{"id":237787,"date":"2025-06-17T15:24:57","date_gmt":"2025-06-17T15:24:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=237787"},"modified":"2025-06-17T15:24:59","modified_gmt":"2025-06-17T15:24:59","slug":"find-the-mass-and-center-of-mass-of-the-lamina-that-occupies-the-region-d-and-has-the-given-density-function","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/find-the-mass-and-center-of-mass-of-the-lamina-that-occupies-the-region-d-and-has-the-given-density-function\/","title":{"rendered":"Find the mass and center of mass of the lamina that occupies the region D and has the given density function"},"content":{"rendered":"\n

Find the mass and center of mass of the lamina that occupies the region D and has the given density function (Enter your answer as an improper fraction:) D = {(T,u)|0 < v < sin (= I\/L) . 0 < =< L}; e (1.y) 19y<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the mass<\/strong> and the center of mass<\/strong> of the lamina over a region DDD with a given density function<\/strong>, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Region<\/strong> D={(x,y)\u22230<x<L,\u00a00<y<sin\u2061(\u03c0xL)}D = \\{(x, y) \\mid 0 < x < L,\\ 0 < y < \\sin\\left(\\frac{\\pi x}{L}\\right)\\}D={(x,y)\u22230<x<L,\u00a00<y<sin(L\u03c0x\u200b)}<\/li>\n\n\n\n
  • Density function<\/strong>: \u03c1(x,y)=19y\\rho(x, y) = 19y\u03c1(x,y)=19y<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 1: Mass of the Lamina<\/strong><\/h3>\n\n\n\n

    The mass mmm of the lamina is given by the double integral: m=\u222cD\u03c1(x,y)\u2009dA=\u222bx=0L\u222by=0sin\u2061(\u03c0xL)19y\u2009dy\u2009dxm = \\iint_D \\rho(x, y)\\, dA = \\int_{x=0}^{L} \\int_{y=0}^{\\sin\\left(\\frac{\\pi x}{L}\\right)} 19y\\, dy\\, dxm=\u222cD\u200b\u03c1(x,y)dA=\u222bx=0L\u200b\u222by=0sin(L\u03c0x\u200b)\u200b19ydydx<\/p>\n\n\n\n

    Evaluate the inner integral:<\/h4>\n\n\n\n

    \u222by=0sin\u2061(\u03c0xL)19y\u2009dy=19[y22]0sin\u2061(\u03c0xL)=192sin\u20612(\u03c0xL)\\int_{y=0}^{\\sin\\left(\\frac{\\pi x}{L}\\right)} 19y\\, dy = 19 \\left[ \\frac{y^2}{2} \\right]_0^{\\sin\\left(\\frac{\\pi x}{L}\\right)} = \\frac{19}{2} \\sin^2\\left(\\frac{\\pi x}{L}\\right)\u222by=0sin(L\u03c0x\u200b)\u200b19ydy=19[2y2\u200b]0sin(L\u03c0x\u200b)\u200b=219\u200bsin2(L\u03c0x\u200b)<\/p>\n\n\n\n

    Now the outer integral:<\/h4>\n\n\n\n

    m=\u222b0L192sin\u20612(\u03c0xL)\u2009dxm = \\int_0^L \\frac{19}{2} \\sin^2\\left(\\frac{\\pi x}{L}\\right)\\, dxm=\u222b0L\u200b219\u200bsin2(L\u03c0x\u200b)dx<\/p>\n\n\n\n

    Use the identity: sin\u20612\u03b8=1\u2212cos\u2061(2\u03b8)2\\sin^2\\theta = \\frac{1 – \\cos(2\\theta)}{2}sin2\u03b8=21\u2212cos(2\u03b8)\u200b m=192\u222b0L1\u2212cos\u2061(2\u03c0xL)2dx=194\u222b0L(1\u2212cos\u2061(2\u03c0xL))\u2009dxm = \\frac{19}{2} \\int_0^L \\frac{1 – \\cos\\left(\\frac{2\\pi x}{L}\\right)}{2} dx = \\frac{19}{4} \\int_0^L \\left(1 – \\cos\\left(\\frac{2\\pi x}{L}\\right)\\right)\\, dxm=219\u200b\u222b0L\u200b21\u2212cos(L2\u03c0x\u200b)\u200bdx=419\u200b\u222b0L\u200b(1\u2212cos(L2\u03c0x\u200b))dx =194[x\u2212L2\u03c0sin\u2061(2\u03c0xL)]0L=194(L\u22120)=19L4= \\frac{19}{4} \\left[ x – \\frac{L}{2\\pi} \\sin\\left(\\frac{2\\pi x}{L}\\right) \\right]_0^L = \\frac{19}{4} \\left( L – 0 \\right) = \\frac{19L}{4}=419\u200b[x\u22122\u03c0L\u200bsin(L2\u03c0x\u200b)]0L\u200b=419\u200b(L\u22120)=419L\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Center of Mass<\/strong><\/h3>\n\n\n\n

