{"id":721,"date":"2025-10-09T15:34:30","date_gmt":"2025-10-09T15:34:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=721"},"modified":"2025-11-13T15:23:27","modified_gmt":"2025-11-13T15:23:27","slug":"center-of-mass-and-moments-of-inertia-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/center-of-mass-and-moments-of-inertia-fresh-take\/","title":{"raw":"Center of Mass and Moments of Inertia: Fresh Take","rendered":"Center of Mass and Moments of Inertia: Fresh Take"},"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<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Finding the center of mass of a flat plate (lamina) means locating its balance point\u2014where it would perfectly balance on a pin. For objects with variable density, this requires calculating three key quantities using double integrals: the mass and two moments that measure how mass is distributed relative to each axis.<\/p>\r\n<strong>The Core Concept<\/strong>\r\n<p class=\"whitespace-normal break-words\">A lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex] (mass per unit area) has:<\/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>Mass:<\/strong> [latex]m = \\iint_R \\rho(x,y) \\, dA[\/latex]<\/li>\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<p class=\"whitespace-normal break-words\">The center of mass is then located at [latex](\\overline{x}, \\overline{y})[\/latex] where:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\frac{M_y}{m} \\quad \\text{and} \\quad \\overline{y} = \\frac{M_x}{m}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Understanding Moments<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Moments measure how mass is distributed. The moment [latex]M_x[\/latex] uses [latex]y[\/latex] because it measures how far mass sits from the [latex]x[\/latex]-axis. Similarly, [latex]M_y[\/latex] uses [latex]x[\/latex] to measure distance from the [latex]y[\/latex]-axis. Dividing each moment by total mass gives the average position in that direction.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Centroid vs. Center of Mass<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When density is constant (or you're finding the geometric center regardless of mass), you're finding the <strong>centroid<\/strong>. The density constant cancels out, giving simpler formulas:<\/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-normal break-words\">For uniform density, the center of mass and centroid coincide. For variable density, the center of mass shifts toward regions of higher density.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol 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\">Sketch the region and identify its boundaries<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Set up the double integral for mass using the given density function<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Set up the double integrals for [latex]M_x[\/latex] and [latex]M_y[\/latex], including the appropriate coordinate ([latex]y[\/latex] or [latex]x[\/latex]) with the density<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate all three integrals<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Divide moments by mass to find [latex]\\overline{x}[\/latex] and [latex]\\overline{y}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"470317685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470317685\"][latex]\\frac{9\\pi}8[\/latex] kg.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"900563247\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"900563247\"][latex]M_x=\\frac{81\\pi}{64}[\/latex] and\u00a0[latex]M_y=\\frac{81\\pi}{64}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"130487486\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130487486\"][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].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"034759837\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"034759837\"][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].[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Moments of Inertia<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Moments of inertia measure how mass is distributed relative to an axis of rotation\u2014they describe an object's resistance to rotational acceleration. While ordinary moments use distance from an axis, moments of inertia use the square of that distance, making them more sensitive to mass located far from the rotation axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Core Formulas<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density [latex]\\rho(x,y)[\/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\"><strong>About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iint_R y^2 \\rho(x,y) \\, dA[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iint_R x^2 \\rho(x,y) \\, dA[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>About the origin (polar moment):<\/strong> [latex]I_0 = I_x + I_y = \\iint_R (x^2 + y^2) \\rho(x,y) \\, dA[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Notice the squared distance terms. For rotation about the [latex]x[\/latex]-axis, we use [latex]y^2[\/latex] because [latex]y[\/latex] measures perpendicular distance from that axis. Similarly, [latex]x[\/latex] measures distance from the [latex]y[\/latex]-axis, so we use [latex]x^2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Moments vs. Moments of Inertia<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Compare the structure of these integrals:<\/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\">Moment: [latex]M_x = \\iint_R y\\rho(x,y) \\, dA[\/latex] (uses [latex]y[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Moment of inertia: [latex]I_x = \\iint_R y^2\\rho(x,y) \\, dA[\/latex] (uses [latex]y^2[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The squared distance in moments of inertia means mass farther from the axis contributes much more to rotational resistance.