{"id":4099,"date":"2022-04-14T18:15:54","date_gmt":"2022-04-14T18:15:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-calculating-centers-of-mass-and-moments-of-inertia\/"},"modified":"2022-11-09T16:41:29","modified_gmt":"2022-11-09T16:41:29","slug":"skills-review-for-calculating-centers-of-mass-and-moments-of-inertia","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-calculating-centers-of-mass-and-moments-of-inertia\/","title":{"raw":"Skills Review for Calculating Centers of Mass and Moments of Inertia","rendered":"Skills Review for Calculating Centers of Mass and Moments of Inertia"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the equation of a line using slope and a point on the line<\/li>\r\n \t<li>Use symmetry to help locate the centroid of a thin plate.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Calculating Centers of Mass and Moments of Inertia section, we will explore how to calculate the center of mass and moment of inertia in two and three dimensions. Here we will review how to write the equation of a line and locate the center of mass of a lamina.\r\n<h2>Write the Equation of a Line<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Parametric Equations and Calculus of Parametric Equations)<\/em><\/strong>\r\nThe <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Slope of a Line<\/h3>\r\nThe slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:\r\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\r\nFind the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].\r\n[reveal-answer q=\"688301\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688301\"]\r\n\r\nWe substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&amp;=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&amp;=\\frac{4}{-7}\\hfill \\\\ \\hfill&amp;=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\r\nThe slope is [latex]-\\frac{4}{7}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nIt does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the slope of the line that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"196055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196055\"]\r\n\r\nslope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]1719[\/ohm_question]\r\n\r\n<\/div>\r\nTo write the equation of a line, the line's slope and a point the line goes through must be known. Perhaps the most familiar form of a linear equation is <strong>slope-intercept form<\/strong> written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. Let us begin with the slope.\r\n\r\nOften, the starting point to writing the equation of a line is to use <strong>point-slope formula<\/strong>.\u00a0Given the slope and one point on a line, we can find the equation of the line using point-slope form shown below.\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\nWe need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Point-Slope Formula<\/h3>\r\nGiven one point and the slope, using point-slope form will lead to the equation of a line:\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n[reveal-answer q=\"201330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201330\"]\r\n\r\nUsing point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\n<\/div>\r\n<div>\r\n\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"634647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634647\"]\r\n\r\n[latex]y=4x - 3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]110942[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794026051\" class=\"bc-section section\">\r\n<h2>Center of Mass of Thin Plates<\/h2>\r\n<p id=\"fs-id1167794143832\">An object that is a two-dimensional \"sheet\" (such as paper) is called a <strong>lamina<\/strong>. A lamina is often represented by a two-dimensional region in a plane. Suppose we have a lamina bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the [latex]x[\/latex]-axis, and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively, as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"261\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213249\/CNX_Calc_Figure_06_06_004.jpg\" alt=\"This image is a graph of y=f(x). It is in the first quadrant. Under the curve is a shaded region labeled \u201cR\u201d. The shaded region is bounded to the left at x=a and to the right at x=b.\" width=\"261\" height=\"234\" \/> A region in the plane representing a lamina.[\/caption]\r\n<p id=\"fs-id1167794094462\">To find the center of mass of a lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the [latex]x[\/latex]- and [latex]y[\/latex]-axes.