{"id":719,"date":"2025-10-09T15:34:24","date_gmt":"2025-10-09T15:34:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=719"},"modified":"2025-11-03T14:57:45","modified_gmt":"2025-11-03T14:57:45","slug":"center-of-mass-and-moments-of-inertia-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/center-of-mass-and-moments-of-inertia-learn-it-1\/","title":{"raw":"Center of Mass and Moments of Inertia: Learn It 1","rendered":"Center of Mass and Moments of Inertia: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find centers of mass in two dimensions<\/li>\r\n \t<li>Calculate moments of inertia in two dimensions<\/li>\r\n \t<li>Find centers of mass in three dimensions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Center of Mass in Two Dimensions<\/h2>\r\n<p class=\"whitespace-normal break-words\">We've already used multiple integrals to find areas, volumes, and average values of functions over bounded regions. Now we'll extend these techniques to find the center of mass and moments of inertia for physical objects. You'll use double integrals for laminas (flat plates) and triple integrals for three-dimensional objects with variable density. When an object has uniform density throughout, we call it <strong>homogeneous<\/strong>. For homogeneous objects, the density is constant.<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>center of mass<\/strong> represents the balance point of an object. In a uniform gravitational field, this is also called the center of gravity. For objects with uniform density, the center of mass is simply the geometric center, which we call the <strong>centroid<\/strong>.<\/p>\r\n<p class=\"whitespace-normal break-words\">Figure 1 shows a lamina perfectly balanced at its center of mass [latex]P[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1407\" align=\"aligncenter\" width=\"452\"]<img class=\"size-full wp-image-1407\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174357\/5-6-1.jpeg\" alt=\"A surface is delicately balanced on a fine point.\" width=\"452\" height=\"233\" \/> Figure 1.\u00a0A lamina is perfectly balanced on a spindle if the lamina\u2019s center of mass sits on the spindle.[\/caption]\r\n<p class=\"whitespace-normal break-words\">To find the coordinates of the center of mass [latex]P(\\overline{x}, \\overline{y})[\/latex] of a lamina, you need three quantities:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The moment [latex]M_x[\/latex] about the [latex]x[\/latex]-axis<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The mass [latex]m[\/latex] of the lamina<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Then the center of mass coordinates are:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}.[\/latex]<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Centers of Mass in One Dimension<\/strong>\r\n<p class=\"whitespace-normal break-words\">For a one-dimensional object like a thin rod with density [latex]\\rho(x)[\/latex] along the [latex]x[\/latex]-axis from [latex]x = a[\/latex] to [latex]x = b[\/latex]:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Mass: [latex]m = \\displaystyle\\int_a^b \\rho(x)dx[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Center of mass: [latex]\\overline{x} = \\frac{1}{m}\\displaystyle\\int_a^b x\\rho(x)dx[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">We'll extend this same idea to two dimensions using double integrals.<\/p>\r\n\r\n<\/section>When the density function is constant, these formulas give the <strong>centroid<\/strong> of the lamina\u2014the geometric center independent of mass distribution.\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Setting Up the Mass Calculation<\/h3>\r\n<p id=\"fs-id1167794052362\">Suppose that the lamina occupies a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane, with density [latex]{\\rho}{({{x},{y}})}[\/latex] (measured in mass per unit area) at any point [latex](x, y)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The density is defined as:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\rho}{({{x},{y}})} = {\\displaystyle\\lim_{{\\Delta}{A}{\\rightarrow}{0}}}{\\frac{{\\Delta}{m}}{{\\Delta}{A}}}[\/latex],<\/p>\r\nwhere [latex]{\\Delta}{m}[\/latex] and [latex]{\\Delta}{A}[\/latex] are the mass and area of a small rectangle containing the point [latex](x, y)[\/latex], and the limit is taken as the rectangle's dimensions approach zero (Figure 2).\r\n<div id=\"CNX_Calc_Figure_15_06_002\" class=\"os-figure\">\r\n\r\n[caption id=\"attachment_1408\" align=\"aligncenter\" width=\"454\"]<img class=\"size-full wp-image-1408\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174437\/5-6-2.jpeg\" alt=\"A lamina R is shown on the x y plane with a point (x, y) surrounded by a small rectangle marked Mass = Delta m and Area = Delta A.\" width=\"454\" height=\"384\" \/> Figure 2.\u00a0The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.[\/caption]\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div>To find the total mass, we partition region [latex]R[\/latex] into tiny rectangles [latex]R_{ij}[\/latex], each with area [latex]\\Delta A[\/latex]. We choose sample points [latex]{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}[\/latex] in each rectangle. The mass [latex]m_{ij}[\/latex] of rectangle [latex]R_{ij}[\/latex] is approximately:<\/div>\r\n<div style=\"text-align: center;\">\u00a0[latex]m_{ij} ={\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex] (Figure 3).