{"id":1179,"date":"2021-06-30T17:02:05","date_gmt":"2021-06-30T17:02:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/mass-and-density\/"},"modified":"2022-03-19T03:36:19","modified_gmt":"2022-03-19T03:36:19","slug":"mass-and-density","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/mass-and-density\/","title":{"raw":"Mass and Density","rendered":"Mass and Density"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the mass of a one-dimensional object from its linear density function<\/li>\r\n \t<li>Determine the mass of a two-dimensional circular object from its radial density function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1167794005165\" class=\"bc-section section\">\r\n<p id=\"fs-id1167793960617\">We can use integration to develop a formula for calculating mass based on a <strong>density function<\/strong>. First we consider a thin rod or wire. Orient the rod so it aligns with the [latex]x\\text{-axis,}[\/latex] with the left end of the rod at [latex]x=a[\/latex] and the right end of the rod at [latex]x=b[\/latex] (Figure 1). 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>\r\n\r\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]\r\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] If the density of the rod is not constant, however, the problem becomes a little more challenging. When the density of the rod varies from point to point, we use a linear density function, [latex]\\rho (x),[\/latex] to denote the density of the rod at any point, [latex]x.[\/latex] Let [latex]\\rho (x)[\/latex] be an integrable linear density function. Now, for [latex]i=0,1,2\\text{,\u2026},n[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Figure 2 shows a representative segment of the rod.<\/p>\r\n\r\n\r\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]\r\n<p id=\"fs-id1167793720021\">The mass [latex]{m}_{i}[\/latex] of the segment of the rod from [latex]{x}_{i-1}[\/latex] to [latex]{x}_{i}[\/latex] is approximated by<\/p>\r\n\r\n<div id=\"fs-id1167793627610\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{i}\\approx \\rho ({x}_{i}^{*})({x}_{i}-{x}_{i-1})=\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\r\n<div><\/div>\r\nAdding the masses of all the segments gives us an approximation for the mass of the entire rod:\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1167793383129\">This is a Riemann sum. Taking the limit as [latex]n\\to \\infty ,[\/latex] we get an expression for the exact mass of the rod:<\/p>\r\n\r\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>\r\n&nbsp;\r\n<p id=\"fs-id1167793411175\">We state this result in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167793953496\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Mass\u2013Density Formula of a One-Dimensional Object<\/h3>\r\n\r\n<hr \/>\r\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>\r\n\r\n<div id=\"fs-id1167794003941\" class=\"equation\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{a}^{b}\\rho (x)dx[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793417236\">We apply this theorem in the next example.<\/p>\r\n\r\n<div id=\"fs-id1167793924292\" class=\"textbook exercises\">\r\n<h3>Example: Calculating Mass from Linear Density<\/h3>\r\nConsider 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?\r\n<div id=\"fs-id1167794004815\" class=\"exercise\">[reveal-answer q=\"fs-id1167793262742\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793262742\"]\r\n<p id=\"fs-id1167793262742\">Applying the mass-density formula directly, we have<\/p>\r\n\r\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>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793363249\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConsider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[1,3\\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)=2{x}^{2}+3,[\/latex] what is the mass of the rod?\r\n\r\n[reveal-answer q=\"718859\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"718859\"]Use the process from the previous example.[\/hidden-answer]\r\n\r\n[reveal-answer q=\"88003546\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88003546\"]\r\n<div id=\"fs-id1167794136385\" class=\"exercise\">\r\n<div class=\"solution\">\r\n<p id=\"fs-id1167794296202\">[latex]\\frac{70}{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793982345\" class=\"commentary\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=134&amp;end=181&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications134to181_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]158684[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793940456\">We now extend this concept to find the mass of a two-dimensional disk of radius [latex]r.[\/latex] As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object. We assume the density is given in terms of mass per unit area (called <span class=\"no-emphasis\"><em>area density<\/em><\/span>), and further assume the density varies only along the disk\u2019s radius (called <span class=\"no-emphasis\"><em>radial density<\/em><\/span>). We orient the disk in the [latex]xy\\text{-plane,}[\/latex] with the center at the origin. Then, the density of the disk can be treated as a function of [latex]x,[\/latex] denoted [latex]\\rho (x).