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Polar Moment of Inertia<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When rotating about the origin rather than a coordinate axis, use the polar moment [latex]I_0[\/latex]. The distance from [latex](x,y)[\/latex] to the origin is [latex]\\sqrt{x^2 + y^2}[\/latex], and squaring gives [latex]x^2 + y^2[\/latex]. Conveniently, [latex]I_0 = I_x + I_y[\/latex], so you can compute it either way. For regions with circular symmetry, polar coordinates simplify the calculation significantly.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Radius of Gyration<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The radius of gyration tells you the effective distance where all the mass could be concentrated while maintaining the same moment of inertia:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R_x = \\sqrt{\\frac{I_x}{m}}, \\quad R_y = \\sqrt{\\frac{I_y}{m}}, \\quad R_0 = \\sqrt{\\frac{I_0}{m}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives intuition about mass distribution\u2014a larger radius of gyration means mass is spread farther from the axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol 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\">Set up the double integral with [latex]x^2[\/latex] or [latex]y^2[\/latex] depending on the axis<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate the moment of inertia integral<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For radius of gyration, also calculate mass and take [latex]\\sqrt{I\/m}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Consider polar coordinates when the region or density has circular symmetry<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Use the triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and the density function [latex]{\\rho}{({x},{y})} = {\\sqrt{xy}}[\/latex]. Find the moments of inertia.[reveal-answer q=\"083745928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"083745928\"][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].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and 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.[reveal-answer q=\"495289348\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"495289348\"][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].[\/hidden-answer]<\/section>\r\n<h2>Center of Mass and Moments of Inertia in Three Dimensions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The formulas for mass, moments, center of mass, and moments of inertia extend naturally from two dimensions to three dimensions. Instead of integrating over an area with [latex]dA[\/latex], you integrate over a volume with [latex]dV[\/latex]. The key is understanding which coordinate measures distance from each plane or axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Mass and Moments<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a solid [latex]Q[\/latex] with density [latex]\\rho(x,y,z)[\/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\"><strong>Mass:<\/strong> [latex]m = \\iiint_Q \\rho(x,y,z) \\, dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Moments about planes:<\/strong> [latex]M_{xy} = \\iiint_Q z\\rho(x,y,z) \\, dV[\/latex], [latex]M_{xz} = \\iiint_Q y\\rho(x,y,z) \\, dV[\/latex], [latex]M_{yz} = \\iiint_Q x\\rho(x,y,z) \\, dV[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The moment subscript tells you which plane, and the missing coordinate is what you multiply by. For [latex]M_{xy}[\/latex] (the [latex]xy[\/latex]-plane), [latex]z[\/latex] measures distance from that plane, so multiply by [latex]z[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Center of Mass<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The center of mass coordinates are:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\frac{M_{yz}}{m}, \\quad \\overline{y} = \\frac{M_{xz}}{m}, \\quad \\overline{z} = \\frac{M_{xy}}{m}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Each coordinate equals its corresponding moment divided by mass. For uniform density (homogeneous solids), this gives the centroid\u2014the geometric center.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Moments of Inertia<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For rotation about each axis, use the sum of squares of the two perpendicular coordinates:<\/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>About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iiint_Q (y^2 + z^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iiint_Q (x^2 + z^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>About the [latex]z[\/latex]-axis:<\/strong> [latex]I_z = \\iiint_Q (x^2 + y^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The pattern: for [latex]I_x[\/latex], use everything except [latex]x[\/latex] (namely [latex]y^2 + z^2[\/latex]), because this gives the squared distance from the [latex]x[\/latex]-axis.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol 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\">Sketch the solid and determine integration bounds<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate mass using [latex]\\iiint_Q \\rho(x,y,z) , dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate the three moments, multiplying by the appropriate coordinate<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Divide moments by mass to find center of mass coordinates<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For moments of inertia, use the sum of squared perpendicular coordinates with density<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1), and use the density function [latex]{\\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}[\/latex].