<\/p>\r\n\r\n<div id=\"fs-id1167793632966\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Center of Mass of a Thin Plate in the <em>xy<\/em>-Plane<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794023295\">Let <em>R<\/em> denote a region bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the [latex]x[\/latex]-axis, and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Let [latex]\\rho [\/latex] denote the density of the associated lamina. Then we can make the following statements:<\/p>\r\n\r\n<ol id=\"fs-id1167794043274\">\r\n \t<li>The mass of the lamina is\r\n<div id=\"fs-id1167793888126\" class=\"equation\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx.[\/latex]<\/div><\/li>\r\n \t<li>The moments [latex]{M}_{x}[\/latex] and [latex]{M}_{y}[\/latex] of the lamina with respect to the [latex]x[\/latex]- and [latex]y[\/latex]-axes, respectively, are\r\n<div id=\"fs-id1167794003845\" class=\"equation\">[latex]{M}_{x}=\\rho {\\displaystyle\\int }_{a}^{b}\\frac{{\\left[f(x)\\right]}^{2}}{2}dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}xf(x)dx.[\/latex]<\/div><\/li>\r\n \t<li>The coordinates of the center of mass [latex](\\overline{x},\\overline{y})[\/latex] are\r\n<div id=\"fs-id1167794218685\" class=\"equation\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{m}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167793479886\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Center of Mass of a Lamina<\/h3>\r\nLet <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=\\sqrt{x}[\/latex] and below by the [latex]x[\/latex]-axis over the interval [latex]\\left[0,4\\right].[\/latex] Find the centroid (center of mass) of the region.\r\n<div id=\"fs-id1167793479888\" class=\"exercise\">[reveal-answer q=\"fs-id1167793960791\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793960791\"]\r\n<p id=\"fs-id1167793960791\">The region is depicted in the following figure.<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213253\/CNX_Calc_Figure_06_06_006.jpg\" alt=\"This figure is the graph of the curve f(x)=squareroot(x). It is an increasing curve in the first quadrant. Under the curve above the x-axis there is a shaded region. It starts at x=0 and is bounded to the right at x=4.\" width=\"266\" height=\"272\" \/>\r\n<p id=\"fs-id1167793881082\">Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant [latex]\\rho [\/latex] cancels out of the calculations eventually. Therefore, for the sake of convenience, let\u2019s assume [latex]\\rho =1.[\/latex]<\/p>\r\n<p id=\"fs-id1167793951832\">First, we need to calculate the total mass:<\/p>\r\n\r\n<div id=\"fs-id1167793720137\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m&amp; =\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx={\\displaystyle\\int }_{0}^{4}\\sqrt{x}dx\\hfill \\\\ &amp; ={\\frac{2}{3}{x}^{3\\text{\/}2}|}_{0}^{4}=\\frac{2}{3}\\left[8-0\\right]=\\frac{16}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794100418\">Next, we compute the moments:<\/p>\r\n\r\n<div id=\"fs-id1167793551987\" 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; ={\\displaystyle\\int }_{0}^{4}\\frac{x}{2}dx={\\frac{1}{4}{x}^{2}|}_{0}^{4}=4\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793271531\">and<\/p>\r\n\r\n<div id=\"fs-id1167793261500\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{y}&amp; =\\rho {\\displaystyle\\int }_{a}^{b}xf(x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{4}x\\sqrt{x}dx={\\displaystyle\\int }_{0}^{4}{x}^{3\\text{\/}2}dx\\hfill \\\\ &amp; ={\\frac{2}{5}{x}^{5\\text{\/}2}|}_{0}^{4}=\\frac{2}{5}\\left[32-0\\right]=\\frac{64}{5}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794077064\">Thus, we have<\/p>\r\n\r\n<div id=\"fs-id1167794077068\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}=\\frac{64\\text{\/}5}{16\\text{\/}3}=\\frac{64}{5}\u00b7\\frac{3}{16}=\\frac{12}{5}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{y}=\\frac{4}{16\\text{\/}3}=4\u00b7\\frac{3}{16}=\\frac{3}{4}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793537336\">The centroid of the region is [latex](12\\text{\/}5,3\\text{\/}4).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793617556\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]R[\/latex] be the region bounded above by the graph of the function [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis over the interval [latex]\\left[0,2\\right].[\/latex]. Find the centroid (center of mass) of the region.