[\/latex] We assume [latex]\\rho (x)[\/latex] is integrable. Because density is a function of [latex]x,[\/latex] we partition the interval from [latex]\\left[0,r\\right][\/latex] along the [latex]x\\text{-axis}.[\/latex] For [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[0,r\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a representative washer are depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"613\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213159\/CNX_Calc_Figure_06_05_003.jpg\" alt=\"This figure has two images. The first is labeled \u201ca\u201d and is a circle with radius r. The center of the circle is labeled 0. The circle also has the positive x-axis beginning at 0, extending through the circle. The second figure is labeled \u201cb\u201d. It has two concentric circles with center at 0 and the x-axis extending out from 0. The concentric circles form a washer. The width of the washer is from xsub(i-1) to xsubi and is labeled delta x.\" width=\"613\" height=\"324\" \/> Figure 3. (a) A thin disk in the xy-plane. (b) A representative washer.[\/caption]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1167793952882\">We now approximate the density and area of the washer to calculate an approximate mass, [latex]{m}_{i}.[\/latex] Note that the area of the washer is given by<\/p>\r\n\r\n<div id=\"fs-id1167793787727\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {A}_{i}&amp; =\\pi {({x}_{i})}^{2}-\\pi {({x}_{i-1})}^{2}\\hfill \\\\ &amp; =\\pi \\left[{x}_{i}^{2}-{x}_{i-1}^{2}\\right]\\hfill \\\\ &amp; =\\pi ({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ &amp; =\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794031090\">You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use [latex]{x}_{i}^{*}\\approx ({x}_{i}+{x}_{i-1})\\text{\/}2[\/latex] to approximate the average radius of the washer. We obtain<\/p>\r\n\r\n<div id=\"fs-id1167793266910\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{i}=\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x\\approx 2\\pi {x}_{i}^{*}\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793930438\">Using [latex]\\rho ({x}_{i}^{*})[\/latex] to approximate the density of the washer, we approximate the mass of the washer by<\/p>\r\n\r\n<div id=\"fs-id1167793918720\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{i}\\approx 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\r\n<div><\/div>\r\nAdding up the masses of the washers, we see the mass [latex]m[\/latex] of the entire disk is approximated by\r\n<div id=\"fs-id1167793384645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794176147\">We again recognize this as a Riemann sum, and take the limit as [latex]n\\to \\infty .[\/latex] This gives us<\/p>\r\n\r\n<div id=\"fs-id1167793930066\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793950100\">We summarize these findings in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167793950103\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Mass\u2013Density Formula of a Circular Object<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793625643\">Let [latex]\\rho (x)[\/latex] be an integrable function representing the radial density of a disk of radius [latex]r.[\/latex] Then the mass of the disk is given by<\/p>\r\n\r\n<div id=\"fs-id1167793469833\" class=\"equation\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793912188\" class=\"textbook exercises\">\r\n<h3>Example: Calculating Mass from Radial Density<\/h3>\r\nLet [latex]\\rho (x)=\\sqrt{x}[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius 4.\r\n<div id=\"fs-id1167793912190\" class=\"exercise\">[reveal-answer q=\"fs-id1167793477594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793477594\"]\r\n<p id=\"fs-id1167793477594\">Applying the formula, we find<\/p>\r\n\r\n<div id=\"fs-id1167793510552\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m&amp; ={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{4}2\\pi x\\sqrt{x}dx=2\\pi {\\displaystyle\\int }_{0}^{4}{x}^{3\\text{\/}2}dx\\hfill \\\\ &amp; =2\\pi {\\frac{2}{5}{x}^{5\\text{\/}2}|}_{0}^{4}=\\frac{4\\pi }{5}\\left[32\\right]=\\frac{128\\pi }{5}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793563675\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]\\rho (x)=3x+2[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius 2.