{"id":3996,"date":"2022-04-12T18:08:52","date_gmt":"2022-04-12T18:08:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3996"},"modified":"2022-11-01T04:37:00","modified_gmt":"2022-11-01T04:37:00","slug":"center-of-mass-and-moments-of-inertia-in-two-dimensions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/center-of-mass-and-moments-of-inertia-in-two-dimensions\/","title":{"raw":"Center of Mass and Moments of Inertia in Two Dimensions","rendered":"Center of Mass and Moments of Inertia in Two Dimensions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3 style=\"text-align: center;\">Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Use double integrals to locate the center of mass of a two-dimensional object.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Use double integrals to find the moment of inertia of a two-dimensional object.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Center of Mass in Two Dimensions<\/h2>\r\n<p id=\"fs-id1167794064730\">The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid.\u00a0Figure 1\u00a0shows a point [latex]P[\/latex] as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.<\/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\r\nTo find the coordinates of the center of mass [latex]P{({\\overline{x}},{\\overline{y}})}[\/latex] of a lamina, we need to find the moment [latex]M_x[\/latex] of the lamina about the [latex]x[\/latex]-axis and the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis. We also need to find the mass [latex]m[\/latex] of the lamina. Then\r\n<p style=\"text-align: center;\">[latex]{\\large{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793393775\">Refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/center-of-mass-and-moments\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"2-6-moments-and-centers-of-mass\">Centers of Mass and Moments<\/a>\u00a0for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.<\/p>\r\n<p id=\"fs-id1167793369776\">If we allow a constant density function, then [latex]{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}[\/latex] give the\u00a0<em data-effect=\"italics\">centroid<\/em>\u00a0of the lamina.<\/p>\r\n<p id=\"fs-id1167794052362\">Suppose that the lamina occupies a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane, and let [latex]{\\rho}{({{x},{y}})}[\/latex] be its density (in units of mass per unit area) at any point [latex](x, y)[\/latex]. Hence, [latex]{\\rho}{({{x},{y}})} = {\\displaystyle\\lim_{{\\Delta}{A}{\\rightarrow}{0}}}{\\frac{{\\Delta}{m}}{{\\Delta}{A}}}[\/latex], 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 dimensions of the rectangle go to 0 (see the following figure).<\/p>\r\n\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>Just as before, we divide the region [latex]R[\/latex] into tiny rectangles [latex]{R}_{ij}[\/latex] with area [latex]{\\Delta}{A}[\/latex] and choose [latex]{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}[\/latex] as sample points. Then the mass [latex]{m}_{ij}[\/latex] of each [latex]{R}_{ij}[\/latex] is equal to [latex]{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex] (Figure 3). Let [latex]k[\/latex] and[latex]l[\/latex] be the number of subintervals in [latex]x[\/latex] and\u00a0[latex]y[\/latex], respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.<\/div>\r\n<div><\/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>Hence, the mass of the lamina is<\/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]\\large{{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<p id=\"fs-id1167793250849\">Let\u2019s see an example now of finding the total mass of a triangular lamina.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the total mass of a lamina<\/h3>\r\nConsider 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.\r\n\r\n[reveal-answer q=\"547923465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547923465\"]\r\n\r\nA sketch of the region [latex]R[\/latex] is always helpful, as shown in the following figure.\r\n\r\n[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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nConsider the same region [latex]R[\/latex] as in the previous example, and use the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the total mass.\u00a0<em data-effect=\"italics\">Hint:<\/em>\u00a0Use trigonometric substitution [latex]{\\sqrt{x}} = {\\sqrt{3}}\\sin{\\theta}[\/latex] and then use the power reducing formulas for trigonometric functions.\r\n\r\n[reveal-answer q=\"470317685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470317685\"]\r\n\r\n[latex]\\frac{9\\pi}8[\/latex] kg.