{"id":594,"date":"2021-02-04T16:32:00","date_gmt":"2021-02-04T16:32:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=594"},"modified":"2022-04-28T18:09:32","modified_gmt":"2022-04-28T18:09:32","slug":"symmetry-and-centroids","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/symmetry-and-centroids\/","title":{"raw":"Symmetry and Centroids","rendered":"Symmetry and Centroids"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply the theorem of Pappus for volume.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n<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\r\n<div id=\"fs-id1167793274939\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Centroid of a Symmetric Region<\/h3>\r\n<div id=\"fs-id1167793274941\" class=\"exercise\">\r\n<p id=\"fs-id1167793274948\">Let <em>R<\/em> 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. Find the centroid of the region.<\/p>\r\n[reveal-answer q=\"fs-id1167793506253\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793506253\"]\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<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\r\n<div id=\"fs-id1167793400814\" class=\"equation unnumbered\" style=\"text-align: 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]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793276857\">Next, we calculate the moments. We only need [latex]{M}_{x}\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793276870\" class=\"equation unnumbered\" style=\"text-align: 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]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793221724\">Then we have<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1167793604226\">The centroid of the region is [latex](0,8\\text{\/}5).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Centroid of a Symmetric Region.\r\n\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\"><\/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\/2.6MomentsAndCentersOfMass1715to1871_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]\r\n<div id=\"fs-id1167793619900\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div id=\"fs-id1167793619904\" class=\"exercise\">\r\n<p id=\"fs-id1167793619908\">Let <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=1-{x}^{2}[\/latex] and below by [latex]x[\/latex]-axis. Find the centroid of the region.<\/p>\r\n[reveal-answer q=\"fs-id1167794075561\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794075561\"]\r\n<p id=\"fs-id1167794075561\">The centroid of the region is [latex](0,2\\text{\/}5).[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167794327222\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793524932\" class=\"textbox tryit\">\r\n<h3>Activity: Engineering The Grand Canyon Skywalk<\/h3>\r\n<p id=\"fs-id1167793524940\">The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"700\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213307\/CNX_Calc_Figure_06_06_011.jpg\" alt=\"This figure is a picture of the Grand Canyon skywalk. It is a building at the edge of the canyon with a walkway extending out over the canyon\" width=\"700\" height=\"526\" \/> Figure 11. The Grand Canyon Skywalk offers magnificent views of the canyon. (credit: 10da_ralta, Wikimedia Commons)[\/caption]\r\n\r\n&nbsp;\r\n\r\nThe Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.\r\n\r\nOne factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.\r\n<p id=\"fs-id1167793502542\">The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors\u2019 center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.<\/p>\r\n<p id=\"fs-id1167793547133\">To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the <em>xy<\/em>-plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted [latex]{R}_{1},[\/latex] consists of the curved part of the U. We model [latex]{R}_{1}[\/latex] as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"700\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213311\/CNX_Calc_Figure_06_06_012.jpg\" alt=\"This figure is a sketch of the Grand Canyon walkway. It is on the xy coordinate system. The walkway is upside-down \u201cu\u201d shaped. It has been divided into three regions. The first region at the top is labeled \u201cRsub1\u201d. It is a semi-circle with outer radius of 35 feet and inner radius of 25 feet. The second region is labeled \u201cRsub2\u201d. It has two rectangles with width of 10 feet each and height of 35 feet. The third region is labeled \u201cRsub3\u201d and is two rectangles. They have a width of 10 feet and height of 48 feet. These represent the part of the walkway inside of the visitor center.\" width=\"700\" height=\"571\" \/> Figure 12. We model the Skywalk with three sub-regions.[\/caption]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1167793630335\">The legs of the platform, extending 35 ft between [latex]{R}_{1}[\/latex] and the canyon wall, comprise the second sub-region, [latex]{R}_{2}.[\/latex] Last, the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, [latex]{R}_{3}.[\/latex] Assume the density of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the visitor center; we will consider that later). Use [latex]g=32{\\text{ft\/sec}}^{2}.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1167793506232\">\r\n \t<li>Compute the area of each of the three sub-regions. Note that the areas of regions [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex] should include the areas of the legs only, not the open space between them. Round answers to the nearest square foot.<\/li>\r\n \t<li>Determine the mass associated with each of the three sub-regions.<\/li>\r\n \t<li>Calculate the center of mass of each of the three sub-regions.<\/li>\r\n \t<li>Now, treat each of the three sub-regions as a point mass located at the center of mass of the corresponding sub-region. Using this representation, calculate the center of mass of the entire platform.<\/li>\r\n \t<li>Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of [latex]{R}_{3}.[\/latex] Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the center of mass change?<\/li>\r\n \t<li>Although the Skywalk was built to limit the number of people on the observation platform to 120, the platform is capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform, and all of them went to the farthest end of the platform, how would the center of gravity of the system be affected? (Include the visitor center in the calculations and represent the people by a point mass located at the farthest edge of the platform, 70 ft from the canyon wall.)