\r\n<div>[reveal-answer q=\"361340\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"361340\"]Use the process from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793617560\" class=\"exercise\">[reveal-answer q=\"fs-id1167793912452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793912452\"]\r\n<p id=\"fs-id1167793912452\">The centroid of the region is [latex](\\frac{3}{2},\\frac{6}{5}).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167794178194\">We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function [latex]f(x),[\/latex] as before, but now, instead of having the lower bound for the region be the [latex]x[\/latex]-axis, suppose the region is bounded below by the graph of a second continuous function, [latex]g(x),[\/latex] as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"225\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213255\/CNX_Calc_Figure_06_06_007.jpg\" alt=\"This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded region labeled \u201cR\u201d. The shaded region is bounded to the left by x=a and to the right by x=b.\" width=\"225\" height=\"203\" \/> A region between two functions.[\/caption]\r\n<p id=\"fs-id1167793369732\">Again, we partition the interval [latex]\\left[a,b\\right][\/latex] and construct rectangles. A representative rectangle is shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"261\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213257\/CNX_Calc_Figure_06_06_008.jpg\" alt=\"This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded rectangle.\" width=\"261\" height=\"234\" \/> A representative rectangle of the region between two functions.[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Center of Mass of a Lamina Bounded by Two Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793778208\">Let <em>R<\/em> denote a region bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the graph of the continuous function [latex]g(x),[\/latex] and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Let [latex]\\rho [\/latex] denote the density of the associated lamina. Then we can make the following statements:<\/p>\r\n\r\n<ol id=\"fs-id1167793984461\">\r\n \t<li>The mass of the lamina is\r\n<div id=\"fs-id1167793309328\" class=\"equation\" style=\"text-align: center;\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div><\/li>\r\n \t<li>The moments [latex]{M}_{x}[\/latex] and [latex]{M}_{y}[\/latex] of the lamina with respect to the [latex]x[\/latex]- and [latex]y[\/latex]-axes, respectively, are\r\n<div id=\"fs-id1167793984368\" class=\"equation\" style=\"text-align: center;\">[latex]{M}_{x}=\\rho {\\displaystyle\\int }_{a}^{b}\\frac{1}{2}({\\left[f(x)\\right]}^{2}-{\\left[g(x)\\right]}^{2})dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div><\/li>\r\n \t<li>The coordinates of the center of mass [latex](\\overline{x},\\overline{y})[\/latex] are\r\n<div id=\"fs-id1167793692892\" class=\"equation\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{m}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167793632957\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Centroid of a Region Bounded by Two Functions<\/h3>\r\nLet <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=1-{x}^{2}[\/latex] and below by the graph of the function [latex]g(x)=x-1.[\/latex] Find the centroid of the region.\r\n<div id=\"fs-id1167793632959\" class=\"exercise\">[reveal-answer q=\"fs-id1167793984385\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793984385\"]\r\n<p id=\"fs-id1167793984385\">The region is depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"273\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213259\/CNX_Calc_Figure_06_06_009.jpg\" alt=\"This figure is a graph. It has two curves. They are labeled f(x)=1-x^2 and g(x)=x-1. In between the curves is a shaded region. The shaded region is bounded to the left by x=a and to the right by x=b.\" width=\"273\" height=\"272\" \/> Finding the centroid of a region between two curves.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1167793616878\">The graphs of the functions intersect at [latex](-2,-3)[\/latex] and [latex](1,0),[\/latex] so we integrate from \u22122 to 1. Once again, for the sake of convenience, assume [latex]\\rho =1.