex]S[\/latex] is a simple closed curve [latex]C[\/latex]. The orientation of [latex]S[\/latex] induces the positive orientation of [latex]C[\/latex] if, as you walk in the positive direction around [latex]C[\/latex] with your head pointing in the direction of [latex]\\mathbf{N}[\/latex], the surface is always on your left.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Stokes&#8217; theorem<\/h3>\n<p id=\"fs-id1167793502349\">Let [latex]S[\/latex] be a piecewise smooth oriented surface with a boundary that is a simple closed curve [latex]C[\/latex] with positive orientation (Figure 1). If [latex]{\\bf{F}}[\/latex] is a vector field with component functions that have continuous partial derivatives on an open region containing [latex]S[\/latex], then:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\n<\/section>\n<figure id=\"attachment_1508\" aria-describedby=\"caption-attachment-1508\" style=\"width: 413px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1508\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279.webp\" alt=\"A diagram of a surface S in three dimensions. The orientation of the curve C around its boundary is positive. Various normals are drawn coming off of the surface.\" width=\"413\" height=\"449\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279.webp 413w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279-276x300.webp 276w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279-65x71.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279-225x245.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17163959\/e525d486fd64d91ca2c5617944964475be95f279-350x381.webp 350w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><figcaption id=\"caption-attachment-1508\" class=\"wp-caption-text\">Figure 1. Stokes\u2019 theorem relates the flux integral over the surface to a line integral around the boundary of the surface. Note that the orientation of the curve is positive.<\/figcaption><\/figure>\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\n<p class=\"whitespace-normal break-words\"><strong>Stokes&#8217; Theorem and Green&#8217;s Theorem<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Suppose surface [latex]S[\/latex] is a flat region in the [latex]xy[\/latex]-plane with upward orientation. Then the unit normal vector is [latex]\\mathbf{k}[\/latex] and surface integral [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex] is actually the double integral [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}{d}A[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">In this special case, Stokes&#8217; theorem gives:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{\\bf{k}}{d}A[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">However, this is the circulation form of Green&#8217;s theorem, which shows us that Green&#8217;s theorem is a special case of Stokes&#8217; theorem. Green&#8217;s theorem can only handle surfaces in a plane, but Stokes&#8217; theorem can handle surfaces in a plane or in space.<\/p>\n<\/section>\n<p>The complete proof of Stokes\u2019 theorem is beyond the scope of this text. We look at an intuitive explanation for the truth of the theorem and then see proof of the theorem in the special case that surface [latex]S[\/latex] is a portion of a graph of a function, and [latex]S[\/latex], the boundary of [latex]S[\/latex], and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are all fairly tame.<\/p>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1167793696474\">First, we look at an informal proof of the theorem. This proof is not rigorous, but it is meant to give a general feeling for why the theorem is true. Let [latex]S[\/latex] be a surface and let [latex]D[\/latex] be a small piece of the surface so that [latex]D[\/latex] does not share any points with the boundary of [latex]S[\/latex]. We choose [latex]D[\/latex] to be small enough so that it can be approximated by an oriented square [latex]E[\/latex]. Let [latex]D[\/latex] inherit its orientation from [latex]S[\/latex], and give [latex]E[\/latex] the same orientation. This square has four sides; denote them [latex]E_l[\/latex], [latex]E_r[\/latex], [latex]E_u[\/latex], and [latex]E_d[\/latex] for the left, right, up, and down sides, respectively. On the square, we can use the flux form of Green\u2019s theorem:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{E_l+E_d+E_r+E_u}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\iint_E\\text{curl }{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_E\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}}[\/latex].