{"id":463,"date":"2025-02-13T19:45:12","date_gmt":"2025-02-13T19:45:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-learn-it-1\/"},"modified":"2025-02-13T19:45:12","modified_gmt":"2025-02-13T19:45:12","slug":"physical-applications-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-learn-it-1\/","title":{"raw":"Physical Applications: Learn It 1","rendered":"Physical Applications: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Calculate the mass of linear and circular objects using their density distributions<\/li>\n\t<li>Compute the work required in various situations, such as pumping fluids or moving objects along a path<\/li>\n\t<li>Determine the force exerted by water on a vertical surface underwater<\/li>\n<\/ul>\n<\/section>\n<h2>Mass and Density<\/h2>\n<h3>Mass\u2013Density Formula of a One-Dimensional Object<\/h3>\n<p id=\"fs-id1167793960617\">We can use integration to calculate the mass of a thin rod based on a <strong>density function<\/strong>. Let's consider a rod oriented along the [latex]x[\/latex]-axis from [latex]x=a[\/latex] to [latex]x=b[\/latex] (Figure 1).&nbsp;<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213154\/CNX_Calc_Figure_06_05_001.jpg\" alt=\"This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b.\" width=\"342\" height=\"197\"> Figure 1. We can calculate the mass of a thin rod oriented along the [latex]x\\text{-axis}[\/latex] by integrating its density function.[\/caption]\n\n<section class=\"textbox proTip\">\n<p>Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.<\/p>\n<\/section>\n<p id=\"fs-id1167793940578\">If the rod has constant density [latex]\\rho ,[\/latex] given in terms of mass per unit length, then the mass of the rod is just the product of the density and the length of the rod: [latex](b-a)\\rho .[\/latex]<\/p>\n<p>If the density varies along the rod, we use a linear density function [latex]\\rho (x)[\/latex]. Let [latex]\\rho (x)[\/latex] be an integrable linear density function. Partition the interval [latex][a,b][\/latex] into [latex]n[\/latex] segments, each of width \u0394x.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213156\/CNX_Calc_Figure_06_05_002.jpg\" alt=\"This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b. The cylinder has been divided into segments. One segment in the middle begins at xsub(i-1) and ends at xsubi.\" width=\"342\" height=\"197\"> Figure 2. A representative segment of the rod.[\/caption]\n\n<p>The mass of a segment [latex][x_{i\u22121},x_{i}][\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]{m}_{i}=\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/p>\n<p>Summing these segments gives an approximation for the total mass:<\/p>\n<div id=\"fs-id1167794223035\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} \\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<p id=\"fs-id1167793383129\">This is a Riemann sum.&nbsp;Taking the limit as [latex]n\\to \\infty ,[\/latex] we get an expression for the exact mass of the rod:<\/p>\n<div id=\"fs-id1167794331133\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\underset{n\\to \\infty }{\\text{lim}}\\displaystyle\\sum_{i=1}^{n} \\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\rho (x)dx.[\/latex]<\/div>\n<p id=\"fs-id1167793411175\">We state this result in the following theorem.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>mass\u2013density formula of a one-dimensional object<\/h3>\n<p id=\"fs-id1167794334422\">Given a thin rod oriented along the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[a,b\\right],[\/latex] let [latex]\\rho (x)[\/latex] denote a linear density function giving the density of the rod at a point [latex]x[\/latex] in the interval. Then the mass of the rod is given by<\/p>\n<div id=\"fs-id1167794003941\" class=\"equation\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{a}^{b}\\rho (x)dx[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[\\frac{\\pi}{2},\\pi \\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)= \\sin x,[\/latex] what is the mass of the rod?<\/p>\n<div id=\"fs-id1167794004815\" class=\"exercise\">[reveal-answer q=\"fs-id1167793262742\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793262742\"]\n\n<p id=\"fs-id1167793262742\">Applying the mass-density formula directly, we have<\/p>\n<div id=\"fs-id1167793269319\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{a}^{b}\\rho (x)dx={\\displaystyle\\int }_{\\pi \\text{\/}2}^{\\pi } \\sin xdx={\\text{\u2212} \\cos x|}_{\\pi \\text{\/}2}^{\\pi }=1.[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]158684[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate the mass of linear and circular objects using their density distributions<\/li>\n<li>Compute the work required in various situations, such as pumping fluids or moving objects along a path<\/li>\n<li>Determine the force exerted by water on a vertical surface underwater<\/li>\n<\/ul>\n<\/section>\n<h2>Mass and Density<\/h2>\n<h3>Mass\u2013Density Formula of a One-Dimensional Object<\/h3>\n<p id=\"fs-id1167793960617\">We can use integration to calculate the mass of a thin rod based on a <strong>density function<\/strong>. Let&#8217;s consider a rod oriented along the [latex]x[\/latex]-axis from [latex]x=a[\/latex] to [latex]x=b[\/latex] (Figure 1).