[\/latex]<\/p>\r\n<p id=\"fs-id1167794027950\">First, we need to calculate the total mass:<\/p>\r\n\r\n<div id=\"fs-id1167794027953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m&amp; =\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-2}^{1}\\left[1-{x}^{2}-(x-1)\\right]dx={\\displaystyle\\int }_{-2}^{1}(2-{x}^{2}-x)dx\\hfill \\\\ &amp; ={\\left[2x-\\frac{1}{3}{x}^{3}-\\frac{1}{2}{x}^{2}\\right]|}_{-2}^{1}=\\left[2-\\frac{1}{3}-\\frac{1}{2}\\right]-\\left[-4+\\frac{8}{3}-2\\right]=\\frac{9}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793880593\">Next, we compute the moments:<\/p>\r\n\r\n<div id=\"fs-id1167793470756\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{x}&amp; =\\rho {\\displaystyle\\int }_{a}^{b}\\frac{1}{2}({\\left[f(x)\\right]}^{2}-{\\left[g(x)\\right]}^{2})dx\\hfill \\\\ &amp; =\\frac{1}{2}{\\displaystyle\\int }_{-2}^{1}({(1-{x}^{2})}^{2}-{(x-1)}^{2})dx=\\frac{1}{2}{\\displaystyle\\int }_{-2}^{1}({x}^{4}-3{x}^{2}+2x)dx\\hfill \\\\ &amp; =\\frac{1}{2}{\\left[\\frac{{x}^{5}}{5}-{x}^{3}+{x}^{2}\\right]|}_{-2}^{1}=-\\frac{27}{10}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793886473\">and<\/p>\r\n\r\n<div id=\"fs-id1167793886476\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{y}&amp; =\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-2}^{1}x\\left[(1-{x}^{2})-(x-1)\\right]dx={\\displaystyle\\int }_{-2}^{1}x\\left[2-{x}^{2}-x\\right]dx={\\displaystyle\\int }_{-2}^{1}(2x-{x}^{4}-{x}^{2})dx\\hfill \\\\ &amp; ={\\left[{x}^{2}-\\frac{{x}^{5}}{5}-\\frac{{x}^{3}}{3}\\right]|}_{-2}^{1}=-\\frac{9}{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793773174\">Therefore, we have<\/p>\r\n\r\n<div id=\"fs-id1167793773177\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}=-\\frac{9}{4}\u00b7\\frac{2}{9}=-\\frac{1}{2}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{y}=-\\frac{27}{10}\u00b7\\frac{2}{9}=-\\frac{3}{5}.[\/latex]<\/div>\r\n<p id=\"fs-id1167794035411\">The centroid of the region is [latex](\\text{\u2212}(1\\text{\/}2),\\text{\u2212}(3\\text{\/}5)).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793424268\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=6-{x}^{2}[\/latex] and below by the graph of the function [latex]g(x)=3-2x.[\/latex] Find the centroid of the region.\r\n\r\n[reveal-answer q=\"738781\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"738781\"]Use the process from the previous example.[\/hidden-answer]\r\n<div id=\"fs-id1167793424272\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1167793308050\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793308050\"]\r\n<p id=\"fs-id1167793308050\">The centroid of the region is [latex](1,\\frac{13}{5}).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5681[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the equation of a line using slope and a point on the line<\/li>\n<li>Use symmetry to help locate the centroid of a thin plate.<\/li>\n<\/ul>\n<\/div>\n<p>In the Calculating Centers of Mass and Moments of Inertia section, we will explore how to calculate the center of mass and moment of inertia in two and three dimensions. Here we will review how to write the equation of a line and locate the center of mass of a lamina.<\/p>\n<h2>Write the Equation of a Line<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Parametric Equations and Calculus of Parametric Equations)<\/em><\/strong><br \/>\nThe <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Slope of a Line<\/h3>\n<p>The slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:<\/p>\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\n<p>Find the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688301\">Show Solution<\/span><\/p>\n<div id=\"q688301\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&=\\frac{4}{-7}\\hfill \\\\ \\hfill&=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\n<p>The slope is [latex]-\\frac{4}{7}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>It does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the slope of the line that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196055\">Show Solution<\/span><\/p>\n<div id=\"q196055\" class=\"hidden-answer\" style=\"display: none\">\n<p>slope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1719\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1719&theme=oea&iframe_resize_id=ohm1719&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>To write the equation of a line, the line&#8217;s slope and a point the line goes through must be known. Perhaps the most familiar form of a linear equation is <strong>slope-intercept form<\/strong> written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. Let us begin with the slope.<\/p>\n<p>Often, the starting point to writing the equation of a line is to use <strong>point-slope formula<\/strong>.\u00a0Given the slope and one point on a line, we can find the equation of the line using point-slope form shown below.