{"id":753,"date":"2025-10-09T15:36:48","date_gmt":"2025-10-09T15:36:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=753"},"modified":"2025-11-13T18:58:02","modified_gmt":"2025-11-13T18:58:02","slug":"line-integrals-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/line-integrals-apply-it\/","title":{"raw":"Line Integrals: Apply It","rendered":"Line Integrals: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate scalar line integrals<\/li>\r\n \t<li>Calculate vector line integrals<\/li>\r\n \t<li>Apply line integrals to compute work<\/li>\r\n \t<li>Describe flux and circulation<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Applications of Line Integrals<\/h2>\r\n<p id=\"fs-id1167793482851\">Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term247\" class=\"no-emphasis\" data-type=\"term\">mass of a wire<\/span>\u00a0using a scalar line integral and the work done by a force using a vector line integral.<\/p>\r\n<p id=\"fs-id1167793693424\">Suppose that a piece of wire is modeled by curve [latex]C[\/latex] in space. The mass per unit length (the linear density) of the wire is a continuous function [latex]{\\rho}{({x}, {y}, {z})}[\/latex]. We can calculate the total mass of the wire using the scalar line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex]. The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by [latex]{\\rho}{({x}^{*}, {y}^{*}, {z}^{*})}{\\Delta}{s}[\/latex] for some point [latex]({x}^{*}, {y}^{*}, {z}^{*})[\/latex] in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the mass of a spring in the shape of a helix parameterized by [latex]\\langle{t},2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq\\frac{\\pi}2[\/latex], with a density function given by [latex]\\rho(x,y,z)=e^x+yz[\/latex] kg\/m.\r\n\r\n[caption id=\"attachment_5175\" align=\"aligncenter\" width=\"421\"]<img class=\"size-full wp-image-5175\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174118\/6.21.jpg\" alt=\"A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi\/2, 0, 2). The arrow on the curve is pointing to the latter endpoint.\" width=\"421\" height=\"522\" \/> Figure 1. The wire from Example \"Calculating the Mass of a Wire\".[\/caption]\r\n\r\n[reveal-answer q=\"394875298\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394875298\"]\r\n<p id=\"fs-id1167793419129\">To calculate the mass of the spring, we must find the value of the scalar line integral [latex]\\displaystyle\\int_c(e^x+yz)ds[\/latex], where [latex]C[\/latex] is the given helix. To calculate this integral, we write it in terms of [latex]t[\/latex]\u00a0using The Scalar Line Integral Calculation Theorem Equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\int_ce^x+yzds&amp;=\\displaystyle\\int_0^{\\pi\/2}\\left((e^t+4\\cos{t}\\sin{t})\\sqrt{1+(-2\\cos{t}^2+(2\\sin{t})^2}\\right)dt \\\\\r\n&amp;=\\displaystyle\\int_0^{\\pi\/2}((e^t+4\\cos{t}\\sin{t})\\sqrt{5})dt \\\\\r\n&amp;=\\sqrt{5}[e^t+2\\sin^2t]_{t=0}^{t=\\pi\/2} \\\\\r\n&amp;=\\sqrt{5}(e^{\\pi\/2}+1)\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167794223613\">Therefore, the mass is [latex]\\sqrt{5}(e^{\\pi\/2}+1)[\/latex] kg.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term248\" class=\"no-emphasis\" data-type=\"term\">work done by a vector field<\/span>\u00a0representing a force is a standard use of vector line integrals. Recall that if an object moves along curve [latex]C[\/latex] in force field\u00a0[latex]{\\bf{F}}[\/latex], then the work required to move the object is given by [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex].\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">How much work is required to move an object in vector force field [latex]{\\bf{F}}=\\langle{y}x,xy,xz\\rangle[\/latex] along path [latex]{\\bf{r}}(t)=\\langle{t}^2,t,t^4\\rangle[\/latex],\u00a0[latex]0\\leq{t}\\leq1[\/latex]?\u00a0 See\u00a0Figure 2.