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167794226024\" class=\"bc-section section\">\r\n<h2>Theorem of Pappus<\/h2>\r\n<p id=\"fs-id1167794226030\">This section ends with a discussion of the <strong>theorem of Pappus for volume<\/strong>, which 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\r\n<div id=\"fs-id1167794140587\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem of Pappus for Volume<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794140594\">Let <em>R<\/em> be a region in the plane and let [latex]l[\/latex] be a line in the plane that does not intersect <em>R<\/em>. Then the volume of the solid of revolution formed by revolving <em>R<\/em> around [latex]l[\/latex] is equal to the area of <em>R<\/em> multiplied by the distance [latex]d[\/latex] traveled by the centroid of <em>R.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793383276\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\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\r\n<div id=\"fs-id1167793372816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m},[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793637985\">where<\/p>\r\n\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\nThen,\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&nbsp;\r\n<p id=\"fs-id1167794138970\">and thus<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1167793720000\">However, using the method of cylindrical shells, we have<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1167793546888\">So,<\/p>\r\n\r\n<div id=\"fs-id1167793546891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=d\u00b7A[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793385031\">and the proof is complete.<\/p>\r\n<p id=\"fs-id1167793385034\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793385038\" class=\"textbook exercises\">\r\n<h3>Example: Using the Theorem of Pappus for Volume<\/h3>\r\n<div id=\"fs-id1167793385040\" class=\"exercise\">\r\n<p id=\"fs-id1167793568525\">Let <em>R<\/em> be a circle of radius 2 centered at [latex](4,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving <em>R<\/em> around the [latex]y[\/latex]-axis.<\/p>\r\n[reveal-answer q=\"fs-id1167794051229\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794051229\"]\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<p id=\"fs-id1167793579510\">The region <em>R<\/em> is a circle of radius 2, so the area of <em>R<\/em> is [latex]A=4\\pi [\/latex] units<sup>2<\/sup>. By the symmetry principle, the centroid of <em>R<\/em> is the center of the circle. The centroid travels around the [latex]y[\/latex]-axis in a circular path of radius 4, 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[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793221560\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet <em>R<\/em> be a circle of radius 1 centered at [latex](3,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving <em>R<\/em> around the [latex]y[\/latex]-axis.\r\n<div id=\"fs-id1167793221565\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1167793504036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793504036\"]\r\n<p id=\"fs-id1167793504036\">[latex]6{\\pi }^{2}[\/latex] units<sup>3<\/sup><\/p>\r\n\r\n<div id=\"fs-id1167793504052\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793419100\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\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\/-XKii-JZrqw?controls=0&amp;start=2076&amp;end=2127&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\/2.6MomentsAndCentersOfMass2076to2127_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]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply the theorem of Pappus for volume.<\/li>\n<\/ul>\n<\/div>\n<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<div id=\"fs-id1167793274939\" class=\"textbook exercises\">\n<h3>Example: Finding the Centroid of a Symmetric Region<\/h3>\n<div id=\"fs-id1167793274941\" class=\"exercise\">\n<p id=\"fs-id1167793274948\">Let <em>R<\/em> 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. Find the centroid of the region.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793506253\">Show Solution<\/span><\/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<div style=\"width: 277px\" 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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10. We can use the symmetry principle to help find the centroid of a symmetric region.<\/p>\n<\/div>\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 id=\"fs-id1167793400814\" class=\"equation unnumbered\" 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>&nbsp;<\/p>\n<p id=\"fs-id1167793276857\">Next, we calculate the moments. We only need [latex]{M}_{x}\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793276870\" class=\"equation unnumbered\" 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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1167793604226\">The centroid of the region is [latex](0,8\\text{\/}5).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Centroid of a Symmetric Region.<\/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\"><\/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\/2.6MomentsAndCentersOfMass1715to1871_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<div id=\"fs-id1167793619900\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<div id=\"fs-id1167793619904\" class=\"exercise\">\n<p id=\"fs-id1167793619908\">Let <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=1-{x}^{2}[\/latex] and below by [latex]x[\/latex]-axis. Find the centroid of the region.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794075561\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794075561\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794075561\">The centroid of the region is [latex](0,2\\text{\/}5).[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167794327222\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793524932\" class=\"textbox tryit\">\n<h3>Activity: Engineering The Grand Canyon Skywalk<\/h3>\n<p id=\"fs-id1167793524940\">The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).<\/p>\n<div style=\"width: 710px\" 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\/11213307\/CNX_Calc_Figure_06_06_011.jpg\" alt=\"This figure is a picture of the Grand Canyon skywalk. It is a building at the edge of the canyon with a walkway extending out over the canyon\" width=\"700\" height=\"526\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11. The Grand Canyon Skywalk offers magnificent views of the canyon. (credit: 10da_ralta, Wikimedia Commons)<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.<\/p>\n<p>One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.<\/p>\n<p id=\"fs-id1167793502542\">The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors\u2019 center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.<\/p>\n<p id=\"fs-id1167793547133\">To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the <em>xy<\/em>-plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted [latex]{R}_{1},[\/latex] consists of the curved part of the U. We model [latex]{R}_{1}[\/latex] as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).