<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<p>We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Point-Slope Formula<\/h3>\n<p>Given one point and the slope, using point-slope form will lead to the equation of a line:<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<\/div>\n<div style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q201330\">Show Solution<\/span><\/p>\n<div id=\"q201330\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<\/div>\n<div>\n<p>Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634647\">Show Solution<\/span><\/p>\n<div id=\"q634647\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=4x - 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm110942\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110942&theme=oea&iframe_resize_id=ohm110942&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1167794026051\" class=\"bc-section section\">\n<h2>Center of Mass of Thin Plates<\/h2>\n<p id=\"fs-id1167794143832\">An object that is a two-dimensional &#8220;sheet&#8221; (such as paper) is called a <strong>lamina<\/strong>. A lamina is often represented by a two-dimensional region in a plane. Suppose we have a lamina bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the [latex]x[\/latex]-axis, and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively, as shown in the following figure.<\/p>\n<div style=\"width: 271px\" 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\/11213249\/CNX_Calc_Figure_06_06_004.jpg\" alt=\"This image is a graph of y=f(x). It is in the first quadrant. Under the curve is a shaded region labeled \u201cR\u201d. The shaded region is bounded to the left at x=a and to the right at x=b.\" width=\"261\" height=\"234\" \/><\/p>\n<p class=\"wp-caption-text\">A region in the plane representing a lamina.<\/p>\n<\/div>\n<p id=\"fs-id1167794094462\">To find the center of mass of a lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-axes.<\/p>\n<div id=\"fs-id1167793632966\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Center of Mass of a Thin Plate in the <em>xy<\/em>-Plane<\/h3>\n<hr \/>\n<p id=\"fs-id1167794023295\">Let <em>R<\/em> denote a region bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the [latex]x[\/latex]-axis, and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Let [latex]\\rho[\/latex] denote the density of the associated lamina. Then we can make the following statements:<\/p>\n<ol id=\"fs-id1167794043274\">\n<li>The mass of the lamina is\n<div id=\"fs-id1167793888126\" class=\"equation\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx.[\/latex]<\/div>\n<\/li>\n<li>The moments [latex]{M}_{x}[\/latex] and [latex]{M}_{y}[\/latex] of the lamina with respect to the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-axes, respectively, are\n<div id=\"fs-id1167794003845\" class=\"equation\">[latex]{M}_{x}=\\rho {\\displaystyle\\int }_{a}^{b}\\frac{{\\left[f(x)\\right]}^{2}}{2}dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}xf(x)dx.[\/latex]<\/div>\n<\/li>\n<li>The coordinates of the center of mass [latex](\\overline{x},\\overline{y})[\/latex] are\n<div id=\"fs-id1167794218685\" class=\"equation\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{m}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167793479886\" class=\"textbook exercises\">\n<h3>Example: Finding the Center of Mass of a Lamina<\/h3>\n<p>Let <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=\\sqrt{x}[\/latex] and below by the [latex]x[\/latex]-axis over the interval [latex]\\left[0,4\\right].[\/latex] Find the centroid (center of mass) of the region.<\/p>\n<div id=\"fs-id1167793479888\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793960791\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793960791\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793960791\">The region is depicted in the following figure.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213253\/CNX_Calc_Figure_06_06_006.jpg\" alt=\"This figure is the graph of the curve f(x)=squareroot(x). It is an increasing curve in the first quadrant. Under the curve above the x-axis there is a shaded region. It starts at x=0 and is bounded to the right at x=4.\" width=\"266\" height=\"272\" \/><\/p>\n<p id=\"fs-id1167793881082\">Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant [latex]\\rho[\/latex] cancels out of the calculations eventually. Therefore, for the sake of convenience, let\u2019s assume [latex]\\rho =1.