{"id":474,"date":"2025-02-13T19:45:18","date_gmt":"2025-02-13T19:45:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/moments-and-centers-of-mass-learn-it-4\/"},"modified":"2026-02-04T15:22:06","modified_gmt":"2026-02-04T15:22:06","slug":"moments-and-centers-of-mass-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/moments-and-centers-of-mass-learn-it-4\/","title":{"raw":"Moments and Centers of Mass: Learn It 4","rendered":"Moments and Centers of Mass: Learn It 4"},"content":{"raw":"<h2>The Symmetry Principle<\/h2>\r\n<p id=\"fs-id1167793367826\">We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1167793274948\">Let [latex]R[\/latex] be the region bounded above by the graph of the function [latex]f(x)=4-{x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis.<\/p>\r\n<p>Find the centroid of the region.<\/p>\r\n<p>[reveal-answer q=\"fs-id1167793506253\"]Show Solution[\/reveal-answer]<br>\r\n[hidden-answer a=\"fs-id1167793506253\"]<\/p>\r\n<p id=\"fs-id1167793506253\">The region is depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"267\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213301\/CNX_Calc_Figure_06_06_010.jpg\" alt=\"This figure is a graph of the function f(x)=4-x^2. It is an upside-down parabola. The region under the parabola above the x-axis is shaded. The curve intersects the x-axis at x=-2 and x=2.\" width=\"267\" height=\"272\"> Figure 10. We can use the symmetry principle to help find the centroid of a symmetric region.[\/caption]\r\n\r\n\r\n<p id=\"fs-id1167793571348\">The region is symmetric with respect to the [latex]y[\/latex]-axis. Therefore, the [latex]x[\/latex]-coordinate of the centroid is zero. We need only calculate [latex]\\overline{y}.[\/latex] Once again, for the sake of convenience, assume [latex]\\rho =1.[\/latex]<\/p>\r\n<p id=\"fs-id1167793420617\">First, we calculate the total mass:<\/p>\r\n<center>[latex]\\begin{array}{cc}\\hfill m&amp; =\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-2}^{2}(4-{x}^{2})dx\\hfill \\\\ &amp; ={\\left[4x-\\frac{{x}^{3}}{3}\\right]|}_{-2}^{2}=\\frac{32}{3}.\\hfill \\end{array}[\/latex] <\/center>\r\nNext, we calculate the moments. We only need [latex]{M}_{x}\\text{:}[\/latex]\r\n\r\n<center>[latex]\\begin{array}{cc}\\hfill {M}_{x}&amp; =\\rho {\\displaystyle\\int }_{a}^{b}\\frac{{\\left[f(x)\\right]}^{2}}{2}dx\\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{-2}^{2}{\\left[4-{x}^{2}\\right]}^{2}dx=\\frac{1}{2}{\\displaystyle\\int }_{-2}^{2}(16-8{x}^{2}+{x}^{4})dx\\hfill \\\\ &amp; =\\frac{1}{2}{\\left[\\frac{{x}^{5}}{5}-\\frac{8{x}^{3}}{3}+16x\\right]|}_{-2}^{2}=\\frac{256}{15}.\\hfill \\end{array}[\/latex]<\/center>\r\n<p id=\"fs-id1167793221724\">Then we have<\/p>\r\n<div id=\"fs-id1167793221728\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{y}=\\frac{{M}_{x}}{y}=\\frac{256}{15}\u00b7\\frac{3}{32}=\\frac{8}{5}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793604226\">The centroid of the region is [latex](0,8\\text{\/}5).[\/latex]<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/-XKii-JZrqw?controls=0&amp;start=1715&amp;end=1871&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.6MomentsAndCentersOfMass1715to1871_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.6 Moments and Centers of Mass\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section><\/div>\r\n\r\n<h2>Theorem of Pappus<\/h2>\r\n<p id=\"fs-id1167794226030\">The <strong>theorem of Pappus for volume <\/strong>allows us to find the volume of particular kinds of solids by using the centroid. There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Theorem of Pappus for volume<\/h3>\r\n<p>Let [latex]R[\/latex] be a region in the plane and let [latex]l[\/latex] be a line in the plane that does not intersect [latex]R[\/latex]. Then the volume of the solid of revolution formed by revolving [latex]R[\/latex] around [latex]l[\/latex] is equal to the area of [latex]R[\/latex] multiplied by the distance [latex]d[\/latex] traveled by the centroid of [latex]R[\/latex]<em>.<\/em><\/p>\r\n<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr>\r\n<p id=\"fs-id1167793383281\">We can prove the case when the region is bounded above by the graph of a function [latex]f(x)[\/latex] and below by the graph of a function [latex]g(x)[\/latex] over an interval [latex]\\left[a,b\\right],[\/latex] and for which the axis of revolution is the [latex]y[\/latex]-axis. In this case, the area of the region is [latex]A={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex] Since the axis of rotation is the [latex]y[\/latex]-axis, the distance traveled by the centroid of the region depends only on the [latex]x[\/latex]-coordinate of the centroid, [latex]\\overline{x},[\/latex] which is<\/p>\r\n<div id=\"fs-id1167793372816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m},[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1167793637985\">where<\/p>\r\n<div id=\"fs-id1167793277642\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\r\n<p>Then,<\/p>\r\n<div id=\"fs-id1167794165426\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]d=2\\pi \\frac{\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx}{\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1167794138970\">and thus<\/p>\r\n<div id=\"fs-id1167794138973\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]d\u00b7A=2\\pi {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1167793720000\">However, using the method of cylindrical shells, we have<\/p>\r\n<div id=\"fs-id1167793720003\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=2\\pi {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1167793546888\">So,<\/p>\r\n<div id=\"fs-id1167793546891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=d\u00b7A[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1167793385031\">and the proof is complete.