<\/p>\n<p id=\"fs-id1167793958564\">To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface (Figure 2). By Green\u2019s theorem, the flux across each approximating square is a line integral over its boundary. Let [latex]F[\/latex] be an approximating square with an orientation inherited from [latex]S[\/latex] and with a right side [latex]E_l[\/latex] (so [latex]F[\/latex] is to the left of [latex]E[\/latex]). Let [latex]F_r[\/latex] denote the right side of [latex]F[\/latex]; then, [latex]E_l=-F_r[\/latex]. In other words, the right side of [latex]F[\/latex] is the same curve as the left side of [latex]E[\/latex], just oriented in the opposite direction. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_{E_l}{\\bf{F}}\\cdot{d}{\\bf{r}}=-\\displaystyle\\int_{F_r}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\n<p id=\"fs-id1167793250345\">As we add up all the fluxes over all the squares approximating surface [latex]S[\/latex], line integrals [latex]\\displaystyle\\int_{E_l}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] and [latex]\\displaystyle\\int_{F_r}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]cancel each other out. The same goes for the line integrals over the other three sides of [latex]E[\/latex]. These three line integrals cancel out with the line integral of the lower side of the square above [latex]E[\/latex], the line integral over the left side of the square to the right of [latex]E[\/latex], and the line integral over the upper side of the square below [latex]E[\/latex] (Figure 3). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of [latex]S[\/latex]. Therefore, the sum of all the fluxes (which, by Green\u2019s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of [latex]S[\/latex]. In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.<\/p>\n<figure id=\"attachment_1510\" aria-describedby=\"caption-attachment-1510\" style=\"width: 403px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1510\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d.webp\" alt=\"A diagram of a surface S sectioned into small pieces in a grid \u2013 they are small enough to be approximated by a square E. The orientation of each square is the same as S, shown with counterclockwise arrows in each square.. The N and F vectors are drawn coming off of one square.\" width=\"403\" height=\"298\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d.webp 403w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d-300x222.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d-65x48.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d-225x166.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164119\/4fadafd8cb8baaa459b20bd9a798cf7060bbb94d-350x259.webp 350w\" sizes=\"auto, (max-width: 403px) 100vw, 403px\" \/><figcaption id=\"caption-attachment-1510\" class=\"wp-caption-text\">Figure 2. Chop the surface into small pieces. The pieces should be small enough that they can be approximated by a square.<\/figcaption><\/figure>\n<figure id=\"attachment_1511\" aria-describedby=\"caption-attachment-1511\" style=\"width: 680px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1511\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362.webp\" alt=\"Two diagrams labeled A and B. A contains two squares, F and E. Both have arrows along the sides in a counterclockwise manner. The right side of F is labeled F_r. The left, right, upper, and lower sides of E are labeled E_l, E_r, E_u, and E_d, respectively. B contains five squares. F and E are drawn as they are in diagram A. Above, to the right, and below E there are three other squares: G, H, and I, respectively. All have arrows along their sides in a counterclockwise manner. The bottom side of G is labeled G_d, the left side of H is labeled H_l, and the upper side of I is labeled I_u.\" width=\"680\" height=\"404\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362.webp 680w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362-300x178.webp 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362-65x39.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362-225x134.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164140\/c29ceb78894cf02969fe9d426037a243f7635362-350x208.webp 350w\" sizes=\"auto, (max-width: 680px) 100vw, 680px\" \/><figcaption id=\"caption-attachment-1511\" class=\"wp-caption-text\">Figure 3. (a) The line integral along [latex]{E_l}[\/latex] cancels out the line integral along [latex]{F_r}[\/latex] because [latex]{E_l}=-{F_r}[\/latex]. (b) The line integral along any of the sides of [latex]E[\/latex] cancels out with the line integral along a side of an adjacent approximating square.<\/figcaption><\/figure>\n<p>Let\u2019s now look at a rigorous proof of the theorem in the special case that [latex]S[\/latex] is the graph of function [latex]z=f(x, y)[\/latex], where [latex]x[\/latex] and [latex]y[\/latex] vary over a bounded, simply connected region [latex]D[\/latex] of finite area (Figure 4). Furthermore, assume that [latex]f[\/latex] has continuous second-order partial derivatives. Let [latex]C[\/latex] denote the boundary of [latex]S[\/latex] and let [latex]C'[\/latex] denote the boundary of [latex]D[\/latex]. Then, [latex]D[\/latex] is the \u201cshadow\u201d of [latex]S[\/latex] in the plane and [latex]C'[\/latex] is the \u201cshadow\u201d of [latex]C[\/latex]. Suppose that [latex]S[\/latex] is oriented upward. The counterclockwise orientation of [latex]C[\/latex] is positive, as is the counterclockwise orientation of [latex]C^\\prime[\/latex]. Let [latex]{\\bf{F}}(x,y,z)=\\langle{P},Q,R\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives.<\/p>\n<figure id=\"attachment_1512\" aria-describedby=\"caption-attachment-1512\" style=\"width: 261px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1512\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae.webp\" alt=\"A diagram of a surface S in three dimensions, where z is a function of x and y notated as z=f(x,y). The normal N is drawn pointing up and away from the surface. D is the shadow, or projection of S in the (x,y)-plane. The curve around S is labeled C, and C\u2019 is the projection of C in the (x,y)-plane. Arrows are drawn on C, the boundary of S, in a counterclockwise manner.\" width=\"261\" height=\"260\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae.webp 261w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae-150x150.webp 150w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae-65x65.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164201\/2269474d805a18f4a65e4b9d751bcc197e6172ae-225x224.webp 225w\" sizes=\"auto, (max-width: 261px) 100vw, 261px\" \/><figcaption id=\"caption-attachment-1512\" class=\"wp-caption-text\">Figure 4. [latex]D[\/latex] is the \u201cshadow,\u201d or projection, of\u00a0[latex]S[\/latex] in the plane and [latex]C'[\/latex] is the projection of [latex]C[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1167794051238\">We take the standard parameterization of [latex]S:x=x,y=y,z=g(x,y)[\/latex]. The tangent vectors are [latex]{\\bf{t}}_x=\\langle1,0,g_x\\rangle[\/latex],<span style=\"font-size: 1rem; text-align: initial;\">and [latex]{\\bf{t}}_y=\\langle0,1,g_y\\rangle[\/latex], and therefore, [latex]{\\bf{t}}_x\\times{\\bf{t}}_y=\\langle-g_x,-g_y,1\\rangle[\/latex]. By the equation to calculate scalar surface integrals<\/span><span style=\"font-size: 1rem; text-align: initial;\">,<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_D[-(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}[\/latex],<\/p>\n<p id=\"fs-id1167793509977\">where the partial derivatives are all evaluated at [latex](x,y,g(x,y))[\/latex], making the integrand depend on [latex]x[\/latex] and [latex]y[\/latex] only. Suppose [latex]\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex] is a parameterization of [latex]C'[\/latex]. Then, a parameterization of [latex]C[\/latex] is [latex]\\langle{x}(t),y(t),g(x(t),y(t))\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Armed with these parameterizations, the Chain rule, and Green\u2019s theorem, and keeping in mind that [latex]P[\/latex],\u00a0[latex]Q[\/latex], and [latex]R[\/latex] are all functions of [latex]x[\/latex] and [latex]y[\/latex], we can evaluate line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_a^b(Px^\\prime(t)+Qy^\\prime(t)+Rz^\\prime(t))dt \\\\  &=\\displaystyle\\int_a^b\\left[Px^\\prime(t)+Qy^\\prime(t)+R\\left(\\frac{\\partial{z}}{\\partial{x}}\\frac{dx}{dt}+\\frac{\\partial{z}}{\\partial{y}}\\frac{dy}{dt}\\right)\\right]dt \\\\  &=\\displaystyle\\int_a^b\\left[\\left((P+R\\frac{\\partial{z}}{\\partial{x}}\\right)x^\\prime(t)+\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)y^\\prime(t)\\right]dt \\\\  &=\\displaystyle\\int_{C^\\prime}\\left(P+R\\frac{\\partial{z}}{\\partial{x}}\\right)dx+\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)dy \\\\  &=\\displaystyle\\iint_D\\left[\\frac{\\partial}{\\partial{x}}\\left(Q+R\\frac{\\partial{z}}{\\partial{y}}\\right)-\\frac{\\partial}{\\partial{y}}\\left(P+R\\frac{\\partial{z}}{\\partial{x}}\\right)\\right]dA \\\\  &=\\displaystyle\\iint_D\\left(\\frac{\\partial{Q}}{\\partial{x}}+\\frac{\\partial{Q}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{x}}+\\frac{\\partial{R}}{\\partial{x}}\\frac{\\partial{z}}{\\partial{y}}+\\frac{\\partial{R}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{x}}\\frac{\\partial{z}}{\\partial{y}}+R\\frac{\\partial^2{z}}{\\partial{x}\\partial{y}}\\right)-\\left(\\frac{\\partial{P}}{\\partial{y}}+\\frac{\\partial{P}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{y}}+\\frac{\\partial{R}}{\\partial{y}}\\frac{\\partial{z}}{\\partial{x}}+\\frac{\\partial{R}}{\\partial{z}}\\frac{\\partial{z}}{\\partial{y}}\\frac{\\partial{z}}{\\partial{x}}+R\\frac{\\partial^2{z}}{\\partial{y}\\partial{x}}\\right)dA  \\end{aligned}}[\/latex].