[\/latex]<\/p>\n<p id=\"fs-id1167793951832\">First, we need to calculate the total mass:<\/p>\n<div id=\"fs-id1167793720137\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m& =\\rho {\\displaystyle\\int }_{a}^{b}f(x)dx={\\displaystyle\\int }_{0}^{4}\\sqrt{x}dx\\hfill \\\\ & ={\\frac{2}{3}{x}^{3\\text{\/}2}|}_{0}^{4}=\\frac{2}{3}\\left[8-0\\right]=\\frac{16}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794100418\">Next, we compute the moments:<\/p>\n<div id=\"fs-id1167793551987\" 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 \\\\ & ={\\displaystyle\\int }_{0}^{4}\\frac{x}{2}dx={\\frac{1}{4}{x}^{2}|}_{0}^{4}=4\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793271531\">and<\/p>\n<div id=\"fs-id1167793261500\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{y}& =\\rho {\\displaystyle\\int }_{a}^{b}xf(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{4}x\\sqrt{x}dx={\\displaystyle\\int }_{0}^{4}{x}^{3\\text{\/}2}dx\\hfill \\\\ & ={\\frac{2}{5}{x}^{5\\text{\/}2}|}_{0}^{4}=\\frac{2}{5}\\left[32-0\\right]=\\frac{64}{5}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794077064\">Thus, we have<\/p>\n<div id=\"fs-id1167794077068\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}=\\frac{64\\text{\/}5}{16\\text{\/}3}=\\frac{64}{5}\u00b7\\frac{3}{16}=\\frac{12}{5}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{y}=\\frac{4}{16\\text{\/}3}=4\u00b7\\frac{3}{16}=\\frac{3}{4}.[\/latex]<\/div>\n<p id=\"fs-id1167793537336\">The centroid of the region is [latex](12\\text{\/}5,3\\text{\/}4).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793617556\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]R[\/latex] be the region bounded above by the graph of the function [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis over the interval [latex]\\left[0,2\\right].[\/latex]. Find the centroid (center of mass) of the region.<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q361340\">Hint<\/span><\/p>\n<div id=\"q361340\" 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-id1167793617560\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793912452\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793912452\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793912452\">The centroid of the region is [latex](\\frac{3}{2},\\frac{6}{5}).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167794178194\">We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function [latex]f(x),[\/latex] as before, but now, instead of having the lower bound for the region be the [latex]x[\/latex]-axis, suppose the region is bounded below by the graph of a second continuous function, [latex]g(x),[\/latex] as shown in the following figure.<\/p>\n<div style=\"width: 235px\" 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\/11213255\/CNX_Calc_Figure_06_06_007.jpg\" alt=\"This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded region labeled \u201cR\u201d. The shaded region is bounded to the left by x=a and to the right by x=b.\" width=\"225\" height=\"203\" \/><\/p>\n<p class=\"wp-caption-text\">A region between two functions.<\/p>\n<\/div>\n<p id=\"fs-id1167793369732\">Again, we partition the interval [latex]\\left[a,b\\right][\/latex] and construct rectangles. A representative rectangle is shown in the following figure.<\/p>\n<div style=\"width: 271px\" 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\/11213257\/CNX_Calc_Figure_06_06_008.jpg\" alt=\"This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded rectangle.\" width=\"261\" height=\"234\" \/><\/p>\n<p class=\"wp-caption-text\">A representative rectangle of the region between two functions.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Center of Mass of a Lamina Bounded by Two Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1167793778208\">Let <em>R<\/em> denote a region bounded above by the graph of a continuous function [latex]f(x),[\/latex] below by the graph of the continuous function [latex]g(x),[\/latex] and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Let [latex]\\rho[\/latex] denote the density of the associated lamina. Then we can make the following statements:<\/p>\n<ol id=\"fs-id1167793984461\">\n<li>The mass of the lamina is\n<div id=\"fs-id1167793309328\" class=\"equation\" style=\"text-align: center;\">[latex]m=\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<\/li>\n<li>The moments [latex]{M}_{x}[\/latex] and [latex]{M}_{y}[\/latex] of the lamina with respect to the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-axes, respectively, are\n<div id=\"fs-id1167793984368\" class=\"equation\" style=\"text-align: center;\">[latex]{M}_{x}=\\rho {\\displaystyle\\int }_{a}^{b}\\frac{1}{2}({\\left[f(x)\\right]}^{2}-{\\left[g(x)\\right]}^{2})dx\\text{ and }{M}_{y}=\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<\/li>\n<li>The coordinates of the center of mass [latex](\\overline{x},\\overline{y})[\/latex] are\n<div id=\"fs-id1167793692892\" class=\"equation\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{m}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167793632957\" class=\"textbook exercises\">\n<h3>Example: Finding the Centroid of a Region Bounded by Two Functions<\/h3>\n<p>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 the graph of the function [latex]g(x)=x-1.