<\/div>\r\n<div>Let [latex]k[\/latex] and [latex]l[\/latex] be the number of subintervals in [latex]x[\/latex] and [latex]y[\/latex] respectively (Figure 3). Note that while we use rectangles for convenience, this limiting process works for any shape, as you've seen in previous sections.<\/div>\r\n<div>[caption id=\"attachment_1410\" align=\"aligncenter\" width=\"454\"]<img class=\"size-full wp-image-1410\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174515\/5-6-3.jpeg\" alt=\"A lamina is shown on the x y plane with a point (x* sub ij, y* sub ij) surrounded by a small rectangle marked R sub ij.\" width=\"454\" height=\"384\" \/> Figure 3. Subdividing the lamina into tiny rectangles\u00a0[latex]R_{ij}[\/latex], each containing a sample point\u00a0[latex](x_{ij}^{\\ast},y_{ij}^{\\ast})[\/latex].[\/caption]<\/div>\r\n<div><\/div>\r\n<div>Taking the limit as the partition becomes infinitely fine gives us the total mass:<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793844477\" data-type=\"equation\">\r\n<div class=\"os-equation-number\" style=\"text-align: center;\">[latex]{m} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}.[\/latex]<\/div>\r\n<\/div>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>mass of a lamina<\/h3>\r\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex], the total mass is:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]m = \\iint_R \\rho(x,y) dA[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Let's see how this works with a concrete example.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Consider a triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {{x}{y}} \\ {\\text{kg\/m}^2}[\/latex]. Find the total mass.[reveal-answer q=\"547923465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547923465\"]A sketch of the region [latex]R[\/latex] is always helpful, as shown in the following figure.[caption id=\"attachment_1411\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-1411\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174558\/5-6-4.jpeg\" alt=\"A triangular lamina is shown on the x y plane bounded by the x and y axes and the line x + y = 3. The point (1, 1) is marked and is surrounded by a small squared marked d m = p(x, y) dA.\" width=\"417\" height=\"347\" \/> Figure 4.\u00a0A lamina in the [latex]xy[\/latex]-plane with density\u00a0[latex]\\rho(x,y)=xy[\/latex].[\/caption]\r\n<p id=\"fs-id1167794049180\">Using the expression developed for mass, we see that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nm &amp; = {\\underset{R}{\\displaystyle\\iint}}{d}{m} = {\\underset{R}{\\displaystyle\\iint}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{xy} \\ {dy} \\ {dx} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\left [{x}{\\frac{y^2}{2}}{\\bigg\\vert}^{{y} = {3} - {x}}_{{y} = {0}}\\right ]}{dx} \\\\\r\n&amp; = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ \\ {\\frac{1}{2}}{x}{({3} - {x})}^{2}{dx} = {\\left [{\\frac{9x^2}{4}} - {{x}^{3}} + {\\frac{x^4}{8}} \\right]}{\\bigg|}^{{x} = {3}}_{{x} = {0}} \\\\\r\n&amp; = {\\frac{27}{8}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793803690\">The computation is straightforward, giving the answer [latex]{m} = {\\frac{27}{8}}[\/latex] kg.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Calculating Moments<\/h3>\r\nNow that we have the mass formula, we can develop expressions for moments. The <strong>moment<\/strong> [latex]M_x[\/latex] about the [latex]x[\/latex]-axis for region [latex]R[\/latex] is the limit of the sums of moments of the rectangles [latex]R_{ij}[\/latex] about the [latex]x[\/latex]-axis:\r\n<p style=\"text-align: center;\">[latex]{{M}_{x}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA}.[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Similarly, the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis is:<\/p>\r\n<p style=\"text-align: center;\">[latex]{{M}_{y}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({x}^{*}_{ij})}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA}.[\/latex]<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>moments of a lamina<\/h3>\r\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex]:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Moment about the [latex]x[\/latex]-axis:<\/strong> [latex]M_x = \\iint_R y\\rho(x,y) dA[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Moment about the [latex]y[\/latex]-axis:<\/strong> [latex]M_y = \\iint_R x\\rho(x,y) dA[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {xy}[\/latex]. Find the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex].[reveal-answer q=\"348723450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348723450\"]\r\n<p id=\"fs-id1167794118097\">Use double integrals for each moment and compute their values:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{M}_{x} = \\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{x}{{y}^{2}}{dy}{dx} = {\\frac{81}{20}}, \\\\\r\n{M}_{y} = \\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{{x}^{2}}{yd}{y}{dx} = {\\frac{81}{20}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167794324398\">The computation is quite straightforward.