{"id":789,"date":"2025-10-09T15:39:26","date_gmt":"2025-10-09T15:39:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=789"},"modified":"2025-12-15T13:36:22","modified_gmt":"2025-12-15T13:36:22","slug":"surface-integrals-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/surface-integrals-apply-it\/","title":{"raw":"Surface Integrals: Apply It","rendered":"Surface Integrals: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Understand and find parametric representations of surfaces<\/li>\r\n \t<li>Calculate surface integrals of scalar functions<\/li>\r\n \t<li>Calculate surface integrals of vector fields<\/li>\r\n \t<li>Use surface integrals to solve real-world problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications: Mass Flux vs. Flow Rate<\/h2>\r\n<p class=\"whitespace-normal break-words\">In an earlier example, we computed the <strong>mass flux<\/strong>, which is the rate of mass flow per unit area. If we want to find the <strong>flow rate<\/strong> (measured in volume per time) instead, we can use flux integral [latex]\\displaystyle\\int\u00a0 \\displaystyle\\int_S{\\bf{v}}\\cdot{\\bf{N}}dS[\/latex], which leaves out the density.<\/p>\r\n<p class=\"whitespace-normal break-words\">Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex] \\text{Flow rate of fluid across } S: \\displaystyle\\int \\displaystyle\\int_S{\\bf{v}}\\cdot{\\bf{N}}dS[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">To compute the flow rate of the fluid in the previous example, we simply remove the density constant, which gives a flow rate of [latex]90\\pi[\/latex] m\u00b3\/sec.<\/p>\r\n<p class=\"whitespace-normal break-words\">Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.<\/p>\r\n\r\n<h2 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Application: Heat Flow<\/h2>\r\n<p class=\"whitespace-normal break-words\">In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point [latex](x, y, z)[\/latex]\u00a0in an object is [latex]T(x,y,z)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the <strong>heat flow<\/strong> is defined as vector field [latex]\\mathbf{F} = -k\\nabla T[\/latex], where the constant [latex]k[\/latex] is the <strong>thermal conductivity<\/strong> of the substance from which the object is made (this constant is determined experimentally).<\/p>\r\n<p class=\"whitespace-normal break-words\">The rate of heat flow across surface [latex]S[\/latex] in the object is given by the flux integral:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_S-k\\nabla{T}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">A cast-iron solid cylinder is given by inequalities [latex]x^2+y^2\\leq1[\/latex], [latex]1\\leq{z}\\leq4[\/latex]. The temperature at point [latex](x, y, z)[\/latex] in a region containing the cylinder is [latex]T(x,y,z)=(x^2+y^2)z[\/latex]. Given that the thermal conductivity of cast iron is [latex]55[\/latex], find the heat flow across the boundary of the solid if this boundary is oriented outward.[reveal-answer q=\"167405820\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"167405820\"]\r\n<p id=\"fs-id1167793546475\">Let [latex]S[\/latex] denote the boundary of the object. To find the heat flow, we need to calculate flux integral [latex]\\displaystyle\\iint_S-k\\nabla{T}\\cdot{d}{\\bf{S}}[\/latex]. Notice that [latex]S[\/latex] is not a smooth surface but is piecewise smooth, since [latex]S[\/latex] is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of [latex]S[\/latex]. Before calculating any integrals, note that the gradient of the temperature is [latex]\\nabla{T}=\\langle2xz,2yz,x^2+y^2\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1167793924511\">First we consider the circular bottom of the object, which we denote [latex]S_1[\/latex]. We can see that [latex]S_1[\/latex] is a circle of radius 1 centered at point [latex](0, 0, 1)[\/latex], sitting in plane [latex]z=1[\/latex]. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle{v}\\cos{u},v\\sin{u}1\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq1[\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u=\\langle-v\\sin{u},v\\cos{u},0\\rangle\\text{ and }{\\bf{t}}_v=\\langle\\cos{u},v\\sin{u},0\\rangle[\/latex],<\/p>\r\n<p id=\"fs-id1167793411048\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,-v\\sin^2u-v\\cos^2u\\rangle=\\langle0,0,-v\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1167793401584\">Since the surface is oriented outward and [latex]S_1[\/latex] is the bottom of the object, it makes sense that this vector points downward. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex], the heat flow across [latex]S_1[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_1}-k\\nabla{T}\\cdot{d}{\\bf{S}}&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},v^2\\cos^2u+v^2\\sin^2u\\rangle\\cdot\\langle0,0,-v\\rangle{d}vdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},v^2\\rangle\\cdot\\langle0,0,-v\\rangle{dvdu} \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1-v^3dvdu=-55\\displaystyle\\int_0^{2\\pi}-\\frac14du=\\frac{55\\pi}2\r\n\\end{aligned}[\/latex].<\/p>\r\nNow let\u2019s consider the circular top of the object, which we denote [latex]S_2[\/latex]. We see that [latex]S_2[\/latex] is a circle of radius 1 centered at point [latex](0, 0, 4)[\/latex], sitting in plane [latex]z=4[\/latex]. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle{v}\\cos{u},v\\sin{u},4\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq1[\/latex]. Therefore,\r\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u=\\langle-v\\sin{u},v\\cos{u},0\\rangle\\text{ and }{\\bf{v}}\\langle\\cos{u},v\\sin{u},0\\rangle[\/latex],<\/p>\r\n<p id=\"fs-id1167793515538\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,-v\\sin^2{u}-v\\cos^2u\\rangle=\\langle0,0,-v\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1167793776582\">Since the surface is oriented outward and [latex]S_1[\/latex] is the top of the object, we instead take vector [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,v\\rangle[\/latex]. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex], the heat flow across [latex]S_1[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_2}-k\\nabla{T}\\bullet{d}{\\bf{S}}&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\bullet({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle8v\\cos{u},8v\\sin{u},v^2\\cos^2u+v^2\\sin^2u\\rangle\\bullet\\langle0,0,v\\rangle{d}vdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle8v\\cos{u},8v\\sin{u},v^2\\rangle\\bullet\\langle0,0,v\\rangle{d}vdu \\\\\r\n&amp;=55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1v^3dvdu=-\\frac{55\\pi}2\r\n\\end{aligned}[\/latex].<\/p>\r\nLast, let\u2019s consider the cylindrical side of the object. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle\\cos{u},\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 1\\leq{v}\\leq4[\/latex]. By\u00a0Example \"Calculating the Surface Integral of a Cylinder\", we know that [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle\\cos{u},\\sin{u},0\\rangle[\/latex]. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex],\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_2}-k\\nabla{T}\\bullet{d}{\\bf{S}}&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\bullet({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_1^4\\langle2v\\cos{u},2v\\sin{u},\\cos^2u+\\sin^2u\\rangle\\bullet\\langle\\cos{u},\\sin{u},0\\rangle{d}vdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},1\\rangle\\bullet\\langle\\cos{u},\\sin{u},0\\rangle{d}vdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(2v\\cos^2u+2v\\sin^2u)dvdu \\\\\r\n&amp;=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^12vdvdu=-55\\displaystyle\\int_0^{2\\pi}du=-110\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, the rate of heat flow across [latex]S[\/latex] is [latex]\\frac{55\\pi}2-\\frac{55\\pi}2-110\\pi=-110\\pi[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Understand and find parametric representations of surfaces<\/li>\n<li>Calculate surface integrals of scalar functions<\/li>\n<li>Calculate surface integrals of vector fields<\/li>\n<li>Use surface integrals to solve real-world problems<\/li>\n<\/ul>\n<\/section>\n<h2>Applications: Mass Flux vs. Flow Rate<\/h2>\n<p class=\"whitespace-normal break-words\">In an earlier example, we computed the <strong>mass flux<\/strong>, which is the rate of mass flow per unit area. If we want to find the <strong>flow rate<\/strong> (measured in volume per time) instead, we can use flux integral [latex]\\displaystyle\\int\u00a0 \\displaystyle\\int_S{\\bf{v}}\\cdot{\\bf{N}}dS[\/latex], which leaves out the density.<\/p>\n<p class=\"whitespace-normal break-words\">Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{Flow rate of fluid across } S: \\displaystyle\\int \\displaystyle\\int_S{\\bf{v}}\\cdot{\\bf{N}}dS[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">To compute the flow rate of the fluid in the previous example, we simply remove the density constant, which gives a flow rate of [latex]90\\pi[\/latex] m\u00b3\/sec.<\/p>\n<p class=\"whitespace-normal break-words\">Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.<\/p>\n<h2 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Application: Heat Flow<\/h2>\n<p class=\"whitespace-normal break-words\">In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point [latex](x, y, z)[\/latex]\u00a0in an object is [latex]T(x,y,z)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the <strong>heat flow<\/strong> is defined as vector field [latex]\\mathbf{F} = -k\\nabla T[\/latex], where the constant [latex]k[\/latex] is the <strong>thermal conductivity<\/strong> of the substance from which the object is made (this constant is determined experimentally).<\/p>\n<p class=\"whitespace-normal break-words\">The rate of heat flow across surface [latex]S[\/latex] in the object is given by the flux integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}{\\bf{S}}=\\displaystyle\\iint_S-k\\nabla{T}\\cdot{d}{\\bf{S}}[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">A cast-iron solid cylinder is given by inequalities [latex]x^2+y^2\\leq1[\/latex], [latex]1\\leq{z}\\leq4[\/latex]. The temperature at point [latex](x, y, z)[\/latex] in a region containing the cylinder is [latex]T(x,y,z)=(x^2+y^2)z[\/latex]. Given that the thermal conductivity of cast iron is [latex]55[\/latex], find the heat flow across the boundary of the solid if this boundary is oriented outward.