\u00a0<span style=\"font-size: 1rem; text-align: initial;\">Find the mass.<\/span>[caption id=\"attachment_1414\" align=\"aligncenter\" width=\"268\"]<img class=\"size-full wp-image-1414\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174918\/5-6-7.jpeg\" alt=\"In x y z space, the solid Q is shown with corners (0, 0, 0), (0, 0, 2), (0, 3, 0), and (6, 0, 0). Alternatively, you could consider the solid as being bounded by the x y, x z, and y z planes and the plane x + 2y + 3z = 6, forming an irregular tetrahedron.\" width=\"268\" height=\"309\" \/> Figure 1. Finding the mass of a three-dimensional solid [latex]Q[\/latex].[\/caption][reveal-answer q=\"685720429\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"685720429\"][latex]\\frac{54}{35}=1.543[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1) and use the density function [latex]{\\rho}{({x},{y},{z})} = xy^{2}z[\/latex]. Find the center of mass.[reveal-answer q=\"039842756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"039842756\"][latex]\\left(\\frac32,\\frac98,\\frac12\\right)[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197113&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=w7ayXgDTBiM&amp;video_target=tpm-plugin-opzew3rd-w7ayXgDTBiM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.41_transcript.html\">transcript for \u201cCP 5.41\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1), and use the density function [latex]{\\rho}{({x},{y},{z})} = {x}{{y}^{2}}{z}[\/latex]. Find the moments of inertia about the three coordinate planes.[reveal-answer q=\"485732209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485732209\"]The moments of inertia of the tetrahedron [latex]Q[\/latex] about the [latex]yz[\/latex]-plane, the [latex]xz[\/latex]-plane, and the [latex]xy[\/latex]-plane are<span style=\"white-space: nowrap;\"> [latex]99\/35[\/latex],\u00a0[latex]36\/7[\/latex], and\u00a0[latex]243\/35[\/latex],\u00a0<\/span>respectively.[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197112&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_EopC8p7DmM&amp;video_target=tpm-plugin-vi6rhg0g-_EopC8p7DmM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.42_transcript.html\">transcript for \u201cCP 5.42\u201d here (opens in new window).<\/a><\/center><\/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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding the center of mass of a flat plate (lamina) means locating its balance point\u2014where it would perfectly balance on a pin. For objects with variable density, this requires calculating three key quantities using double integrals: the mass and two moments that measure how mass is distributed relative to each axis.<\/p>\n<p><strong>The Core Concept<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A lamina occupying region [latex]R[\/latex] with density function [latex]\\rho(x,y)[\/latex] (mass per unit area) has:<\/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>Mass:<\/strong> [latex]m = \\iint_R \\rho(x,y) \\, dA[\/latex]<\/li>\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<p class=\"whitespace-normal break-words\">The center of mass is then located at [latex](\\overline{x}, \\overline{y})[\/latex] where:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\frac{M_y}{m} \\quad \\text{and} \\quad \\overline{y} = \\frac{M_x}{m}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Understanding Moments<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Moments measure how mass is distributed. The moment [latex]M_x[\/latex] uses [latex]y[\/latex] because it measures how far mass sits from the [latex]x[\/latex]-axis. Similarly, [latex]M_y[\/latex] uses [latex]x[\/latex] to measure distance from the [latex]y[\/latex]-axis. Dividing each moment by total mass gives the average position in that direction.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Centroid vs. Center of Mass<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When density is constant (or you&#8217;re finding the geometric center regardless of mass), you&#8217;re finding the <strong>centroid<\/strong>. The density constant cancels out, giving simpler formulas:<\/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-normal break-words\">For uniform density, the center of mass and centroid coincide. For variable density, the center of mass shifts toward regions of higher density.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol 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\">Sketch the region and identify its boundaries<\/li>\n<li class=\"whitespace-normal break-words\">Set up the double integral for mass using the given density function<\/li>\n<li class=\"whitespace-normal break-words\">Set up the double integrals for [latex]M_x[\/latex] and [latex]M_y[\/latex], including the appropriate coordinate ([latex]y[\/latex] or [latex]x[\/latex]) with the density<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate all three integrals<\/li>\n<li class=\"whitespace-normal break-words\">Divide moments by mass to find [latex]\\overline{x}[\/latex] and [latex]\\overline{y}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q470317685\">Show Solution<\/button><\/p>\n<div id=\"q470317685\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{9\\pi}8[\/latex] kg.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q900563247\">Show Solution<\/button><\/p>\n<div id=\"q900563247\" class=\"hidden-answer\" style=\"display: none\">[latex]M_x=\\frac{81\\pi}{64}[\/latex] and\u00a0[latex]M_y=\\frac{81\\pi}{64}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q130487486\">Show Solution<\/button><\/p>\n<div id=\"q130487486\" class=\"hidden-answer\" style=\"display: none\">[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].