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793444617\">Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment [latex]M_x[\/latex] about the [latex]x[\/latex]-axis for [latex]R[\/latex] is the limit of the sums of moments of the regions [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis. Hence<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{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\nSimilarly, the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis for [latex]R[\/latex] is the limit of the sums of moments of the regions [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis. Hence\r\n<p style=\"text-align: center;\">[latex]\\large{{{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<div class=\"textbox exercises\">\r\n<h3>Example: finding moments<\/h3>\r\nConsider 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].\r\n\r\n[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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nConsider the same lamina [latex]R[\/latex] as above, and use the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex].\r\n\r\n[reveal-answer q=\"900563247\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"900563247\"]\r\n\r\n[latex]M_x=\\frac{81\\pi}{64}[\/latex] and\u00a0[latex]M_y=\\frac{81\\pi}{64}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793240540\">Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the [latex]x[\/latex]-coordinate of the center of mass by [latex]{\\overline{x}}[\/latex] and the [latex]y[\/latex]-coordinate by [latex]{\\overline{y}}[\/latex]. Specifically,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\overline{x}} = {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA}}{\\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA}}{\\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}}}}.[\/latex]<\/p>\r\n\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the center of mass<\/h3>\r\nAgain 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.\r\n\r\n[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 class=\"os-title-label\">Solution<\/span><\/h2>\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1167794136665\">Using the formulas we developed, we have<\/p>\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<h2 id=\"9\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nAgain use the same region [latex]R[\/latex] as above and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the center of mass.\r\n\r\n[reveal-answer q=\"384264209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384264209\"]\r\n\r\n[latex]\\overline{x}=\\frac{M_y}m=\\frac{81\\pi\/63}{9\\pi\/8}=\\frac98[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{81\\pi\/63}{9\\pi\/8}=\\frac98[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793462607\">Once again, based on the comments at the end of\u00a0Example \"Finding the Center of Mass\",\u00a0we have expressions for the centroid of a region on the plane:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{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}}}}.[\/latex]<\/p>\r\nWe should use these formulas and verify the centroid of the triangular region [latex]R[\/latex] referred to in the last three examples.\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding Mass, Moments, and Center of Mass<\/h3>\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the mass, moments, and the center of mass of the region between the curves [latex]{y} = {x}[\/latex] and [latex]{y} = {{x}^{2}}[\/latex] with the density function [latex]{\\rho}{({x},{y})} = {x}[\/latex] in the interval [latex]{0} \\leq {x} \\leq {1}[\/latex].\r\n\r\n[reveal-answer q=\"130487486\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130487486\"]\r\n\r\n[latex]\\overline{x}=\\frac{M_y}m=\\frac{1\/20}{1\/12}=\\frac35[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{1\/24}{1\/12}=\\frac12[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197110&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=P_3DRbSem_w&amp;video_target=tpm-plugin-pdh1flyj-P_3DRbSem_w\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.36_transcript.html\">transcript for \u201cCP 5.36\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a centroid<\/h3>\r\nFind 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).\r\n\r\n[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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the centroid of the region between the curves [latex]y=x[\/latex] and [latex]{y} = {\\sqrt{x}}[\/latex] with uniform density in the interval [latex]{0} \\leq {x} \\leq {1}[\/latex].