\r\n<div data-type=\"note\">\r\n\r\n[reveal-answer q=\"218730126\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218730126\"]\r\n<p id=\"fs-id1167793631774\">Let [latex]C[\/latex] denote the given path. We need to find the value of [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex]. To do this, we use the equation to compute vector line 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_0^1(\\langle{t}^5,t^3,t^6\\rangle\\cdot\\langle2t,1,4t^3\\rangle)dt\\\\\r\n&amp;=\\displaystyle\\int_0^1(2t^6+t^3+4t^9) \\ dt \\\\\r\n&amp;=\\left[\\frac{2t^7}7+\\frac{t^4}4+\\frac{2t^{10}}5\\right]_{t=0}^{t=1}=\\frac{131}{140}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_5178\" align=\"aligncenter\" width=\"316\"]<img class=\"size-full wp-image-5178\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174320\/6.22.jpg\" alt=\"A three dimensional diagram of the curve and vector field for the example. The curve is an increasing concave up curve starting close to the origin and above the x axis. As the curve goes left above the (x,y) plane, the height also increases. The arrows in the vector field get longer as the z component becomes larger.\" width=\"316\" height=\"597\" \/> Figure 2. The curve and vector field for Example \"Calculating Work\".[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate the mass of a spring in the shape of a helix parameterized by [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex], with a density function given by\u00a0[latex]\\rho(x,y,z)=x+y+z[\/latex] kg\/m.[reveal-answer q=\"904921475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"904921475\"][latex]18\\sqrt2\\pi^2[\/latex] kg.[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250316&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=WwobFvL33Jc&amp;video_target=tpm-plugin-k9958w5c-WwobFvL33Jc\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP6.20_transcript.html\">transcript for \u201cCP 6.20\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate scalar line integrals<\/li>\n<li>Calculate vector line integrals<\/li>\n<li>Apply line integrals to compute work<\/li>\n<li>Describe flux and circulation<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Applications of Line Integrals<\/h2>\n<p id=\"fs-id1167793482851\">Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term247\" class=\"no-emphasis\" data-type=\"term\">mass of a wire<\/span>\u00a0using a scalar line integral and the work done by a force using a vector line integral.<\/p>\n<p id=\"fs-id1167793693424\">Suppose that a piece of wire is modeled by curve [latex]C[\/latex] in space. The mass per unit length (the linear density) of the wire is a continuous function [latex]{\\rho}{({x}, {y}, {z})}[\/latex]. We can calculate the total mass of the wire using the scalar line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex]. The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by [latex]{\\rho}{({x}^{*}, {y}^{*}, {z}^{*})}{\\Delta}{s}[\/latex] for some point [latex]({x}^{*}, {y}^{*}, {z}^{*})[\/latex] in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral [latex]{\\displaystyle\\int_{C}}{\\rho}{({x}, {y}, {z})}{ds}[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the mass of a spring in the shape of a helix parameterized by [latex]\\langle{t},2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq\\frac{\\pi}2[\/latex], with a density function given by [latex]\\rho(x,y,z)=e^x+yz[\/latex] kg\/m.<\/p>\n<figure id=\"attachment_5175\" aria-describedby=\"caption-attachment-5175\" style=\"width: 421px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5175\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174118\/6.21.jpg\" alt=\"A three dimensional diagram. An increasing, then slightly decreasing concave down curve is drawn from (0,2,0) to (pi\/2, 0, 2). The arrow on the curve is pointing to the latter endpoint.\" width=\"421\" height=\"522\" \/><figcaption id=\"caption-attachment-5175\" class=\"wp-caption-text\">Figure 1. The wire from Example &#8220;Calculating the Mass of a Wire&#8221;.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q394875298\">Show Solution<\/button><\/p>\n<div id=\"q394875298\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793419129\">To calculate the mass of the spring, we must find the value of the scalar line integral [latex]\\displaystyle\\int_c(e^x+yz)ds[\/latex], where [latex]C[\/latex] is the given helix. To calculate this integral, we write it in terms of [latex]t[\/latex]\u00a0using The Scalar Line Integral Calculation Theorem Equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_ce^x+yzds&=\\displaystyle\\int_0^{\\pi\/2}\\left((e^t+4\\cos{t}\\sin{t})\\sqrt{1+(-2\\cos{t}^2+(2\\sin{t})^2}\\right)dt \\\\  &=\\displaystyle\\int_0^{\\pi\/2}((e^t+4\\cos{t}\\sin{t})\\sqrt{5})dt \\\\  &=\\sqrt{5}[e^t+2\\sin^2t]_{t=0}^{t=\\pi\/2} \\\\  &=\\sqrt{5}(e^{\\pi\/2}+1)  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167794223613\">Therefore, the mass is [latex]\\sqrt{5}(e^{\\pi\/2}+1)[\/latex] kg.