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q034759837\">Show Solution<\/button><\/p>\n<div id=\"q034759837\" class=\"hidden-answer\" style=\"display: none\">[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].<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Moments of Inertia<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Moments of inertia measure how mass is distributed relative to an axis of rotation\u2014they describe an object&#8217;s resistance to rotational acceleration. While ordinary moments use distance from an axis, moments of inertia use the square of that distance, making them more sensitive to mass located far from the rotation axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Core Formulas<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a lamina occupying region [latex]R[\/latex] with density [latex]\\rho(x,y)[\/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\"><strong>About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iint_R y^2 \\rho(x,y) \\, dA[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iint_R x^2 \\rho(x,y) \\, dA[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>About the origin (polar moment):<\/strong> [latex]I_0 = I_x + I_y = \\iint_R (x^2 + y^2) \\rho(x,y) \\, dA[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Notice the squared distance terms. For rotation about the [latex]x[\/latex]-axis, we use [latex]y^2[\/latex] because [latex]y[\/latex] measures perpendicular distance from that axis. Similarly, [latex]x[\/latex] measures distance from the [latex]y[\/latex]-axis, so we use [latex]x^2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Moments vs. Moments of Inertia<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Compare the structure of these integrals:<\/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\">Moment: [latex]M_x = \\iint_R y\\rho(x,y) \\, dA[\/latex] (uses [latex]y[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Moment of inertia: [latex]I_x = \\iint_R y^2\\rho(x,y) \\, dA[\/latex] (uses [latex]y^2[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The squared distance in moments of inertia means mass farther from the axis contributes much more to rotational resistance.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Polar Moment of Inertia<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When rotating about the origin rather than a coordinate axis, use the polar moment [latex]I_0[\/latex]. The distance from [latex](x,y)[\/latex] to the origin is [latex]\\sqrt{x^2 + y^2}[\/latex], and squaring gives [latex]x^2 + y^2[\/latex]. Conveniently, [latex]I_0 = I_x + I_y[\/latex], so you can compute it either way. For regions with circular symmetry, polar coordinates simplify the calculation significantly.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Radius of Gyration<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The radius of gyration tells you the effective distance where all the mass could be concentrated while maintaining the same moment of inertia:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R_x = \\sqrt{\\frac{I_x}{m}}, \\quad R_y = \\sqrt{\\frac{I_y}{m}}, \\quad R_0 = \\sqrt{\\frac{I_0}{m}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives intuition about mass distribution\u2014a larger radius of gyration means mass is spread farther from the axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol 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\">Set up the double integral with [latex]x^2[\/latex] or [latex]y^2[\/latex] depending on the axis<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the moment of inertia integral<\/li>\n<li class=\"whitespace-normal break-words\">For radius of gyration, also calculate mass and take [latex]\\sqrt{I\/m}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Consider polar coordinates when the region or density has circular symmetry<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Use the triangular region [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] 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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q083745928\">Show Solution<\/button><\/p>\n<div id=\"q083745928\" class=\"hidden-answer\" style=\"display: none\">[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].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same triangular lamina [latex]R[\/latex] with vertices [latex](0, 0)[\/latex], [latex](2, 2)[\/latex], and [latex](2, 0)[\/latex] and 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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q495289348\">Show Solution<\/button><\/p>\n<div id=\"q495289348\" class=\"hidden-answer\" style=\"display: none\">[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].<\/div>\n<\/div>\n<\/section>\n<h2>Center of Mass and Moments of Inertia in Three Dimensions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The formulas for mass, moments, center of mass, and moments of inertia extend naturally from two dimensions to three dimensions. Instead of integrating over an area with [latex]dA[\/latex], you integrate over a volume with [latex]dV[\/latex]. The key is understanding which coordinate measures distance from each plane or axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Mass and Moments<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a solid [latex]Q[\/latex] with density [latex]\\rho(x,y,z)[\/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\"><strong>Mass:<\/strong> [latex]m = \\iiint_Q \\rho(x,y,z) \\, dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Moments about planes:<\/strong> [latex]M_{xy} = \\iiint_Q z\\rho(x,y,z) \\, dV[\/latex], [latex]M_{xz} = \\iiint_Q y\\rho(x,y,z) \\, dV[\/latex], [latex]M_{yz} = \\iiint_Q x\\rho(x,y,z) \\, dV[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The moment subscript tells you which plane, and the missing coordinate is what you multiply by. For [latex]M_{xy}[\/latex] (the [latex]xy[\/latex]-plane), [latex]z[\/latex] measures distance from that plane, so multiply by [latex]z[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Center of Mass<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The center of mass coordinates are:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\overline{x} = \\frac{M_{yz}}{m}, \\quad \\overline{y} = \\frac{M_{xz}}{m}, \\quad \\overline{z} = \\frac{M_{xy}}{m}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Each coordinate equals its corresponding moment divided by mass. For uniform density (homogeneous solids), this gives the centroid\u2014the geometric center.