\r\n\r\n[reveal-answer q=\"034759837\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"034759837\"]\r\n\r\n[latex]\\overline{x}=\\frac{M_y}m=\\frac{1\/15}{1\/6}=\\frac25[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{1\/12}{1\/6}=\\frac12[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Moments of Inertia<\/h2>\r\n<p id=\"fs-id1167793805130\">For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. The moment of inertia of a particle of mass [latex]m[\/latex] about an axis is [latex]mr^{2}[\/latex] where [latex]r[\/latex] is the distance of the particle from the axis. We can see from\u00a0Figure 3\u00a0that the moment of inertia of the subrectangle [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis is [latex]{({y}^{*}_{ij})^{2}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex]. Similarly, the moment of inertia of the subrectangle [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis is [latex]{({x}^{*}_{ij})^{2}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex]. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.<\/p>\r\n<p id=\"fs-id1167794327004\">The moment of inertia [latex]I_x[\/latex] about the [latex]x[\/latex]-axis for the region [latex]R[\/latex] is the limit of the sum of moments of inertia of the regions [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis. Hence<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{x}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {y}^{*}_{ij} \\right )}^{2}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {y}^{*}_{ij} \\right )}^{2}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = {\\underset{R}{\\displaystyle\\iint}}{{y}^{2}}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\r\nSimilarly, the moment of inertia [latex]I_y[\/latex] about the [latex]y[\/latex]-axis for [latex]R[\/latex] is the limit of the sum of moments of inertia of the regions [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis. Hence\r\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{y}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {x}^{*}_{ij} \\right )}^{2}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {x}^{*}_{ij} \\right )}^{2}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = {\\underset{R}{\\displaystyle\\iint}}{{x}^{2}}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793219176\">Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by [latex]I_0[\/latex] and obtain it by adding the moments of inertia [latex]I_x[\/latex] and [latex]I_y[\/latex]. Hence<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{0}} = {{I}_{x}} + {{I}_{y}} = {\\underset{R}{\\displaystyle\\iint}}{({x^{2}} + {y^{2}})}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793883659\">All these expressions can be written in polar coordinates by substituting [latex]{x} = {r} \\ {\\cos} \\ {\\theta}, {y} = {r} \\ {\\sin} \\ {\\theta},[\/latex] and [latex]{dA} = {r} \\ {dr} \\ {d}{\\theta}[\/latex]. For example, [latex]{{I}_{0}} = {\\underset{R}{\\displaystyle\\iint}}{{r}^{2}}{\\rho}{({{r}{\\cos}{\\theta}},{{r}{\\sin}{\\theta}})}{dA}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding moments of inertia for a triangular lamina<\/h3>\r\nUse the triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {x}{y}[\/latex] as in previous examples. Find the moments of inertia.\r\n\r\n[reveal-answer q=\"567926450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567926450\"]\r\n<p id=\"fs-id1167793419956\">Using the expressions established above for the moments of inertia, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{{I}_{x}} = &amp; {\\underset{R}{\\displaystyle\\iint}}{{y}^{2}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {x}}_{{y} = {0}}}{x}{{y}^{3}}{dy} \\ {dx} = {\\frac{8}{3}}, \\\\\r\n{{I}_{y}} = &amp; {\\underset{R}{\\displaystyle\\iint}}{{x}^{2}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {x}}_{{y} = {0}}}{{x}^{3}}{y} \\ {dy} \\ {dx} = {\\frac{16}{3}}, \\\\\r\n{{I}_{0}} = &amp; {\\underset{R}{\\displaystyle\\iint}}{({{x}^{2}} + {{y}^{2}})} \\ {\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{2}_{0}}{\\displaystyle\\int^{x}_{0}}{({{x}^{2}} + {{y}^{2}})} \\ {xy} \\ {dy} \\ {dx} \\\\\r\n= &amp; {{I}_{x}} + {{I}_{y}} = {8}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nAgain use the same region [latex]R[\/latex] as above and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the moments of inertia.