<\/p>\n<div style=\"width: 710px\" 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\/11213311\/CNX_Calc_Figure_06_06_012.jpg\" alt=\"This figure is a sketch of the Grand Canyon walkway. It is on the xy coordinate system. The walkway is upside-down \u201cu\u201d shaped. It has been divided into three regions. The first region at the top is labeled \u201cRsub1\u201d. It is a semi-circle with outer radius of 35 feet and inner radius of 25 feet. The second region is labeled \u201cRsub2\u201d. It has two rectangles with width of 10 feet each and height of 35 feet. The third region is labeled \u201cRsub3\u201d and is two rectangles. They have a width of 10 feet and height of 48 feet. These represent the part of the walkway inside of the visitor center.\" width=\"700\" height=\"571\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12. We model the Skywalk with three sub-regions.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793630335\">The legs of the platform, extending 35 ft between [latex]{R}_{1}[\/latex] and the canyon wall, comprise the second sub-region, [latex]{R}_{2}.[\/latex] Last, the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, [latex]{R}_{3}.[\/latex] Assume the density of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the visitor center; we will consider that later). Use [latex]g=32{\\text{ft\/sec}}^{2}.[\/latex]<\/p>\n<ol id=\"fs-id1167793506232\">\n<li>Compute the area of each of the three sub-regions. Note that the areas of regions [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex] should include the areas of the legs only, not the open space between them. Round answers to the nearest square foot.<\/li>\n<li>Determine the mass associated with each of the three sub-regions.<\/li>\n<li>Calculate the center of mass of each of the three sub-regions.<\/li>\n<li>Now, treat each of the three sub-regions as a point mass located at the center of mass of the corresponding sub-region. Using this representation, calculate the center of mass of the entire platform.<\/li>\n<li>Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of [latex]{R}_{3}.[\/latex] Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the center of mass change?<\/li>\n<li>Although the Skywalk was built to limit the number of people on the observation platform to 120, the platform is capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform, and all of them went to the farthest end of the platform, how would the center of gravity of the system be affected? (Include the visitor center in the calculations and represent the people by a point mass located at the farthest edge of the platform, 70 ft from the canyon wall.)<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167794226024\" class=\"bc-section section\">\n<h2>Theorem of Pappus<\/h2>\n<p id=\"fs-id1167794226030\">This section ends with a discussion of the <strong>theorem of Pappus for volume<\/strong>, which 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<div id=\"fs-id1167794140587\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem of Pappus for Volume<\/h3>\n<hr \/>\n<p id=\"fs-id1167794140594\">Let <em>R<\/em> be a region in the plane and let [latex]l[\/latex] be a line in the plane that does not intersect <em>R<\/em>. Then the volume of the solid of revolution formed by revolving <em>R<\/em> around [latex]l[\/latex] is equal to the area of <em>R<\/em> multiplied by the distance [latex]d[\/latex] traveled by the centroid of <em>R.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1167793383276\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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<div id=\"fs-id1167793385038\" class=\"textbook exercises\">\n<h3>Example: Using the Theorem of Pappus for Volume<\/h3>\n<div id=\"fs-id1167793385040\" class=\"exercise\">\n<p id=\"fs-id1167793568525\">Let <em>R<\/em> be a circle of radius 2 centered at [latex](4,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving <em>R<\/em> around the [latex]y[\/latex]-axis.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794051229\">Show Solution<\/span><\/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<div style=\"width: 718px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<p id=\"fs-id1167793579510\">The region <em>R<\/em> is a circle of radius 2, so the area of <em>R<\/em> is [latex]A=4\\pi[\/latex] units<sup>2<\/sup>. By the symmetry principle, the centroid of <em>R<\/em> is the center of the circle. The centroid travels around the [latex]y[\/latex]-axis in a circular path of radius 4, 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<\/div>\n<\/div>\n<div id=\"fs-id1167793221560\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let <em>R<\/em> be a circle of radius 1 centered at [latex](3,0).[\/latex] Use the theorem of Pappus for volume to find the volume of the torus generated by revolving <em>R<\/em> around the [latex]y[\/latex]-axis.<\/p>\n<div id=\"fs-id1167793221565\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793504036\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793504036\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793504036\">[latex]6{\\pi }^{2}[\/latex] units<sup>3<\/sup><\/p>\n<div id=\"fs-id1167793504052\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793419100\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\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\/-XKii-JZrqw?controls=0&amp;start=2076&amp;end=2127&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\/2.6MomentsAndCentersOfMass2076to2127_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\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-594\">\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>2.6 Moments and Center of Mass. <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 1. <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\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/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-1\/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":24,"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\":\"2.6 Moments and Center of Mass\",\"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-594","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/594","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/594\/revisions"}],"predecessor-version":[{"id":4979,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/594\/revisions\/4979"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/65"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/594\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=594"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=594"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=594"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=594"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}