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the\u00a0<span id=\"34b090a5-96d8-48dc-97e5-4aab175adbb5_term248\" class=\"no-emphasis\" data-type=\"term\">work done by a vector field<\/span>\u00a0representing a force is a standard use of vector line integrals. Recall that if an object moves along curve [latex]C[\/latex] in force field\u00a0[latex]{\\bf{F}}[\/latex], then the work required to move the object is given by [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">How much work is required to move an object in vector force field [latex]{\\bf{F}}=\\langle{y}x,xy,xz\\rangle[\/latex] along path [latex]{\\bf{r}}(t)=\\langle{t}^2,t,t^4\\rangle[\/latex],\u00a0[latex]0\\leq{t}\\leq1[\/latex]?\u00a0 See\u00a0Figure 2.<\/p>\n<div data-type=\"note\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q218730126\">Show Solution<\/button><\/p>\n<div id=\"q218730126\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793631774\">Let [latex]C[\/latex] denote the given path. We need to find the value of [latex]\\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}[\/latex]. To do this, we use the equation to compute vector line integrals:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\int_c{\\bf{F}}\\cdot \\ d\\bf{r}&=\\displaystyle\\int_0^1(\\langle{t}^5,t^3,t^6\\rangle\\cdot\\langle2t,1,4t^3\\rangle)dt\\\\  &=\\displaystyle\\int_0^1(2t^6+t^3+4t^9) \\ dt \\\\  &=\\left[\\frac{2t^7}7+\\frac{t^4}4+\\frac{2t^{10}}5\\right]_{t=0}^{t=1}=\\frac{131}{140}.  \\end{aligned}[\/latex]<\/p>\n<figure id=\"attachment_5178\" aria-describedby=\"caption-attachment-5178\" style=\"width: 316px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5178\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/31174320\/6.22.jpg\" alt=\"A three dimensional diagram of the curve and vector field for the example. The curve is an increasing concave up curve starting close to the origin and above the x axis. As the curve goes left above the (x,y) plane, the height also increases. The arrows in the vector field get longer as the z component becomes larger.\" width=\"316\" height=\"597\" \/><figcaption id=\"caption-attachment-5178\" class=\"wp-caption-text\">Figure 2. The curve and vector field for Example &#8220;Calculating Work&#8221;.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the mass of a spring in the shape of a helix parameterized by [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq6\\pi[\/latex], with a density function given by\u00a0[latex]\\rho(x,y,z)=x+y+z[\/latex] kg\/m.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q904921475\">Show Solution<\/button><\/p>\n<div id=\"q904921475\" class=\"hidden-answer\" style=\"display: none\">[latex]18\\sqrt2\\pi^2[\/latex] kg.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250316&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=WwobFvL33Jc&amp;video_target=tpm-plugin-k9958w5c-WwobFvL33Jc\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP6.20_transcript.html\">transcript for \u201cCP 6.20\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":14,"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-753","chapter","type-chapter","status-publish","hentry"],"part":436,"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\/753","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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/753\/revisions"}],"predecessor-version":[{"id":2227,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/753\/revisions\/2227"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/436"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/753\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=753"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=753"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=753"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=753"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}