\r\n\r\n[reveal-answer q=\"083745928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"083745928\"]\r\n\r\n[latex]I_x=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}y^2\\sqrt{xy}dy \\ dx = \\frac{64}{35}[\/latex] and\u00a0[latex]I_y=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}x^2\\sqrt{xy}dy \\ dx = \\frac{64}{15}[\/latex]. Also,\u00a0[latex]I_0=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}(x^2+y^2)\\sqrt{xy}dy \\ dx = \\frac{128}{21}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197111&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=H3dKuZxGpxw&amp;video_target=tpm-plugin-obobsoz2-H3dKuZxGpxw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.38_transcript.html\">transcript for \u201cCP 5.38\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167793432456\">As mentioned earlier, the moment of inertia of a particle of mass [latex]m[\/latex] about an axis is [latex]mr^{2}[\/latex] where [latex]r[\/latex] is the distance of the particle from the axis, also known as the\u00a0<strong><span id=\"43fd2933-285f-4a11-85d9-8913abc213df_term224\" data-type=\"term\">radius of gyration<\/span>.<\/strong><\/p>\r\n<p id=\"fs-id1167793280692\">Hence the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{R}_{x}} = {\\sqrt{\\frac{{I}_{x}}{m}}}, {{R}_{y}} = {\\sqrt{\\frac{{I}_{y}}{m}}}, \\ {\\text{and}} \\ {{R}_{0}} = {\\sqrt{\\frac{{I}_{0}}{m}}}}, [\/latex]<\/p>\r\n<p id=\"fs-id1167793276987\">respectively. In each case, the radius of gyration tells us how far (perpendicular distance) from the axis of rotation the entire mass of an object might be concentrated. The moments of an object are useful for finding information on the balance and torque of the object about an axis, but radii of gyration are used to describe the distribution of mass around its centroidal axis. There are many applications in engineering and physics. Sometimes it is necessary to find the radius of gyration, as in the next example.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the radius of gyration for a triangular lamina<\/h3>\r\nConsider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {x}{y}[\/latex] as in previous examples. Find the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin.\r\n\r\n[reveal-answer q=\"634752734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634752734\"]\r\n<p id=\"fs-id1167794145591\">If we compute the mass of this region we find that [latex]m=2[\/latex]. We found the moments of inertia of this lamina in\u00a0Example \"Finding Mass, Moments, and Center of Mass\". From these data, the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin are, respectively,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{{R}_{x}} = &amp; {\\sqrt{\\frac{{I}_{x}}{m}}} = {\\sqrt{\\frac{8\/3}{2}}} = {\\sqrt{\\frac{8}{6}}} = {\\frac{{2}{\\sqrt{3}}}{3}}, \\\\\r\n{{R}_{y}} = &amp; {\\sqrt{\\frac{{I}_{y}}{m}}} = {\\sqrt{\\frac{16\/3}{2}}} = {\\sqrt{\\frac{8}{3}}} = {\\frac{{2}{\\sqrt{6}}}{3}}, \\\\\r\n{{R}_{0}} = &amp; {\\sqrt{\\frac{{I}_{0}}{m}}} = {\\sqrt{\\frac{8}{2}}} = {\\sqrt{4}} = {2}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse the same region [latex]R[\/latex] from\u00a0Example \"Finding the Radius of Gyration for a Triangular Lamina\" and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin.\r\n\r\n[reveal-answer q=\"495289348\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"495289348\"]\r\n\r\n[latex]R_x=\\frac{6\\sqrt{35}}{35}[\/latex],\u00a0[latex]R_y=\\frac{6\\sqrt{15}}{15}[\/latex], and\u00a0[latex]R_0=\\frac{4\\sqrt{42}}{7}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3 style=\"text-align: center;\">Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Use double integrals to locate the center of mass of a two-dimensional object.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Use double integrals to find the moment of inertia of a two-dimensional object.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Center of Mass in Two Dimensions<\/h2>\n<p id=\"fs-id1167794064730\">The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid.\u00a0Figure 1\u00a0shows a point [latex]P[\/latex] as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.<\/p>\n<div id=\"attachment_1407\" style=\"width: 462px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1407\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<p>To find the coordinates of the center of mass [latex]P{({\\overline{x}},{\\overline{y}})}[\/latex] of a lamina, we need to find the moment [latex]M_x[\/latex] of the lamina about the [latex]x[\/latex]-axis and the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis. We also need to find the mass [latex]m[\/latex] of the lamina. Then<\/p>\n<p style=\"text-align: center;\">[latex]{\\large{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}}.