<\/p>\r\n<p id=\"fs-id1167793385034\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1167793568525\">Let [latex]R[\/latex] be a circle of radius [latex]2[\/latex] centered at [latex](4,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving [latex]R[\/latex] around the [latex]y[\/latex]-axis.<\/p>\r\n<p>[reveal-answer q=\"fs-id1167794051229\"]Show Solution[\/reveal-answer]<br>\r\n[hidden-answer a=\"fs-id1167794051229\"]<\/p>\r\n<p id=\"fs-id1167794051229\">The region and torus are depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"708\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213314\/CNX_Calc_Figure_06_06_013.jpg\" alt=\"This figure has two graphs. The first is the x y coordinate system with a circle centered on the x-axis at x=4. The radius is 2. The second figure is the x y coordinate system. The circle from the first image has been revolved about the y-axis to form a torus.\" width=\"708\" height=\"386\"> Figure 13. Determining the volume of a torus by using the theorem of Pappus. (a) A circular region R in the plane; (b) the torus generated by revolving R about the y-axis.[\/caption]\r\n\r\n\r\n<p id=\"fs-id1167793579510\">The region [latex]R[\/latex] is a circle of radius [latex]2[\/latex], so the area of [latex]R[\/latex] is [latex]A=4\\pi [\/latex] units<sup>2<\/sup>. By the symmetry principle, the centroid of [latex]R[\/latex] is the center of the circle. The centroid travels around the [latex]y[\/latex]-axis in a circular path of radius [latex]4[\/latex], so the centroid travels [latex]d=8\\pi [\/latex] units. Then, the volume of the torus is [latex]A\u00b7d=32{\\pi }^{2}[\/latex] units<sup>3<\/sup>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>The Symmetry Principle<\/h2>\n<p id=\"fs-id1167793367826\">We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793274948\">Let [latex]R[\/latex] be the region bounded above by the graph of the function [latex]f(x)=4-{x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis.<\/p>\n<p>Find the centroid of the region.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793506253\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793506253\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793506253\">The region is depicted in the following figure.<\/p>\n<figure style=\"width: 267px\" 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\/11213301\/CNX_Calc_Figure_06_06_010.jpg\" alt=\"This figure is a graph of the function f(x)=4-x^2. It is an upside-down parabola. The region under the parabola above the x-axis is shaded. The curve intersects the x-axis at x=-2 and x=2.\" width=\"267\" height=\"272\" \/><figcaption class=\"wp-caption-text\">Figure 10. We can use the symmetry principle to help find the centroid of a symmetric region.<\/figcaption><\/figure>\n<p id=\"fs-id1167793571348\">The region is symmetric with respect to the [latex]y[\/latex]-axis. Therefore, the [latex]x[\/latex]-coordinate of the centroid is zero. We need only calculate [latex]\\overline{y}.[\/latex] Once again, for the sake of convenience, assume [latex]\\rho =1.[\/latex]<\/p>\n<p id=\"fs-id1167793420617\">First, we calculate the total mass:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m& =\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{-2}^{2}(4-{x}^{2})dx\\hfill \\\\ & ={\\left[4x-\\frac{{x}^{3}}{3}\\right]|}_{-2}^{2}=\\frac{32}{3}.\\hfill \\end{array}[\/latex] <\/div>\n<p>Next, we calculate the moments. We only need [latex]{M}_{x}\\text{:}[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{x}& =\\rho {\\displaystyle\\int }_{a}^{b}\\frac{{\\left[f(x)\\right]}^{2}}{2}dx\\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{-2}^{2}{\\left[4-{x}^{2}\\right]}^{2}dx=\\frac{1}{2}{\\displaystyle\\int }_{-2}^{2}(16-8{x}^{2}+{x}^{4})dx\\hfill \\\\ & =\\frac{1}{2}{\\left[\\frac{{x}^{5}}{5}-\\frac{8{x}^{3}}{3}+16x\\right]|}_{-2}^{2}=\\frac{256}{15}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793221724\">Then we have<\/p>\n<div id=\"fs-id1167793221728\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{y}=\\frac{{M}_{x}}{y}=\\frac{256}{15}\u00b7\\frac{3}{32}=\\frac{8}{5}.[\/latex]<\/div>\n<p id=\"fs-id1167793604226\">The centroid of the region is [latex](0,8\\text{\/}5).