[\/latex] Find the centroid of the region.<\/p>\n<div id=\"fs-id1167793632959\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793984385\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793984385\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793984385\">The region is depicted in the following figure.<\/p>\n<div style=\"width: 283px\" 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\/11213259\/CNX_Calc_Figure_06_06_009.jpg\" alt=\"This figure is a graph. It has two curves. They are labeled f(x)=1-x^2 and g(x)=x-1. In between the curves is a shaded region. The shaded region is bounded to the left by x=a and to the right by x=b.\" width=\"273\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">Finding the centroid of a region between two curves.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1167793616878\">The graphs of the functions intersect at [latex](-2,-3)[\/latex] and [latex](1,0),[\/latex] so we integrate from \u22122 to 1. Once again, for the sake of convenience, assume [latex]\\rho =1.[\/latex]<\/p>\n<p id=\"fs-id1167794027950\">First, we need to calculate the total mass:<\/p>\n<div id=\"fs-id1167794027953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill m& =\\rho {\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ & ={\\displaystyle\\int }_{-2}^{1}\\left[1-{x}^{2}-(x-1)\\right]dx={\\displaystyle\\int }_{-2}^{1}(2-{x}^{2}-x)dx\\hfill \\\\ & ={\\left[2x-\\frac{1}{3}{x}^{3}-\\frac{1}{2}{x}^{2}\\right]|}_{-2}^{1}=\\left[2-\\frac{1}{3}-\\frac{1}{2}\\right]-\\left[-4+\\frac{8}{3}-2\\right]=\\frac{9}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793880593\">Next, we compute the moments:<\/p>\n<div id=\"fs-id1167793470756\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{x}& =\\rho {\\displaystyle\\int }_{a}^{b}\\frac{1}{2}({\\left[f(x)\\right]}^{2}-{\\left[g(x)\\right]}^{2})dx\\hfill \\\\ & =\\frac{1}{2}{\\displaystyle\\int }_{-2}^{1}({(1-{x}^{2})}^{2}-{(x-1)}^{2})dx=\\frac{1}{2}{\\displaystyle\\int }_{-2}^{1}({x}^{4}-3{x}^{2}+2x)dx\\hfill \\\\ & =\\frac{1}{2}{\\left[\\frac{{x}^{5}}{5}-{x}^{3}+{x}^{2}\\right]|}_{-2}^{1}=-\\frac{27}{10}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793886473\">and<\/p>\n<div id=\"fs-id1167793886476\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {M}_{y}& =\\rho {\\displaystyle\\int }_{a}^{b}x\\left[f(x)-g(x)\\right]dx\\hfill \\\\ & ={\\displaystyle\\int }_{-2}^{1}x\\left[(1-{x}^{2})-(x-1)\\right]dx={\\displaystyle\\int }_{-2}^{1}x\\left[2-{x}^{2}-x\\right]dx={\\displaystyle\\int }_{-2}^{1}(2x-{x}^{4}-{x}^{2})dx\\hfill \\\\ & ={\\left[{x}^{2}-\\frac{{x}^{5}}{5}-\\frac{{x}^{3}}{3}\\right]|}_{-2}^{1}=-\\frac{9}{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793773174\">Therefore, we have<\/p>\n<div id=\"fs-id1167793773177\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\overline{x}=\\frac{{M}_{y}}{m}=-\\frac{9}{4}\u00b7\\frac{2}{9}=-\\frac{1}{2}\\text{ and }\\overline{y}=\\frac{{M}_{x}}{y}=-\\frac{27}{10}\u00b7\\frac{2}{9}=-\\frac{3}{5}.[\/latex]<\/div>\n<p id=\"fs-id1167794035411\">The centroid of the region is [latex](\\text{\u2212}(1\\text{\/}2),\\text{\u2212}(3\\text{\/}5)).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793424268\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let <em>R<\/em> be the region bounded above by the graph of the function [latex]f(x)=6-{x}^{2}[\/latex] and below by the graph of the function [latex]g(x)=3-2x.[\/latex] 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=\"q738781\">Hint<\/span><\/p>\n<div id=\"q738781\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<div id=\"fs-id1167793424272\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793308050\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793308050\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793308050\">The centroid of the region is [latex](1,\\frac{13}{5}).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5681\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5681&theme=oea&iframe_resize_id=ohm5681&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-4099\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div 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