&nbsp;<\/p>\n<figure style=\"width: 342px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213154\/CNX_Calc_Figure_06_05_001.jpg\" alt=\"This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b.\" width=\"342\" height=\"197\" \/><figcaption class=\"wp-caption-text\">Figure 1. We can calculate the mass of a thin rod oriented along the [latex]x\\text{-axis}[\/latex] by integrating its density function.<\/figcaption><\/figure>\n<section class=\"textbox proTip\">\n<p>Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.<\/p>\n<\/section>\n<p id=\"fs-id1167793940578\">If the rod has constant density [latex]\\rho ,[\/latex] given in terms of mass per unit length, then the mass of the rod is just the product of the density and the length of the rod: [latex](b-a)\\rho .[\/latex]<\/p>\n<p>If the density varies along the rod, we use a linear density function [latex]\\rho (x)[\/latex]. Let [latex]\\rho (x)[\/latex] be an integrable linear density function. Partition the interval [latex][a,b][\/latex] into [latex]n[\/latex] segments, each of width \u0394x.<\/p>\n<figure style=\"width: 342px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213156\/CNX_Calc_Figure_06_05_002.jpg\" alt=\"This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b. The cylinder has been divided into segments. One segment in the middle begins at xsub(i-1) and ends at xsubi.\" width=\"342\" height=\"197\" \/><figcaption class=\"wp-caption-text\">Figure 2. A representative segment of the rod.<\/figcaption><\/figure>\n<p>The mass of a segment [latex][x_{i\u22121},x_{i}][\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]{m}_{i}=\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/p>\n<p>Summing these segments gives an approximation for the total mass:<\/p>\n<div id=\"fs-id1167794223035\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} \\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<p id=\"fs-id1167793383129\">This is a Riemann sum.&nbsp;Taking the limit as [latex]n\\to \\infty ,[\/latex] we get an expression for the exact mass of the rod:<\/p>\n<div id=\"fs-id1167794331133\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\underset{n\\to \\infty }{\\text{lim}}\\displaystyle\\sum_{i=1}^{n} \\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\rho (x)dx.[\/latex]<\/div>\n<p id=\"fs-id1167793411175\">We state this result in the following theorem.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>mass\u2013density formula of a one-dimensional object<\/h3>\n<p id=\"fs-id1167794334422\">Given a thin rod oriented along the [latex]x\\text{-axis}[\/latex] over the interval [latex]\\left[a,b\\right],[\/latex] let [latex]\\rho (x)[\/latex] denote a linear density function giving the density of the rod at a point [latex]x[\/latex] in the interval. Then the mass of the rod is given by<\/p>\n<div id=\"fs-id1167794003941\" class=\"equation\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{a}^{b}\\rho (x)dx[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[\\frac{\\pi}{2},\\pi \\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)= \\sin x,[\/latex] what is the mass of the rod?<\/p>\n<div id=\"fs-id1167794004815\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793262742\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793262742\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793262742\">Applying the mass-density formula directly, we have<\/p>\n<div id=\"fs-id1167793269319\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{a}^{b}\\rho (x)dx={\\displaystyle\\int }_{\\pi \\text{\/}2}^{\\pi } \\sin xdx={\\text{\u2212} \\cos x|}_{\\pi \\text{\/}2}^{\\pi }=1.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm158684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=158684&theme=lumen&iframe_resize_id=ohm158684&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.5 Physical Applications\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":450,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"2.5 Physical Applications","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/463"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/463\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/450"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/463\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=463"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=463"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=463"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=463"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}