<\/p>\n<p>By Clairaut\u2019s theorem, [latex]\\frac{\\partial^2{z}}{\\partial{x}\\partial{y}}=\\frac{\\partial^2{z}}{\\partial{y}\\partial{x}}[\/latex]. Therefore, four of the terms disappear from this double integral, and we are left with<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\iint_D[-R(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}[\/latex],<\/p>\n<p id=\"fs-id1167793372539\">which equals [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">We have shown that Stokes&#8217; theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field [latex]\\mathbf{F}[\/latex] is conservative.<\/p>\n<p class=\"whitespace-normal break-words\">If [latex]\\mathbf{F}[\/latex] is conservative, the curl of [latex]\\mathbf{F}[\/latex] is zero, so [latex]\\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}=0[\/latex]. Since the boundary of [latex]S[\/latex] is a closed curve,[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] is also zero.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Verify that Stokes\u2019 theorem is true for vector field [latex]{\\bf{F}}(x,y,z)=\\langle{y},2z,x^2\\rangle[\/latex] and surface [latex]S[\/latex], where [latex]S[\/latex] is the paraboloid [latex]z=4-x^{2}-y^{2}[\/latex]. Assume the surface is outward oriented and [latex]z\\geq0[\/latex]<\/p>\n<figure id=\"attachment_1514\" aria-describedby=\"caption-attachment-1514\" style=\"width: 450px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1514\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1.webp\" alt=\"A diagram in three dimensions of a hemisphere in a vector field. The arrows of the vector field follow the shape of the hemisphere, which is located in quadrants 2 and 3 of the (x, y) plane and stretches up and down into the z-plane. The center of the hemisphere is at the origin. The normal N is drawn stretching up and away from the hemisphere.\" width=\"450\" height=\"453\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1.webp 450w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1-298x300.webp 298w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1-150x150.webp 150w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1-65x65.webp 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1-225x227.webp 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17164335\/56c82fa38d06956b2bb70b5d382a8b8b71458ec1-350x352.webp 350w\" sizes=\"auto, (max-width: 450px) 100vw, 450px\" \/><figcaption id=\"caption-attachment-1514\" class=\"wp-caption-text\">Figure 5.\u00a0Verifying Stokes\u2019 theorem for a hemisphere in a vector field.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q384759728\">Show Solution<\/button><\/p>\n<div id=\"q384759728\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793371929\">As a surface integral, you have [latex]g(x, y)=4-x^{2}-y^{2}[\/latex], [latex]g_x=-2y[\/latex], and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\text{curl }{\\bf{F}}=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ \\frac{\\partial}{\\partial{x}}&\\frac{\\partial}{\\partial{y}}&\\frac{\\partial}{\\partial{z}} \\\\ y&2z&x^2\\end{vmatrix}=\\langle-2,-2x,-1\\rangle}[\/latex].<\/p>\n<p id=\"fs-id1167793918257\">By the equation to calculate scalar surface integrals,<\/p>\n<div id=\"fs-id1167793400648\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_S\\text{curl }{\\bf{F}}\\cdot{d}{\\bf{S}}&=\\displaystyle\\iint_S\\text{curl }{\\bf{F}}({\\bf{r}}(\\phi,\\theta))\\cdot{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta)dA \\\\  &=\\displaystyle\\iint_D\\langle-2,-2x,-1\\rangle\\cdot\\langle2x,2y,1\\rangle{d}A \\\\  &=\\displaystyle\\int_{-2}^2\\displaystyle\\int_{\\sqrt{4-x^2}}^{\\sqrt{4-x^2}}(-4x-4xy-1)dydx \\\\  &=\\displaystyle\\int_{-2}^2\\left(8x\\sqrt{4-x^2}-2\\sqrt{4-x^2}\\right){d}x \\\\  &=4\\pi  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793831988\">As a line integral, you can parameterize [latex]C[\/latex] by [latex]{\\bf{r}}(t)=\\langle2\\cos{t},2\\sin{t},0\\rangle, \\ 0\\leq{t}\\leq2\\pi[\/latex]. By the equation to calculate scalar surface integrals,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^{2\\pi}\\langle2\\sin{t},0,4\\cos^2t\\rangle\\cdot\\langle-2\\sin{t},2\\cos{t},0\\rangle{d}t \\\\  &=\\displaystyle\\int_0^{2\\pi}-4\\sin^2tdt=-4\\pi  \\end{aligned}[\/latex].<\/p>\n<p id=\"eip-idm922876032\">Therefore, we have verified Stokes&#8217; theorem for this example.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-791","chapter","type-chapter","status-publish","hentry"],"part":438,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/791","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/791\/revisions"}],"predecessor-version":[{"id":2105,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/791\/revisions\/2105"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/438"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/791\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=791"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=791"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=791"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=791"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}