    Let (x\u02c9,y\u02c9)(\\bar{x}, \\bar{y})(x\u02c9,y\u02c9\u200b) be the center of mass.<\/p>\n\n\n\n

    x\u02c9=1m\u222cDx\u03c1(x,y)\u2009dA\\bar{x} = \\frac{1}{m} \\iint_D x \\rho(x, y)\\, dAx\u02c9=m1\u200b\u222cD\u200bx\u03c1(x,y)dA<\/h4>\n\n\n\n

    x\u02c9=1m\u222b0L\u222b0sin\u2061(\u03c0xL)x(19y)\u2009dy\u2009dx\\bar{x} = \\frac{1}{m} \\int_0^L \\int_0^{\\sin\\left(\\frac{\\pi x}{L}\\right)} x(19y)\\, dy\\, dxx\u02c9=m1\u200b\u222b0L\u200b\u222b0sin(L\u03c0x\u200b)\u200bx(19y)dydx =1m\u222b0L19x[y22]0sin\u2061(\u03c0xL)dx=1m\u222b0L19x2sin\u20612(\u03c0xL)dx= \\frac{1}{m} \\int_0^L 19x \\left[\\frac{y^2}{2} \\right]_0^{\\sin\\left(\\frac{\\pi x}{L}\\right)} dx = \\frac{1}{m} \\int_0^L \\frac{19x}{2} \\sin^2\\left(\\frac{\\pi x}{L}\\right) dx=m1\u200b\u222b0L\u200b19x[2y2\u200b]0sin(L\u03c0x\u200b)\u200bdx=m1\u200b\u222b0L\u200b219x\u200bsin2(L\u03c0x\u200b)dx<\/p>\n\n\n\n

    Use substitution u=\u03c0xL\u21d2dx=L\u03c0du, x=Lu\u03c0u = \\frac{\\pi x}{L} \\Rightarrow dx = \\frac{L}{\\pi} du,\\ x = \\frac{Lu}{\\pi}u=L\u03c0x\u200b\u21d2dx=\u03c0L\u200bdu, x=\u03c0Lu\u200b x\u02c9=119L4\u22c5192\u22c5\u222b0Lxsin\u20612(\u03c0xL)dx=2L\u222b0Lxsin\u20612(\u03c0xL)dx\\bar{x} = \\frac{1}{\\frac{19L}{4}} \\cdot \\frac{19}{2} \\cdot \\int_0^L x \\sin^2\\left(\\frac{\\pi x}{L}\\right) dx = \\frac{2}{L} \\int_0^L x \\sin^2\\left(\\frac{\\pi x}{L}\\right) dxx\u02c9=419L\u200b1\u200b\u22c5219\u200b\u22c5\u222b0L\u200bxsin2(L\u03c0x\u200b)dx=L2\u200b\u222b0L\u200bxsin2(L\u03c0x\u200b)dx<\/p>\n\n\n\n

    Using integration by parts or known result: \u222b0Lxsin\u20612(\u03c0xL)dx=L24\\int_0^L x \\sin^2\\left(\\frac{\\pi x}{L}\\right) dx = \\frac{L^2}{4}\u222b0L\u200bxsin2(L\u03c0x\u200b)dx=4L2\u200b<\/p>\n\n\n\n

    So: x\u02c9=2L\u22c5L24=L2\\bar{x} = \\frac{2}{L} \\cdot \\frac{L^2}{4} = \\frac{L}{2}x\u02c9=L2\u200b\u22c54L2\u200b=2L\u200b<\/p>\n\n\n\n

    \n\n\n\n

    y\u02c9=1m\u222cDy\u03c1(x,y)\u2009dA=1m\u222b0L\u222b0sin\u2061(\u03c0xL)19y2\u2009dy\u2009dx\\bar{y} = \\frac{1}{m} \\iint_D y \\rho(x, y)\\, dA = \\frac{1}{m} \\int_0^L \\int_0^{\\sin\\left(\\frac{\\pi x}{L}\\right)} 19y^2\\, dy\\, dxy\u02c9\u200b=m1\u200b\u222cD\u200by\u03c1(x,y)dA=m1\u200b\u222b0L\u200b\u222b0sin(L\u03c0x\u200b)\u200b19y2dydx<\/h4>\n\n\n\n

    \u222b0sin\u2061(\u03c0x\/L)19y2\u2009dy=19[y33]0sin\u2061(\u03c0x\/L)=193sin\u20613(\u03c0xL)\\int_0^{\\sin(\\pi x\/L)} 19y^2\\, dy = 19 \\left[ \\frac{y^3}{3} \\right]_0^{\\sin(\\pi x\/L)} = \\frac{19}{3} \\sin^3\\left(\\frac{\\pi x}{L}\\right)\u222b0sin(\u03c0x\/L)\u200b19y2dy=19[3y3\u200b]0sin(\u03c0x\/L)\u200b=319\u200bsin3(L\u03c0x\u200b) y\u02c9=119L4\u22c5\u222b0L193sin\u20613(\u03c0xL)\u2009dx=4L\u22c513\u22c5\u222b0Lsin\u20613(\u03c0xL)\u2009dx\\bar{y} = \\frac{1}{\\frac{19L}{4}} \\cdot \\int_0^L \\frac{19}{3} \\sin^3\\left(\\frac{\\pi x}{L}\\right)\\, dx = \\frac{4}{L} \\cdot \\frac{1}{3} \\cdot \\int_0^L \\sin^3\\left(\\frac{\\pi x}{L}\\right)\\, dxy\u02c9\u200b=419L\u200b1\u200b\u22c5\u222b0L\u200b319\u200bsin3(L\u03c0x\u200b)dx=L4\u200b\u22c531\u200b\u22c5\u222b0L\u200bsin3(L\u03c0x\u200b)dx<\/p>\n\n\n\n