[\/latex]<\/p>\n<p id=\"fs-id1167793393775\">Refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/center-of-mass-and-moments\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"2-6-moments-and-centers-of-mass\">Centers of Mass and Moments<\/a>\u00a0for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.<\/p>\n<p id=\"fs-id1167793369776\">If we allow a constant density function, then [latex]{\\overline{x}} = {\\frac{{M}_{y}}{m}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}}[\/latex] give the\u00a0<em data-effect=\"italics\">centroid<\/em>\u00a0of the lamina.<\/p>\n<p id=\"fs-id1167794052362\">Suppose that the lamina occupies a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane, and let [latex]{\\rho}{({{x},{y}})}[\/latex] be its density (in units of mass per unit area) at any point [latex](x, y)[\/latex]. Hence, [latex]{\\rho}{({{x},{y}})} = {\\displaystyle\\lim_{{\\Delta}{A}{\\rightarrow}{0}}}{\\frac{{\\Delta}{m}}{{\\Delta}{A}}}[\/latex], 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 dimensions of the rectangle go to 0 (see the following figure).<\/p>\n<div id=\"CNX_Calc_Figure_15_06_002\" class=\"os-figure\">\n<div id=\"attachment_1408\" style=\"width: 464px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1408\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div>Just as before, we divide the region [latex]R[\/latex] into tiny rectangles [latex]{R}_{ij}[\/latex] with area [latex]{\\Delta}{A}[\/latex] and choose [latex]{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}[\/latex] as sample points. Then the mass [latex]{m}_{ij}[\/latex] of each [latex]{R}_{ij}[\/latex] is equal to [latex]{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex] (Figure 3). Let [latex]k[\/latex] and[latex]l[\/latex] be the number of subintervals in [latex]x[\/latex] and\u00a0[latex]y[\/latex], respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.<\/div>\n<div><\/div>\n<div>\n<div id=\"attachment_1410\" style=\"width: 464px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1410\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div>Hence, the mass of the lamina is<\/div>\n<div><\/div>\n<div id=\"fs-id1167793844477\" data-type=\"equation\">\n<div class=\"os-equation-number\" style=\"text-align: center;\">[latex]\\large{{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<p id=\"fs-id1167793250849\">Let\u2019s see an example now of finding the total mass of a triangular lamina.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the total mass of a lamina<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q547923465\">Show Solution<\/span><\/p>\n<div id=\"q547923465\" class=\"hidden-answer\" style=\"display: none\">\n<p>A sketch of the region [latex]R[\/latex] is always helpful, as shown in the following figure.<\/p>\n<div id=\"attachment_1411\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1411\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Consider the same region [latex]R[\/latex] as in the previous example, and use the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the total mass.\u00a0<em data-effect=\"italics\">Hint:<\/em>\u00a0Use trigonometric substitution [latex]{\\sqrt{x}} = {\\sqrt{3}}\\sin{\\theta}[\/latex] and then use the power reducing formulas for trigonometric functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q470317685\">Show Solution<\/span><\/p>\n<div id=\"q470317685\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{9\\pi}8[\/latex] kg.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793444617\">Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment [latex]M_x[\/latex] about the [latex]x[\/latex]-axis for [latex]R[\/latex] is the limit of the sums of moments of the regions [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis. Hence<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{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>Similarly, the moment [latex]M_y[\/latex] about the [latex]y[\/latex]-axis for [latex]R[\/latex] is the limit of the sums of moments of the regions [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis. Hence<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{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<div class=\"textbox exercises\">\n<h3>Example: finding moments<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348723450\">Show Solution<\/span><\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Consider the same lamina [latex]R[\/latex] as above, and use the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the