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Center of Mass<\/h3>\r\n<p class=\"whitespace-normal break-words\">We can now express the center of mass coordinates in terms of integrals. We denote the [latex]x[\/latex]-coordinate by [latex]\\overline{x}[\/latex] and the [latex]y[\/latex]-coordinate by [latex]\\overline{y}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\dfrac{M_y}{m} = \\dfrac{\\iint_R x\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA} \\quad \\text{and} \\quad \\overline{y} = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>center of mass formulas<\/h3>\r\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density [latex]\\rho(x,y)[\/latex], the center of mass is at [latex](\\overline{x}, \\overline{y})[\/latex] where:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\dfrac{M_y}{m} = \\dfrac{\\iint_R x\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\overline{y} = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Again consider the same triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density function [latex]{\\rho}{({x},{y})} = {xy}[\/latex]. Find the center of mass.[reveal-answer q=\"204975845\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204975845\"]\r\n<div id=\"fs-id1167793371150\" class=\" ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\"><section class=\"ui-body\" role=\"alert\">\r\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">Using the formulas we developed, we have<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n\r\n[latex]\\hspace{6cm}\\large{\\begin{align}\r\n\r\n\\overline{x}&amp;=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x\\rho(x,y) \\ dA}{\\underset{R}{\\displaystyle\\iint}\\rho(x,y) \\ dA}=\\frac{81\/20}{27\/8}=\\frac65 \\\\\r\n\r\n\\overline{y}&amp;=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y\\rho(x,y)dA}{\\underset{R}{\\displaystyle\\iint}\\rho(x,y) \\ dA}=\\frac{81\/20}{27\/8}=\\frac65\r\n\r\n\\end{align}}[\/latex]\r\n<p id=\"fs-id1167793879558\">Therefore, the center of mass is the point [latex]\\left(\\frac65,\\frac65\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1167794005633\" data-type=\"commentary\">\r\n<p id=\"9\" data-type=\"commentary-title\"><strong><span class=\"os-title-label\">Analysis<\/span><\/strong><\/p>\r\n<p id=\"fs-id1167793277849\">If we choose the density [latex]{\\rho}{({x},{y})}[\/latex] instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{{x}_{c}} = {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{x} \\ {dA}}{\\underset{R}{\\displaystyle\\iint}{dA}}} = {\\frac{9\/2}{9\/2}} = 1, \\\\\r\n{{y}_{c}} = {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{y} \\ {dA}}{\\underset{R}{\\displaystyle\\iint}{dA}}} = {\\frac{9\/2}{9\/2}} = 1.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167794040344\">Notice that the center of mass [latex]{\\left ({\\frac{6}{5}},{\\frac{6}{5}} \\right )}[\/latex] is not exactly the same as the centroid [latex](1, 1)[\/latex] of the triangular region. This is due to the variable density of [latex]R[\/latex]. If the density is constant, then we just use [latex]{\\rho}{({x},{y})} = {c}[\/latex] (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Centroid Formulas<\/h3>\r\nBased on the observation in the previous example, when density is constant (or we're simply finding the geometric center), we have simplified formulas for the <strong>centroid<\/strong> of a region:\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]x_c = \\frac{M_y}{m} = \\dfrac{\\iint_R x , dA}{\\iint_R dA} \\quad \\text{and} \\quad y_c = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y , dA}{\\iint_R dA}[\/latex]<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>centroid formulas<\/h3>\r\n<p class=\"whitespace-normal break-words\">For a region [latex]R[\/latex] in the plane, the centroid (geometric center) is at [latex](x_c, y_c)[\/latex] where:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]x_c = \\dfrac{\\iint_R x , dA}{\\iint_R dA} \\quad \\text{and} \\quad y_c = \\dfrac{\\iint_R y , dA}{\\iint_R dA}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Note: [latex]\\iint_R dA[\/latex] gives the area of region [latex]R[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Center of Mass vs. Centroid<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Centroid<\/strong>: Geometric center of a region (constant density)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Center of Mass<\/strong>: Balance point accounting for variable density<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">When density is uniform, these two points coincide!<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793264086\">Find the mass, moments, and the center of mass of the lamina of density [latex]{\\rho}{({x},{y})} = {x} + {y}[\/latex] occupying the region [latex]R[\/latex] under the curve [latex]y=x^{2}[\/latex] in the interval [latex]{0} \\leq {x} \\leq {2}[\/latex] (see the following figure).<\/p>\r\n\r\n[caption id=\"attachment_1412\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-1412\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174729\/5-6-5.jpeg\" alt=\"A lamina R is shown on the x y plane bounded by the x axis, the line x = 2, and the line y = x squared. The corners of the shape are (0, 0), (2, 0), and (2, 4).\" width=\"342\" height=\"459\" \/> Figure 5.\u00a0Locating the center of mass of a lamina\u00a0[latex]R[\/latex] with density\u00a0[latex]\\rho(x,y)=x+y[\/latex].