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Moments of Inertia<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For rotation about each axis, use the sum of squares of the two perpendicular coordinates:<\/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>About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iiint_Q (y^2 + z^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iiint_Q (x^2 + z^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>About the [latex]z[\/latex]-axis:<\/strong> [latex]I_z = \\iiint_Q (x^2 + y^2) \\rho(x,y,z) \\, dV[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The pattern: for [latex]I_x[\/latex], use everything except [latex]x[\/latex] (namely [latex]y^2 + z^2[\/latex]), because this gives the squared distance from the [latex]x[\/latex]-axis.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol 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\">Sketch the solid and determine integration bounds<\/li>\n<li class=\"whitespace-normal break-words\">Calculate mass using [latex]\\iiint_Q \\rho(x,y,z) , dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Calculate the three moments, multiplying by the appropriate coordinate<\/li>\n<li class=\"whitespace-normal break-words\">Divide moments by mass to find center of mass coordinates<\/li>\n<li class=\"whitespace-normal break-words\">For moments of inertia, use the sum of squared perpendicular coordinates with density<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1), and use the density function [latex]{\\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}[\/latex].\u00a0<span style=\"font-size: 1rem; text-align: initial;\">Find the mass.<\/span><\/p>\n<figure id=\"attachment_1414\" aria-describedby=\"caption-attachment-1414\" style=\"width: 268px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1414\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174918\/5-6-7.jpeg\" alt=\"In x y z space, the solid Q is shown with corners (0, 0, 0), (0, 0, 2), (0, 3, 0), and (6, 0, 0). Alternatively, you could consider the solid as being bounded by the x y, x z, and y z planes and the plane x + 2y + 3z = 6, forming an irregular tetrahedron.\" width=\"268\" height=\"309\" \/><figcaption id=\"caption-attachment-1414\" class=\"wp-caption-text\">Figure 1. Finding the mass of a three-dimensional solid [latex]Q[\/latex].<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q685720429\">Show Solution<\/button><\/p>\n<div id=\"q685720429\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{54}{35}=1.543[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1) and use the density function [latex]{\\rho}{({x},{y},{z})} = xy^{2}z[\/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=\"q039842756\">Show Solution<\/button><\/p>\n<div id=\"q039842756\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(\\frac32,\\frac98,\\frac12\\right)[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197113&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=w7ayXgDTBiM&amp;video_target=tpm-plugin-opzew3rd-w7ayXgDTBiM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.41_transcript.html\">transcript for \u201cCP 5.41\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Consider the same region [latex]Q[\/latex] (Figure 1), and use the density function [latex]{\\rho}{({x},{y},{z})} = {x}{{y}^{2}}{z}[\/latex]. Find the moments of inertia about the three coordinate planes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q485732209\">Show Solution<\/button><\/p>\n<div id=\"q485732209\" class=\"hidden-answer\" style=\"display: none\">The moments of inertia of the tetrahedron [latex]Q[\/latex] about the [latex]yz[\/latex]-plane, the [latex]xz[\/latex]-plane, and the [latex]xy[\/latex]-plane are<span style=\"white-space: nowrap;\"> [latex]99\/35[\/latex],\u00a0[latex]36\/7[\/latex], and\u00a0[latex]243\/35[\/latex],\u00a0<\/span>respectively.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197112&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_EopC8p7DmM&amp;video_target=tpm-plugin-vi6rhg0g-_EopC8p7DmM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.42_transcript.html\">transcript for \u201cCP 5.42\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":21,"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-721","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"fresh_take","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\/721","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":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/721\/revisions"}],"predecessor-version":[{"id":2211,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/721\/revisions\/2211"}],"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\/721\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=721"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=721"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=721"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=721"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}