moments [latex]M_x[\/latex] and\u00a0[latex]M_y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q900563247\">Show Solution<\/span><\/p>\n<div id=\"q900563247\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]M_x=\\frac{81\\pi}{64}[\/latex] and\u00a0[latex]M_y=\\frac{81\\pi}{64}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793240540\">Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the [latex]x[\/latex]-coordinate of the center of mass by [latex]{\\overline{x}}[\/latex] and the [latex]y[\/latex]-coordinate by [latex]{\\overline{y}}[\/latex]. Specifically,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\overline{x}} = {\\frac{{M}_{y}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{x}{\\rho}{({x},{y})}{dA}}{\\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}}} \\ {\\text{and}} \\ {\\overline{y}} = {\\frac{{M}_{x}}{m}} = {\\dfrac{\\underset{R}{\\displaystyle\\iint}{y}{\\rho}{({x},{y})}{dA}}{\\underset{R}{\\displaystyle\\iint}{\\rho}{({x},{y})}{dA}}}}.[\/latex]<\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the center of mass<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204975845\">Show Solution<\/span><\/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 class=\"os-title-label\">Solution<\/span><\/h2>\n<div class=\"os-solution-container\">\n<p id=\"fs-id1167794136665\">Using the formulas we developed, we have<\/p>\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<h2 id=\"9\" data-type=\"commentary-title\"><span class=\"os-title-label\">Analysis<\/span><\/h2>\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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Again use the same region [latex]R[\/latex] as above and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the center of mass.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384264209\">Show Solution<\/span><\/p>\n<div id=\"q384264209\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\overline{x}=\\frac{M_y}m=\\frac{81\\pi\/63}{9\\pi\/8}=\\frac98[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{81\\pi\/63}{9\\pi\/8}=\\frac98[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793462607\">Once again, based on the comments at the end of\u00a0Example &#8220;Finding the Center of Mass&#8221;,\u00a0we have expressions for the centroid of a region on the plane:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{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}}}}.[\/latex]<\/p>\n<p>We should use these formulas and verify the centroid of the triangular region [latex]R[\/latex] referred to in the last three examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding Mass, Moments, and Center of Mass<\/h3>\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<div id=\"attachment_1412\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1412\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324187023\">Show Solution<\/span><\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the mass, moments, and the center of mass of the region between the curves [latex]{y} = {x}[\/latex] and [latex]{y} = {{x}^{2}}[\/latex] with the density function [latex]{\\rho}{({x},{y})} = {x}[\/latex] in the interval [latex]{0} \\leq {x} \\leq {1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q130487486\">Show Solution<\/span><\/p>\n<div id=\"q130487486\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\overline{x}=\\frac{M_y}m=\\frac{1\/20}{1\/12}=\\frac35[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{1\/24}{1\/12}=\\frac12[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197110&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=P_3DRbSem_w&amp;video_target=tpm-plugin-pdh1flyj-P_3DRbSem_w\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.36_transcript.html\">transcript for \u201cCP 5.36\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding a centroid<\/h3>\n<p>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<div id=\"attachment_1413\" style=\"width: 389px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1413\" 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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q452983465\">Show Solution<\/span><\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the centroid of the region between the curves [latex]y=x[\/latex] and [latex]{y} = {\\sqrt{x}}[\/latex] with uniform density in the interval [latex]{0} \\leq {x} \\leq {1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q034759837\">Show Solution<\/span><\/p>\n<div