[\/caption][reveal-answer q=\"324187023\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324187023\"]\r\n<p id=\"fs-id1167793453338\">First we compute the mass [latex]m[\/latex]. We need to describe the region between the graph of [latex]y=x^{2}[\/latex] and the vertical lines [latex]x=0[\/latex] and\u00a0[latex]x=2[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{m} &amp; = {\\underset{R}{\\displaystyle\\iint}}{dm} = {\\underset{R}{\\displaystyle\\iint}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{({x} + {y})}{dy} \\ {dx} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}}{\\left [{xy} + {\\frac{{y}^{2}}{2}}{\\bigg\\vert}^{{y} = {{x}^{2}}}_{{y} = {0}} \\right ]}{dx} \\\\\r\n&amp; = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\left [{{x}^{3}} + {\\frac{{x}^{4}}{2}} \\right ]}{dx} = {\\left [ {\\frac{{x}^{4}}{4}} + {\\frac{{x}^{5}}{10}} \\right ]}{\\bigg\\vert}^{{x} = {2}}_{{x} = {0}} = {\\frac{36}{5}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167794041327\">Now compute the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{M}_{x} = &amp; {\\underset{R}{\\displaystyle\\iint}}{y}{\\rho}{({x},{y})}{d}{A} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{y}{({x} + {y})}{dy} \\ {dx} = {\\frac{80}{7}}, \\\\\r\n{M}_{y} = &amp; {\\underset{R}{\\displaystyle\\iint}}{x}{\\rho}{({x},{y})}{d}{A} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{x}{({x} + {y})}{dy} \\ {dx} = {\\frac{176}{15}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793268216\">Finally, evaluate the center of mass,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\overline{x}} = &amp; {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint} \\ {x}{\\rho}{({x},{y})}{d}{A}}{\\underset{R}{\\displaystyle\\iint} \\ {\\rho}{({x},{y})}{d}{A}}} = {\\frac{176\/15}{36\/5}} = {\\frac{44}{27}}, \\\\\r\n{\\overline{y}} = &amp; {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint} \\ {y}{\\rho}{({x},{y})}{d}{A}}{\\underset{R}{\\displaystyle\\iint} \\ {\\rho}{({x},{y})}{d}{A}}} = {\\frac{80\/7}{36\/5}} = {\\frac{100}{63}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167794040056\">Hence the center of mass is [latex]{({\\overline{x}},{\\overline{y}})} = {\\left ({\\frac{44}{27}},{\\frac{100}{63}} \\right )}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the centroid of the region under the curve [latex]{y} = {{e}^{x}}[\/latex] over the interval [latex]{1} \\leq {x} \\leq {3}[\/latex] (see the following figure).[caption id=\"attachment_1413\" align=\"aligncenter\" width=\"379\"]<img class=\"size-full wp-image-1413\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174809\/5-6-6.jpeg\" alt=\"On the x y plane the curve y = e to the x is shown from x = 0 to x = 3 (3, e cubed). The points (1, 0) and (3, 0) are marked on the x axes. A dashed line rises from (1, 0) marked x = 1; similarly, a solid line rises from (3, 0) marked x = 3.\" width=\"379\" height=\"384\" \/> Figure 6.\u00a0Finding a centroid of a region below the curve\u00a0[latex]y=e^{x}[\/latex][\/caption][reveal-answer q=\"452983465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"452983465\"]\r\n<p id=\"fs-id1167793266369\">To compute the centroid, we assume that the density function is constant and hence it cancels out:<\/p>\r\n[latex]\\hspace{4cm}\\begin{align}\r\nx_c&amp;=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}\\text{ and }y_c=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}, \\\\\r\nx_c&amp;=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}=\\dfrac{\\displaystyle\\int^{x=3}_{x=1}\\displaystyle\\int^{y=e^x}_{y=0}x \\ dy \\ dx}{\\displaystyle\\int^{x=3}_{x=1}\\displaystyle\\int^{y=e^x}_{y=0}dy \\ dx}=\\dfrac{\\displaystyle\\int^{x=3}_{x=1}xe^x{dx}}{\\displaystyle\\int^{x=3}_{x=1}e^x{dx}}=\\frac{2e^3}{e^3-e}=\\frac{2e^2}{e^2-1}, \\\\\r\n\r\ny_c&amp;=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}=\\dfrac{\\displaystyle\\int_{x=1}^{x=3}\\displaystyle\\int_{y=0}^{y=e^x}y \\ dy \\ dx}{\\displaystyle\\int_{x=1}^{x=3}\\displaystyle\\int_{y=0}^{y=e^x}dy \\ dx}=\\dfrac{\\displaystyle\\int_{x=1}^{x=3}\\frac{e^{2x}}2dx}{\\displaystyle\\int_{x=1}^{x=3}e^xdx}=\\frac{\\frac14e^2(e^4-1)}{e(e^2-1)}=\\frac14e(e^2+1).\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167794228258\">Thus the centroid of the region is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{({{x}_{c}},{{y}_{c}})} = {\\left ( {\\frac{{2}{e}^{2}}{{e}^{2} - {1}}, {\\frac{1}{4}}{e} \\ {\\left ( {e}^{2} + {1} \\right )}} \\right )}}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find centers of mass in two dimensions<\/li>\n<li>Calculate moments of inertia in two dimensions<\/li>\n<li>Find centers of mass in three dimensions<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Center of Mass in Two Dimensions<\/h2>\n<p class=\"whitespace-normal break-words\">We&#8217;ve already used multiple integrals to find areas, volumes, and average values of functions over bounded regions. Now we&#8217;ll extend these techniques to find the center of mass and moments of inertia for physical objects. You&#8217;ll use double integrals for laminas (flat plates) and triple integrals for three-dimensional objects with variable density. When an object has uniform density throughout, we call it <strong>homogeneous<\/strong>. For homogeneous objects, the density is constant.