\r\n<div>[reveal-answer q=\"242187\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"242187\"]Use the process from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793563678\" class=\"exercise\">[reveal-answer q=\"fs-id1167793547373\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793547373\"]\r\n<p id=\"fs-id1167793547373\">[latex]24\\pi [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=397&amp;end=477&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications397to477_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the mass of a one-dimensional object from its linear density function<\/li>\n<li>Determine the mass of a two-dimensional circular object from its radial density function<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1167794005165\" class=\"bc-section section\">\n<p id=\"fs-id1167793960617\">We can use integration to develop a formula for calculating mass based on a <strong>density function<\/strong>. First we consider a thin rod or wire. Orient the rod so it aligns with the [latex]x\\text{-axis,}[\/latex] with the left end of the rod at [latex]x=a[\/latex] and the right end of the rod at [latex]x=b[\/latex] (Figure 1). 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<div style=\"width: 352px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\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] If the density of the rod is not constant, however, the problem becomes a little more challenging. When the density of the rod varies from point to point, we use a linear density function, [latex]\\rho (x),[\/latex] to denote the density of the rod at any point, [latex]x.[\/latex] Let [latex]\\rho (x)[\/latex] be an integrable linear density function. Now, for [latex]i=0,1,2\\text{,\u2026},n[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Figure 2 shows a representative segment of the rod.<\/p>\n<div style=\"width: 352px\" 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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A representative segment of the rod.<\/p>\n<\/div>\n<p id=\"fs-id1167793720021\">The mass [latex]{m}_{i}[\/latex] of the segment of the rod from [latex]{x}_{i-1}[\/latex] to [latex]{x}_{i}[\/latex] is approximated by<\/p>\n<div id=\"fs-id1167793627610\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{i}\\approx \\rho ({x}_{i}^{*})({x}_{i}-{x}_{i-1})=\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<div><\/div>\n<p>Adding the masses of all the segments gives us an approximation for the mass of the entire rod:<\/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>&nbsp;<\/p>\n<p id=\"fs-id1167793383129\">This is a Riemann sum. 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>&nbsp;<\/p>\n<p id=\"fs-id1167793411175\">We state this result in the following theorem.<\/p>\n<div id=\"fs-id1167793953496\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Mass\u2013Density Formula of a One-Dimensional Object<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793417236\">We apply this theorem in the next example.<\/p>\n<div id=\"fs-id1167793924292\" class=\"textbook exercises\">\n<h3>Example: Calculating Mass from Linear Density<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793262742\">Show Solution<\/span><\/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<\/div>\n<div id=\"fs-id1167793363249\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Consider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[1,3\\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)=2{x}^{2}+3,[\/latex] what is the mass of the rod?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q718859\">Hint<\/span><\/p>\n<div id=\"q718859\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88003546\">Show Solution<\/span><\/p>\n<div id=\"q88003546\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794136385\" class=\"exercise\">\n<div class=\"solution\">\n<p id=\"fs-id1167794296202\">[latex]\\frac{70}{3}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1167793982345\" class=\"commentary\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=134&amp;end=181&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications134to181_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm158684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=158684&theme=oea&iframe_resize_id=ohm158684&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167793940456\">We now extend this concept to find the mass of a two-dimensional disk of radius [latex]r.[\/latex] As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object. We assume the density is given in terms of mass per unit area (called <span class=\"no-emphasis\"><em>area density<\/em><\/span>), and further assume the density varies only along the disk\u2019s radius (called <span class=\"no-emphasis\"><em>radial density<\/em><\/span>). We orient the disk in the [latex]xy\\text{-plane,}[\/latex] with the center at the origin. Then, the density of the disk can be treated as a function of [latex]x,[\/latex] denoted [latex]\\rho (x).[\/latex] We assume [latex]\\rho (x)[\/latex] is integrable. Because density is a function of [latex]x,[\/latex] we partition the interval from [latex]\\left[0,r\\right][\/latex] along the [latex]x\\text{-axis}.[\/latex] For [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[0,r\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a representative washer are depicted in the following figure.