id=\"q034759837\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\overline{x}=\\frac{M_y}m=\\frac{1\/15}{1\/6}=\\frac25[\/latex] and\u00a0[latex]\\overline{y}=\\frac{M_x}m\\frac{1\/12}{1\/6}=\\frac12[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Moments of Inertia<\/h2>\n<p id=\"fs-id1167793805130\">For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. The moment of inertia of a particle of mass [latex]m[\/latex] about an axis is [latex]mr^{2}[\/latex] where [latex]r[\/latex] is the distance of the particle from the axis. We can see from\u00a0Figure 3\u00a0that the moment of inertia of the subrectangle [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis is [latex]{({y}^{*}_{ij})^{2}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex]. Similarly, the moment of inertia of the subrectangle [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis is [latex]{({x}^{*}_{ij})^{2}}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A}[\/latex]. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.<\/p>\n<p id=\"fs-id1167794327004\">The moment of inertia [latex]I_x[\/latex] about the [latex]x[\/latex]-axis for the region [latex]R[\/latex] is the limit of the sum of moments of inertia of the regions [latex]{R}_{ij}[\/latex] about the [latex]x[\/latex]-axis. Hence<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{x}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {y}^{*}_{ij} \\right )}^{2}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {y}^{*}_{ij} \\right )}^{2}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = {\\underset{R}{\\displaystyle\\iint}}{{y}^{2}}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\n<p>Similarly, the moment of inertia [latex]I_y[\/latex] about the [latex]y[\/latex]-axis for [latex]R[\/latex] is the limit of the sum of moments of inertia of the regions [latex]{R}_{ij}[\/latex] about the [latex]y[\/latex]-axis. Hence<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{y}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {x}^{*}_{ij} \\right )}^{2}{{m}_{ij}} = {\\displaystyle\\lim_{{k},{l}{\\rightarrow}{\\infty}}}{\\displaystyle\\sum^{k}_{{i} = {1}}}{\\displaystyle\\sum^{l}_{{j} = {1}}}{\\left ( {x}^{*}_{ij} \\right )}^{2}{\\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\\Delta}{A} = {\\underset{R}{\\displaystyle\\iint}}{{x}^{2}}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\n<p id=\"fs-id1167793219176\">Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by [latex]I_0[\/latex] and obtain it by adding the moments of inertia [latex]I_x[\/latex] and [latex]I_y[\/latex]. Hence<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{I}_{0}} = {{I}_{x}} + {{I}_{y}} = {\\underset{R}{\\displaystyle\\iint}}{({x^{2}} + {y^{2}})}{\\rho}{({x},{y})}{dA}}.[\/latex]<\/p>\n<p id=\"fs-id1167793883659\">All these expressions can be written in polar coordinates by substituting [latex]{x} = {r} \\ {\\cos} \\ {\\theta}, {y} = {r} \\ {\\sin} \\ {\\theta},[\/latex] and [latex]{dA} = {r} \\ {dr} \\ {d}{\\theta}[\/latex]. For example, [latex]{{I}_{0}} = {\\underset{R}{\\displaystyle\\iint}}{{r}^{2}}{\\rho}{({{r}{\\cos}{\\theta}},{{r}{\\sin}{\\theta}})}{dA}.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding moments of inertia for a triangular lamina<\/h3>\n<p>Use the triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {x}{y}[\/latex] as in previous examples. Find the moments of inertia.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567926450\">Show Solution<\/span><\/p>\n<div id=\"q567926450\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793419956\">Using the expressions established above for the moments of inertia, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {{I}_{x}} = & {\\underset{R}{\\displaystyle\\iint}}{{y}^{2}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {x}}_{{y} = {0}}}{x}{{y}^{3}}{dy} \\ {dx} = {\\frac{8}{3}}, \\\\  {{I}_{y}} = & {\\underset{R}{\\displaystyle\\iint}}{{x}^{2}}{\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{{x} = {2}}_{{x} = {0}}} \\ {\\displaystyle\\int^{{y} = {x}}_{{y} = {0}}}{{x}^{3}}{y} \\ {dy} \\ {dx} = {\\frac{16}{3}}, \\\\  {{I}_{0}} = & {\\underset{R}{\\displaystyle\\iint}}{({{x}^{2}} + {{y}^{2}})} \\ {\\rho}{({x},{y})}{dA} = {\\displaystyle\\int^{2}_{0}}{\\displaystyle\\int^{x}_{0}}{({{x}^{2}} + {{y}^{2}})} \\ {xy} \\ {dy} \\ {dx} \\\\  = & {{I}_{x}} + {{I}_{y}} = {8}.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Again use the same region [latex]R[\/latex] as above and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the moments of inertia.