<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>center of mass<\/strong> represents the balance point of an object. In a uniform gravitational field, this is also called the center of gravity. For objects with uniform density, the center of mass is simply the geometric center, which we call the <strong>centroid<\/strong>.<\/p>\n<p class=\"whitespace-normal break-words\">Figure 1 shows a lamina perfectly balanced at its center of mass [latex]P[\/latex].<\/p>\n<figure id=\"attachment_1407\" aria-describedby=\"caption-attachment-1407\" style=\"width: 452px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1407\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174357\/5-6-1.jpeg\" alt=\"A surface is delicately balanced on a fine point.\" width=\"452\" height=\"233\" \/><figcaption id=\"caption-attachment-1407\" class=\"wp-caption-text\">Figure 1.\u00a0A lamina is perfectly balanced on a spindle if the lamina\u2019s center of mass sits on the spindle.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">To find the coordinates of the center of mass [latex]P(\\overline{x}, \\overline{y})[\/latex] of a lamina, you need three quantities:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The moment [latex]M_x[\/latex] about the [latex]x[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">The moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">The mass [latex]m[\/latex] of the lamina<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Then the center of mass coordinates are:<\/p>\n<p style=\"text-align: center;\">[latex]{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}.[\/latex]<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Centers of Mass in One Dimension<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a one-dimensional object like a thin rod with density [latex]\\rho(x)[\/latex] along the [latex]x[\/latex]-axis from [latex]x = a[\/latex] to [latex]x = b[\/latex]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Mass: [latex]m = \\displaystyle\\int_a^b \\rho(x)dx[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Center of mass: [latex]\\overline{x} = \\frac{1}{m}\\displaystyle\\int_a^b x\\rho(x)dx[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">We&#8217;ll extend this same idea to two dimensions using double integrals.<\/p>\n<\/section>\n<p>When the density function is constant, these formulas give the <strong>centroid<\/strong> of the lamina\u2014the geometric center independent of mass distribution.<\/p>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Setting Up the Mass Calculation<\/h3>\n<p id=\"fs-id1167794052362\">Suppose that the lamina occupies a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane, with density [latex]{\\rho}{({{x},{y}})}[\/latex] (measured in mass per unit area) at any point [latex](x, y)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The density is defined as:<\/p>\n<p style=\"text-align: center;\">[latex]{\\rho}{({{x},{y}})} = {\\displaystyle\\lim_{{\\Delta}{A}{\\rightarrow}{0}}}{\\frac{{\\Delta}{m}}{{\\Delta}{A}}}[\/latex],<\/p>\n<p>where [latex]{\\Delta}{m}[\/latex] and [latex]{\\Delta}{A}[\/latex] are the mass and area of a small rectangle containing the point [latex](x, y)[\/latex], and the limit is taken as the rectangle&#8217;s dimensions approach zero (Figure 2).<\/p>\n<div id=\"CNX_Calc_Figure_15_06_002\" class=\"os-figure\">\n<figure id=\"attachment_1408\" aria-describedby=\"caption-attachment-1408\" style=\"width: 454px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1408\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174437\/5-6-2.jpeg\" alt=\"A lamina R is shown on the x y plane with a point (x, y) surrounded by a small rectangle marked Mass = Delta m and Area = Delta A.\" width=\"454\" height=\"384\" \/><figcaption id=\"caption-attachment-1408\" class=\"wp-caption-text\">Figure 2.\u00a0The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.<\/figcaption><\/figure>\n<\/div>\n<div><\/div>\n<div>To find the total mass, we partition region [latex]R[\/latex] into tiny rectangles [latex]R_{ij}[\/latex], each with area [latex]\\Delta A[\/latex]. We choose sample points [latex]{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}[\/latex] in each rectangle. The mass [latex]m_{ij}[\/latex] of rectangle [latex]R_{ij}[\/latex] is approximately:<\/div>\n<div style=\"text-align: center;\">\u00a0[latex]m_{ij} ={\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex] (Figure 3).<\/div>\n<div>Let [latex]k[\/latex] and [latex]l[\/latex] be the number of subintervals in [latex]x[\/latex] and [latex]y[\/latex] respectively (Figure 3). Note that while we use rectangles for convenience, this limiting process works for any shape, as you&#8217;ve seen in previous sections.<\/div>\n<div>\n<figure id=\"attachment_1410\" aria-describedby=\"caption-attachment-1410\" style=\"width: 454px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1410\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174515\/5-6-3.jpeg\" alt=\"A lamina is shown on the x y plane with a point (x* sub ij, y* sub ij) surrounded by a small rectangle marked R sub ij.\" width=\"454\" height=\"384\" \/><figcaption id=\"caption-attachment-1410\" class=\"wp-caption-text\">Figure 3. Subdividing the lamina into tiny rectangles\u00a0[latex]R_{ij}[\/latex], each containing a sample point\u00a0[latex](x_{ij}^{\\ast},y_{ij}^{\\ast})[\/latex].