    Use the identity: sin\u20613\u03b8=3sin\u2061\u03b8\u2212sin\u2061(3\u03b8)4\\sin^3\\theta = \\frac{3\\sin\\theta – \\sin(3\\theta)}{4}sin3\u03b8=43sin\u03b8\u2212sin(3\u03b8)\u200b<\/p>\n\n\n\n

    So: \u222b0Lsin\u20613(\u03c0xL)\u2009dx=L\u03c0\u222b0\u03c0sin\u20613(u)\u2009du=L\u03c0\u22c543\\int_0^L \\sin^3\\left(\\frac{\\pi x}{L}\\right)\\, dx = \\frac{L}{\\pi} \\int_0^\\pi \\sin^3(u)\\, du = \\frac{L}{\\pi} \\cdot \\frac{4}{3}\u222b0L\u200bsin3(L\u03c0x\u200b)dx=\u03c0L\u200b\u222b0\u03c0\u200bsin3(u)du=\u03c0L\u200b\u22c534\u200b y\u02c9=4L\u22c513\u22c54L3\u03c0=169\u03c0\\bar{y} = \\frac{4}{L} \\cdot \\frac{1}{3} \\cdot \\frac{4L}{3\\pi} = \\frac{16}{9\\pi}y\u02c9\u200b=L4\u200b\u22c531\u200b\u22c53\u03c04L\u200b=9\u03c016\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Final Answer:<\/strong><\/h3>\n\n\n\n
      \n
    • Mass<\/strong> m=19L4m = \\boxed{\\frac{19L}{4}}m=419L\u200b\u200b<\/li>\n\n\n\n
    • Center of Mass<\/strong>:
      \u2003\u2003x\u02c9=L2,\u00a0y\u02c9=169\u03c0\\bar{x} = \\boxed{\\frac{L}{2}},\\ \\bar{y} = \\boxed{\\frac{16}{9\\pi}}x\u02c9=2L\u200b\u200b,\u00a0y\u02c9\u200b=9\u03c016\u200b\u200b<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Explanation<\/strong><\/h3>\n\n\n\n

      To find the mass and center of mass of a lamina with variable density, we apply double integrals over the defined region. In this case, the region DDD is bounded below by the x-axis and above by the curve y=sin\u2061(\u03c0xL)y = \\sin\\left(\\frac{\\pi x}{L}\\right)y=sin(L\u03c0x\u200b), for xxx ranging from 0 to LLL. The density function given is \u03c1(x,y)=19y\\rho(x, y) = 19y\u03c1(x,y)=19y, which means density increases with height.<\/p>\n\n\n\n

      To calculate mass, we integrate the density over the area. The integration begins with respect to yyy, from 0 up to the sine curve. This gives a function of xxx, which we then integrate from 0 to LLL. Using trigonometric identities simplifies the integral, resulting in a mass of 19L4\\frac{19L}{4}419L\u200b.<\/p>\n\n\n\n

      The x-coordinate of the center of mass (x\u02c9\\bar{x}x\u02c9) is found by integrating x\u22c5\u03c1(x,y)x \\cdot \\rho(x, y)x\u22c5\u03c1(x,y). Because of symmetry and the nature of the sine function, we find that x\u02c9\\bar{x}x\u02c9 equals L2\\frac{L}{2}2L\u200b, indicating that the mass is balanced horizontally at the midpoint of the interval.<\/p>\n\n\n\n

      For the y-coordinate (y\u02c9\\bar{y}y\u02c9\u200b), we integrate y\u22c5\u03c1(x,y)y \\cdot \\rho(x, y)y\u22c5\u03c1(x,y). This requires evaluating the cube of the sine function. Using known trigonometric identities and substitutions, we simplify the integral to find y\u02c9=169\u03c0\\bar{y} = \\frac{16}{9\\pi}y\u02c9\u200b=9\u03c016\u200b. This lower value reflects that most mass lies closer to the x-axis since density increases with yyy.<\/p>\n\n\n\n

      This process applies the principles of multivariable calculus and emphasizes the role of both density and shape in determining mass distribution.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Find the mass and center of mass of the lamina that occupies the region D and has the given density function (Enter your answer as an improper fraction:) D = {(T,u)|0 < v < sin (= I\/L) . 0 < =< L}; e (1.y) 19y The Correct Answer and Explanation is: To find the mass […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-237787","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237787","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=237787"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237787\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=237787"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=237787"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=237787"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}