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q083745928\">Show Solution<\/span><\/p>\n<div id=\"q083745928\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]I_x=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}y^2\\sqrt{xy}dy \\ dx = \\frac{64}{35}[\/latex] and\u00a0[latex]I_y=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}x^2\\sqrt{xy}dy \\ dx = \\frac{64}{15}[\/latex]. Also,\u00a0[latex]I_0=\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=0}^{y=x}(x^2+y^2)\\sqrt{xy}dy \\ dx = \\frac{128}{21}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197111&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=H3dKuZxGpxw&amp;video_target=tpm-plugin-obobsoz2-H3dKuZxGpxw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.38_transcript.html\">transcript for \u201cCP 5.38\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167793432456\">As mentioned earlier, the moment of inertia of a particle of mass [latex]m[\/latex] about an axis is [latex]mr^{2}[\/latex] where [latex]r[\/latex] is the distance of the particle from the axis, also known as the\u00a0<strong><span id=\"43fd2933-285f-4a11-85d9-8913abc213df_term224\" data-type=\"term\">radius of gyration<\/span>.<\/strong><\/p>\n<p id=\"fs-id1167793280692\">Hence the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{R}_{x}} = {\\sqrt{\\frac{{I}_{x}}{m}}}, {{R}_{y}} = {\\sqrt{\\frac{{I}_{y}}{m}}}, \\ {\\text{and}} \\ {{R}_{0}} = {\\sqrt{\\frac{{I}_{0}}{m}}}},[\/latex]<\/p>\n<p id=\"fs-id1167793276987\">respectively. In each case, the radius of gyration tells us how far (perpendicular distance) from the axis of rotation the entire mass of an object might be concentrated. The moments of an object are useful for finding information on the balance and torque of the object about an axis, but radii of gyration are used to describe the distribution of mass around its centroidal axis. There are many applications in engineering and physics. Sometimes it is necessary to find the radius of gyration, as in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the radius of gyration for a triangular lamina<\/h3>\n<p>Consider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and with density [latex]{\\rho}{({x},{y})} = {x}{y}[\/latex] as in previous examples. Find the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634752734\">Show Solution<\/span><\/p>\n<div id=\"q634752734\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794145591\">If we compute the mass of this region we find that [latex]m=2[\/latex]. We found the moments of inertia of this lamina in\u00a0Example &#8220;Finding Mass, Moments, and Center of Mass&#8221;. From these data, the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin are, respectively,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {{R}_{x}} = & {\\sqrt{\\frac{{I}_{x}}{m}}} = {\\sqrt{\\frac{8\/3}{2}}} = {\\sqrt{\\frac{8}{6}}} = {\\frac{{2}{\\sqrt{3}}}{3}}, \\\\  {{R}_{y}} = & {\\sqrt{\\frac{{I}_{y}}{m}}} = {\\sqrt{\\frac{16\/3}{2}}} = {\\sqrt{\\frac{8}{3}}} = {\\frac{{2}{\\sqrt{6}}}{3}}, \\\\  {{R}_{0}} = & {\\sqrt{\\frac{{I}_{0}}{m}}} = {\\sqrt{\\frac{8}{2}}} = {\\sqrt{4}} = {2}.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use the same region [latex]R[\/latex] from\u00a0Example &#8220;Finding the Radius of Gyration for a Triangular Lamina&#8221; and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the radii of gyration with respect to the [latex]x[\/latex]-axis, the [latex]y[\/latex]-axis, and the origin.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q495289348\">Show Solution<\/span><\/p>\n<div id=\"q495289348\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]R_x=\\frac{6\\sqrt{35}}{35}[\/latex],\u00a0[latex]R_y=\\frac{6\\sqrt{15}}{15}[\/latex], and\u00a0[latex]R_0=\\frac{4\\sqrt{42}}{7}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3996\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 5.36. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 5.38. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":25,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.36\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 5.38\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3996","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3996","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3996\/revisions"}],"predecessor-version":[{"id":6373,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3996\/revisions\/6373"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3996\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3996"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3996"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3996"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}