<\/figcaption><\/figure>\n<\/div>\n<div><\/div>\n<div>Taking the limit as the partition becomes infinitely fine gives us the total mass:<\/div>\n<div><\/div>\n<div id=\"fs-id1167793844477\" data-type=\"equation\">\n<div class=\"os-equation-number\" style=\"text-align: center;\">[latex]{m} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}.[\/latex]<\/div>\n<\/div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>mass of a lamina<\/h3>\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex], the total mass is:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]m = \\iint_R \\rho(x,y) dA[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Let&#8217;s see how this works with a concrete example.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Consider a triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {{x}{y}} \\ {\\text{kg\/m}^2}[\/latex]. Find the total mass.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q547923465\">Show Solution<\/button><\/p>\n<div id=\"q547923465\" class=\"hidden-answer\" style=\"display: none\">A sketch of the region [latex]R[\/latex] is always helpful, as shown in the following figure.<\/p>\n<figure id=\"attachment_1411\" aria-describedby=\"caption-attachment-1411\" style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1411\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174558\/5-6-4.jpeg\" alt=\"A triangular lamina is shown on the x y plane bounded by the x and y axes and the line x + y = 3. The point (1, 1) is marked and is surrounded by a small squared marked d m = p(x, y) dA.\" width=\"417\" height=\"347\" \/><figcaption id=\"caption-attachment-1411\" class=\"wp-caption-text\">Figure 4.\u00a0A lamina in the [latex]xy[\/latex]-plane with density\u00a0[latex]\\rho(x,y)=xy[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1167794049180\">Using the expression developed for mass, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  m & = {\\underset{R}{\\displaystyle\\iint}}{d}{m} = {\\underset{R}{\\displaystyle\\iint}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{xy} \\ {dy} \\ {dx} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\left [{x}{\\frac{y^2}{2}}{\\bigg\\vert}^{{y} = {3} - {x}}_{{y} = {0}}\\right ]}{dx} \\\\  & = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ \\ {\\frac{1}{2}}{x}{({3} - {x})}^{2}{dx} = {\\left [{\\frac{9x^2}{4}} - {{x}^{3}} + {\\frac{x^4}{8}} \\right]}{\\bigg|}^{{x} = {3}}_{{x} = {0}} \\\\  & = {\\frac{27}{8}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793803690\">The computation is straightforward, giving the answer [latex]{m} = {\\frac{27}{8}}[\/latex] kg.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Calculating Moments<\/h3>\n<p>Now that we have the mass formula, we can develop expressions for moments. The <strong>moment<\/strong> [latex]M_x[\/latex] about the [latex]x[\/latex]-axis for region [latex]R[\/latex] is the limit of the sums of moments of the rectangles [latex]R_{ij}[\/latex] about the [latex]x[\/latex]-axis:<\/p>\n<p style=\"text-align: center;\">[latex]{{M}_{x}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA}.[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Similarly, the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis is:<\/p>\n<p style=\"text-align: center;\">[latex]{{M}_{y}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({x}^{*}_{ij})}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = \\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA}.[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>moments of a lamina<\/h3>\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex]:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Moment about the [latex]x[\/latex]-axis:<\/strong> [latex]M_x = \\iint_R y\\rho(x,y) dA[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Moment about the [latex]y[\/latex]-axis:<\/strong> [latex]M_y = \\iint_R x\\rho(x,y) dA[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {xy}[\/latex]. Find the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q348723450\">Show Solution<\/button><\/p>\n<div id=\"q348723450\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794118097\">Use double integrals for each moment and compute their values:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {M}_{x} = \\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{x}{{y}^{2}}{dy}{dx} = {\\frac{81}{20}}, \\\\  {M}_{y} = \\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {3}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {3} - {x}}_{{y} = {0}}}{{x}^{2}}{yd}{y}{dx} = {\\frac{81}{20}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167794324398\">The computation is quite straightforward.