<\/p>\n<div style=\"width: 623px\" 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\/11213159\/CNX_Calc_Figure_06_05_003.jpg\" alt=\"This figure has two images. The first is labeled \u201ca\u201d and is a circle with radius r. The center of the circle is labeled 0. The circle also has the positive x-axis beginning at 0, extending through the circle. The second figure is labeled \u201cb\u201d. It has two concentric circles with center at 0 and the x-axis extending out from 0. The concentric circles form a washer. The width of the washer is from xsub(i-1) to xsubi and is labeled delta x.\" width=\"613\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. (a) A thin disk in the xy-plane. (b) A representative washer.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793952882\">We now approximate the density and area of the washer to calculate an approximate mass, [latex]{m}_{i}.[\/latex] Note that the area of the washer is given by<\/p>\n<div id=\"fs-id1167793787727\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {A}_{i}& =\\pi {({x}_{i})}^{2}-\\pi {({x}_{i-1})}^{2}\\hfill \\\\ & =\\pi \\left[{x}_{i}^{2}-{x}_{i-1}^{2}\\right]\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794031090\">You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use [latex]{x}_{i}^{*}\\approx ({x}_{i}+{x}_{i-1})\\text{\/}2[\/latex] to approximate the average radius of the washer. We obtain<\/p>\n<div id=\"fs-id1167793266910\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{i}=\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x\\approx 2\\pi {x}_{i}^{*}\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793930438\">Using [latex]\\rho ({x}_{i}^{*})[\/latex] to approximate the density of the washer, we approximate the mass of the washer by<\/p>\n<div id=\"fs-id1167793918720\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{i}\\approx 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<div><\/div>\n<p>Adding up the masses of the washers, we see the mass [latex]m[\/latex] of the entire disk is approximated by<\/p>\n<div id=\"fs-id1167793384645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794176147\">We again recognize this as a Riemann sum, and take the limit as [latex]n\\to \\infty .[\/latex] This gives us<\/p>\n<div id=\"fs-id1167793930066\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793950100\">We summarize these findings in the following theorem.<\/p>\n<div id=\"fs-id1167793950103\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Mass\u2013Density Formula of a Circular Object<\/h3>\n<hr \/>\n<p id=\"fs-id1167793625643\">Let [latex]\\rho (x)[\/latex] be an integrable function representing the radial density of a disk of radius [latex]r.[\/latex] Then the mass of the disk is given by<\/p>\n<div id=\"fs-id1167793469833\" class=\"equation\" style=\"text-align: center;\">[latex]m={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1167793912188\" class=\"textbook exercises\">\n<h3>Example: Calculating Mass from Radial Density<\/h3>\n<p>Let [latex]\\rho (x)=\\sqrt{x}[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius 4.<\/p>\n<div id=\"fs-id1167793912190\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793477594\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793477594\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793477594\">Applying the formula, we find<\/p>\n<div id=\"fs-id1167793510552\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m& ={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{4}2\\pi x\\sqrt{x}dx=2\\pi {\\displaystyle\\int }_{0}^{4}{x}^{3\\text{\/}2}dx\\hfill \\\\ & =2\\pi {\\frac{2}{5}{x}^{5\\text{\/}2}|}_{0}^{4}=\\frac{4\\pi }{5}\\left[32\\right]=\\frac{128\\pi }{5}.\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793563675\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]\\rho (x)=3x+2[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius 2.<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q242187\">Hint<\/span><\/p>\n<div id=\"q242187\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793563678\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793547373\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793547373\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793547373\">[latex]24\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=397&amp;end=477&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications397to477_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1179\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>2.5 Physical Applications. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.5 Physical Applications\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1179","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1179","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1179\/revisions"}],"predecessor-version":[{"id":1389,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1179\/revisions\/1389"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1179\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1179"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1179"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1179"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}