[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/-XKii-JZrqw?controls=0&amp;start=1715&amp;end=1871&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/2.6MomentsAndCentersOfMass1715to1871_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.6 Moments and Centers of Mass&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<h2>Theorem of Pappus<\/h2>\n<p id=\"fs-id1167794226030\">The <strong>theorem of Pappus for volume <\/strong>allows us to find the volume of particular kinds of solids by using the centroid. There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Theorem of Pappus for volume<\/h3>\n<p>Let [latex]R[\/latex] be a region in the plane and let [latex]l[\/latex] be a line in the plane that does not intersect [latex]R[\/latex]. Then the volume of the solid of revolution formed by revolving [latex]R[\/latex] around [latex]l[\/latex] is equal to the area of [latex]R[\/latex] multiplied by the distance [latex]d[\/latex] traveled by the centroid of [latex]R[\/latex]<em>.<\/em><\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1167793383281\">We can prove the case when the region is bounded above by the graph of a function [latex]f(x)[\/latex] and below by the graph of a function [latex]g(x)[\/latex] over an interval [latex]\\left[a,b\\right],[\/latex] and for which the axis of revolution is the [latex]y[\/latex]-axis. In this case, the area of the region is [latex]A={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex] Since the axis of rotation is the [latex]y[\/latex]-axis, the distance traveled by the centroid of the region depends only on the [latex]x[\/latex]-coordinate of the centroid, [latex]\\overline{x},[\/latex] which is<\/p>\n<div id=\"fs-id1167793372816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793637985\">where<\/p>\n<div id=\"fs-id1167793277642\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<p>Then,<\/p>\n<div id=\"fs-id1167794165426\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]d=2\\pi \\frac{\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx}{\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794138970\">and thus<\/p>\n<div id=\"fs-id1167794138973\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]d\u00b7A=2\\pi {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793720000\">However, using the method of cylindrical shells, we have<\/p>\n<div id=\"fs-id1167793720003\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=2\\pi {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793546888\">So,<\/p>\n<div id=\"fs-id1167793546891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=d\u00b7A[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793385031\">and the proof is complete.<\/p>\n<p id=\"fs-id1167793385034\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793568525\">Let [latex]R[\/latex] be a circle of radius [latex]2[\/latex] centered at [latex](4,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving [latex]R[\/latex] around the [latex]y[\/latex]-axis.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794051229\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794051229\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794051229\">The region and torus are depicted in the following figure.<\/p>\n<figure style=\"width: 708px\" 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\/11213314\/CNX_Calc_Figure_06_06_013.jpg\" alt=\"This figure has two graphs. The first is the x y coordinate system with a circle centered on the x-axis at x=4. The radius is 2. The second figure is the x y coordinate system. The circle from the first image has been revolved about the y-axis to form a torus.\" width=\"708\" height=\"386\" \/><figcaption class=\"wp-caption-text\">Figure 13. Determining the volume of a torus by using the theorem of Pappus. (a) A circular region R in the plane; (b) the torus generated by revolving R about the y-axis.<\/figcaption><\/figure>\n<p id=\"fs-id1167793579510\">The region [latex]R[\/latex] is a circle of radius [latex]2[\/latex], so the area of [latex]R[\/latex] is [latex]A=4\\pi[\/latex] units<sup>2<\/sup>. By the symmetry principle, the centroid of [latex]R[\/latex] is the center of the circle. The centroid travels around the [latex]y[\/latex]-axis in a circular path of radius [latex]4[\/latex], so the centroid travels [latex]d=8\\pi[\/latex] units. Then, the volume of the torus is [latex]A\u00b7d=32{\\pi }^{2}[\/latex] units<sup>3<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"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":[],"part":450,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/474"}],"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":2,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/474\/revisions"}],"predecessor-version":[{"id":2544,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/474\/revisions\/2544"}],"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\/474\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=474"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=474"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=474"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=474"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}