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Center of Mass<\/h3>\n<p class=\"whitespace-normal break-words\">We can now express the center of mass coordinates in terms of integrals. We denote the [latex]x[\/latex]-coordinate by [latex]\\overline{x}[\/latex] and the [latex]y[\/latex]-coordinate by [latex]\\overline{y}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\dfrac{M_y}{m} = \\dfrac{\\iint_R x\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA} \\quad \\text{and} \\quad \\overline{y} = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>center of mass formulas<\/h3>\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density [latex]\\rho(x,y)[\/latex], the center of mass is at [latex](\\overline{x}, \\overline{y})[\/latex] where:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\dfrac{M_y}{m} = \\dfrac{\\iint_R x\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\overline{y} = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y\\rho(x,y) dA}{\\iint_R \\rho(x,y) dA}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Again consider the same triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](0, 3)[\/latex], [latex](3, 0)[\/latex] and with density function [latex]{\\rho}{({x},{y})} = {xy}[\/latex]. Find the center of mass.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q204975845\">Show Solution<\/button><\/p>\n<div id=\"q204975845\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793371150\" class=\"ui-solution-visible\" data-type=\"solution\" aria-label=\"hide solution\" aria-expanded=\"true\">\n<section class=\"ui-body\" role=\"alert\">\n<h2 data-type=\"solution-title\"><span style=\"font-size: 16px; font-weight: 400;\">Using the formulas we developed, we have<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p>[latex]\\hspace{6cm}\\large{\\begin{align}    \\overline{x}&=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x\\rho(x,y) \\ dA}{\\underset{R}{\\displaystyle\\iint}\\rho(x,y) \\ dA}=\\frac{81\/20}{27\/8}=\\frac65 \\\\    \\overline{y}&=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y\\rho(x,y)dA}{\\underset{R}{\\displaystyle\\iint}\\rho(x,y) \\ dA}=\\frac{81\/20}{27\/8}=\\frac65    \\end{align}}[\/latex]<\/p>\n<p id=\"fs-id1167793879558\">Therefore, the center of mass is the point [latex]\\left(\\frac65,\\frac65\\right)[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1167794005633\" data-type=\"commentary\">\n<p id=\"9\" data-type=\"commentary-title\"><strong><span class=\"os-title-label\">Analysis<\/span><\/strong><\/p>\n<p id=\"fs-id1167793277849\">If we choose the density [latex]{\\rho}{({x},{y})}[\/latex] instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {{x}_{c}} = {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{x} \\ {dA}}{\\underset{R}{\\displaystyle\\iint}{dA}}} = {\\frac{9\/2}{9\/2}} = 1, \\\\  {{y}_{c}} = {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{y} \\ {dA}}{\\underset{R}{\\displaystyle\\iint}{dA}}} = {\\frac{9\/2}{9\/2}} = 1.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167794040344\">Notice that the center of mass [latex]{\\left ({\\frac{6}{5}},{\\frac{6}{5}} \\right )}[\/latex] is not exactly the same as the centroid [latex](1, 1)[\/latex] of the triangular region. This is due to the variable density of [latex]R[\/latex]. If the density is constant, then we just use [latex]{\\rho}{({x},{y})} = {c}[\/latex] (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3>Centroid Formulas<\/h3>\n<p>Based on the observation in the previous example, when density is constant (or we&#8217;re simply finding the geometric center), we have simplified formulas for the <strong>centroid<\/strong> of a region:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]x_c = \\frac{M_y}{m} = \\dfrac{\\iint_R x , dA}{\\iint_R dA} \\quad \\text{and} \\quad y_c = \\dfrac{M_x}{m} = \\dfrac{\\iint_R y , dA}{\\iint_R dA}[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>centroid formulas<\/h3>\n<p class=\"whitespace-normal break-words\">For a region [latex]R[\/latex] in the plane, the centroid (geometric center) is at [latex](x_c, y_c)[\/latex] where:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]x_c = \\dfrac{\\iint_R x , dA}{\\iint_R dA} \\quad \\text{and} \\quad y_c = \\dfrac{\\iint_R y , dA}{\\iint_R dA}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Note: [latex]\\iint_R dA[\/latex] gives the area of region [latex]R[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Center of Mass vs. Centroid<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Centroid<\/strong>: Geometric center of a region (constant density)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Center of Mass<\/strong>: Balance point accounting for variable density<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">When density is uniform, these two points coincide!<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793264086\">Find the mass, moments, and the center of mass of the lamina of density [latex]{\\rho}{({x},{y})} = {x} + {y}[\/latex] occupying the region [latex]R[\/latex] under the curve [latex]y=x^{2}[\/latex] in the interval [latex]{0} \\leq {x} \\leq {2}[\/latex] (see the following figure).<\/p>\n<figure id=\"attachment_1412\" aria-describedby=\"caption-attachment-1412\" style=\"width: 342px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1412\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174729\/5-6-5.jpeg\" alt=\"A lamina R is shown on the x y plane bounded by the x axis, the line x = 2, and the line y = x squared. The corners of the shape are (0, 0), (2, 0), and (2, 4).\" width=\"342\" height=\"459\" \/><figcaption id=\"caption-attachment-1412\" class=\"wp-caption-text\">Figure 5.\u00a0Locating the center of mass of a lamina\u00a0[latex]R[\/latex] with density\u00a0[latex]\\rho(x,y)=x+y[\/latex].