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q167405820\">Show Solution<\/button><\/p>\n<div id=\"q167405820\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793546475\">Let [latex]S[\/latex] denote the boundary of the object. To find the heat flow, we need to calculate flux integral [latex]\\displaystyle\\iint_S-k\\nabla{T}\\cdot{d}{\\bf{S}}[\/latex]. Notice that [latex]S[\/latex] is not a smooth surface but is piecewise smooth, since [latex]S[\/latex] is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of [latex]S[\/latex]. Before calculating any integrals, note that the gradient of the temperature is [latex]\\nabla{T}=\\langle2xz,2yz,x^2+y^2\\rangle[\/latex].<\/p>\n<p id=\"fs-id1167793924511\">First we consider the circular bottom of the object, which we denote [latex]S_1[\/latex]. We can see that [latex]S_1[\/latex] is a circle of radius 1 centered at point [latex](0, 0, 1)[\/latex], sitting in plane [latex]z=1[\/latex]. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle{v}\\cos{u},v\\sin{u}1\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq1[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u=\\langle-v\\sin{u},v\\cos{u},0\\rangle\\text{ and }{\\bf{t}}_v=\\langle\\cos{u},v\\sin{u},0\\rangle[\/latex],<\/p>\n<p id=\"fs-id1167793411048\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,-v\\sin^2u-v\\cos^2u\\rangle=\\langle0,0,-v\\rangle[\/latex].<\/p>\n<p id=\"fs-id1167793401584\">Since the surface is oriented outward and [latex]S_1[\/latex] is the bottom of the object, it makes sense that this vector points downward. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex], the heat flow across [latex]S_1[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_{S_1}-k\\nabla{T}\\cdot{d}{\\bf{S}}&=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},v^2\\cos^2u+v^2\\sin^2u\\rangle\\cdot\\langle0,0,-v\\rangle{d}vdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},v^2\\rangle\\cdot\\langle0,0,-v\\rangle{dvdu} \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1-v^3dvdu=-55\\displaystyle\\int_0^{2\\pi}-\\frac14du=\\frac{55\\pi}2  \\end{aligned}[\/latex].<\/p>\n<p>Now let\u2019s consider the circular top of the object, which we denote [latex]S_2[\/latex]. We see that [latex]S_2[\/latex] is a circle of radius 1 centered at point [latex](0, 0, 4)[\/latex], sitting in plane [latex]z=4[\/latex]. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle{v}\\cos{u},v\\sin{u},4\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq1[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u=\\langle-v\\sin{u},v\\cos{u},0\\rangle\\text{ and }{\\bf{v}}\\langle\\cos{u},v\\sin{u},0\\rangle[\/latex],<\/p>\n<p id=\"fs-id1167793515538\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,-v\\sin^2{u}-v\\cos^2u\\rangle=\\langle0,0,-v\\rangle[\/latex].<\/p>\n<p id=\"fs-id1167793776582\">Since the surface is oriented outward and [latex]S_1[\/latex] is the top of the object, we instead take vector [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle0,0,v\\rangle[\/latex]. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex], the heat flow across [latex]S_1[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_{S_2}-k\\nabla{T}\\bullet{d}{\\bf{S}}&=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\bullet({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle8v\\cos{u},8v\\sin{u},v^2\\cos^2u+v^2\\sin^2u\\rangle\\bullet\\langle0,0,v\\rangle{d}vdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle8v\\cos{u},8v\\sin{u},v^2\\rangle\\bullet\\langle0,0,v\\rangle{d}vdu \\\\  &=55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1v^3dvdu=-\\frac{55\\pi}2  \\end{aligned}[\/latex].<\/p>\n<p>Last, let\u2019s consider the cylindrical side of the object. This surface has parameterization [latex]{\\bf{r}}(u,v)=\\langle\\cos{u},\\sin{u},v\\rangle[\/latex], [latex]0\\leq{u}\\leq2\\pi, \\ 1\\leq{v}\\leq4[\/latex]. By\u00a0Example &#8220;Calculating the Surface Integral of a Cylinder&#8221;, we know that [latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle\\cos{u},\\sin{u},0\\rangle[\/latex]. By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\iint_{S_2}-k\\nabla{T}\\bullet{d}{\\bf{S}}&=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\nabla{T}(u,v)\\bullet({\\bf{t}}_u\\times{\\bf{t}}_v)dvdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_1^4\\langle2v\\cos{u},2v\\sin{u},\\cos^2u+\\sin^2u\\rangle\\bullet\\langle\\cos{u},\\sin{u},0\\rangle{d}vdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1\\langle2v\\cos{u},2v\\sin{u},1\\rangle\\bullet\\langle\\cos{u},\\sin{u},0\\rangle{d}vdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^1(2v\\cos^2u+2v\\sin^2u)dvdu \\\\  &=-55\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^12vdvdu=-55\\displaystyle\\int_0^{2\\pi}du=-110\\pi  \\end{aligned}[\/latex].<\/p>\n<p>Therefore, the rate of heat flow across [latex]S[\/latex] is [latex]\\frac{55\\pi}2-\\frac{55\\pi}2-110\\pi=-110\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":15,"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-789","chapter","type-chapter","status-publish","hentry"],"part":438,"module-header":"apply_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\/789","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":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/789\/revisions"}],"predecessor-version":[{"id":2803,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/789\/revisions\/2803"}],"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\/789\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=789"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=789"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=789"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}