<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q324187023\">Show Solution<\/button><\/p>\n<div id=\"q324187023\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793453338\">First we compute the mass [latex]m[\/latex]. We need to describe the region between the graph of [latex]y=x^{2}[\/latex] and the vertical lines [latex]x=0[\/latex] and\u00a0[latex]x=2[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {m} & = {\\underset{R}{\\displaystyle\\iint}}{dm} = {\\underset{R}{\\displaystyle\\iint}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{({x} + {y})}{dy} \\ {dx} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}}{\\left [{xy} + {\\frac{{y}^{2}}{2}}{\\bigg\\vert}^{{y} = {{x}^{2}}}_{{y} = {0}} \\right ]}{dx} \\\\  & = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\left [{{x}^{3}} + {\\frac{{x}^{4}}{2}} \\right ]}{dx} = {\\left [ {\\frac{{x}^{4}}{4}} + {\\frac{{x}^{5}}{10}} \\right ]}{\\bigg\\vert}^{{x} = {2}}_{{x} = {0}} = {\\frac{36}{5}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167794041327\">Now compute the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {M}_{x} = & {\\underset{R}{\\displaystyle\\iint}}{y}{\\rho}{({x},{y})}{d}{A} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{y}{({x} + {y})}{dy} \\ {dx} = {\\frac{80}{7}}, \\\\  {M}_{y} = & {\\underset{R}{\\displaystyle\\iint}}{x}{\\rho}{({x},{y})}{d}{A} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{x}{({x} + {y})}{dy} \\ {dx} = {\\frac{176}{15}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793268216\">Finally, evaluate the center of mass,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\overline{x}} = & {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint} \\ {x}{\\rho}{({x},{y})}{d}{A}}{\\underset{R}{\\displaystyle\\iint} \\ {\\rho}{({x},{y})}{d}{A}}} = {\\frac{176\/15}{36\/5}} = {\\frac{44}{27}}, \\\\  {\\overline{y}} = & {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint} \\ {y}{\\rho}{({x},{y})}{d}{A}}{\\underset{R}{\\displaystyle\\iint} \\ {\\rho}{({x},{y})}{d}{A}}} = {\\frac{80\/7}{36\/5}} = {\\frac{100}{63}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167794040056\">Hence the center of mass is [latex]{({\\overline{x}},{\\overline{y}})} = {\\left ({\\frac{44}{27}},{\\frac{100}{63}} \\right )}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the centroid of the region under the curve [latex]{y} = {{e}^{x}}[\/latex] over the interval [latex]{1} \\leq {x} \\leq {3}[\/latex] (see the following figure).<\/p>\n<figure id=\"attachment_1413\" aria-describedby=\"caption-attachment-1413\" style=\"width: 379px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1413\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174809\/5-6-6.jpeg\" alt=\"On the x y plane the curve y = e to the x is shown from x = 0 to x = 3 (3, e cubed). The points (1, 0) and (3, 0) are marked on the x axes. A dashed line rises from (1, 0) marked x = 1; similarly, a solid line rises from (3, 0) marked x = 3.\" width=\"379\" height=\"384\" \/><figcaption id=\"caption-attachment-1413\" class=\"wp-caption-text\">Figure 6.\u00a0Finding a centroid of a region below the curve\u00a0[latex]y=e^{x}[\/latex]<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q452983465\">Show Solution<\/button><\/p>\n<div id=\"q452983465\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793266369\">To compute the centroid, we assume that the density function is constant and hence it cancels out:<\/p>\n<p>[latex]\\hspace{4cm}\\begin{align}  x_c&=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}\\text{ and }y_c=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}, \\\\  x_c&=\\frac{M_y}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}x \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}=\\dfrac{\\displaystyle\\int^{x=3}_{x=1}\\displaystyle\\int^{y=e^x}_{y=0}x \\ dy \\ dx}{\\displaystyle\\int^{x=3}_{x=1}\\displaystyle\\int^{y=e^x}_{y=0}dy \\ dx}=\\dfrac{\\displaystyle\\int^{x=3}_{x=1}xe^x{dx}}{\\displaystyle\\int^{x=3}_{x=1}e^x{dx}}=\\frac{2e^3}{e^3-e}=\\frac{2e^2}{e^2-1}, \\\\    y_c&=\\frac{M_x}m=\\dfrac{\\underset{R}{\\displaystyle\\iint}y \\ dA}{\\underset{R}{\\displaystyle\\iint}dA}=\\dfrac{\\displaystyle\\int_{x=1}^{x=3}\\displaystyle\\int_{y=0}^{y=e^x}y \\ dy \\ dx}{\\displaystyle\\int_{x=1}^{x=3}\\displaystyle\\int_{y=0}^{y=e^x}dy \\ dx}=\\dfrac{\\displaystyle\\int_{x=1}^{x=3}\\frac{e^{2x}}2dx}{\\displaystyle\\int_{x=1}^{x=3}e^xdx}=\\frac{\\frac14e^2(e^4-1)}{e(e^2-1)}=\\frac14e(e^2+1).  \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167794228258\">Thus the centroid of the region is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{({{x}_{c}},{{y}_{c}})} = {\\left ( {\\frac{{2}{e}^{2}}{{e}^{2} - {1}}, {\\frac{1}{4}}{e} \\ {\\left ( {e}^{2} + {1} \\right )}} \\right )}}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-719","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"learn_it","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/719","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/719\/revisions"}],"predecessor-version":[{"id":1947,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/719\/revisions\/1947"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/719\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=719"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=719"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=719"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=719"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}