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Lecture Notes (Chapter 2.5 Application of Multiple Integral)

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  1 Chapter 2: Multiple Integrals    Apart from finding the area of region, surface area and volume by using double integrals, its application can be extended further in finding the mass  , moment  , center of mass ( or centroid  ) , moment inertia for lamina  .      Any flat object with negligible thickness is called a lamina . Some examples of lamina are shown as follows: (a)   Homogeneous Lamina (b) Non-homogeneous Lamina    A lamina with regular shape and made from same material is called homogeneous lamina, and its density function, ( , )  x y   is constant k  .    A lamina with irregular shape is called non-homogeneous lamina, and its density function, ( , )  x y   is expressed in terms of  x  and  y .    If a lamina with continuous density function ( , )  x y   occupies a region  R  in the  xy - plane , its total mass m  is given by:   , ( , )  R  Mass m x y dA       Application of Double Integrals Application of Double Integrals: (a) Calculating the Mass of a Lamina  2 Chapter 2: Multiple Integrals Example 1 A lamina bounded by  x -axis,  x  = 1 and the curve 2  y x  has density ( , )  x y x y       . Find its total mass. Solution , ( , )  R  Mass m x y dA       10 01 20 01 201 32015 220 2( )02225 42 15 41320  y x x x y x x x x x x x  x ydydx y xy dx x x x dx x x dx x x                     Example 2 A triangular lamina with vertices (0,0),(0,1) and (1,0) has density ( , )  x y xy      . Find its total mass. Solution Equation of line:  y mx c     1 0, 10 1  slope m      and  y -intercept, c  = 1 Hence, the equation of line which connects the points (0,1) and (1,0) is given by: 1  y x        2  y x    R  3 Chapter 2: Multiple Integrals , ( , )  R  Mass m x y dA       110 011 20 01 201 2013 2014 3 20 2( 1)02( 2 1)21221 22 4 3 21 1 2 102 4 3 2124  y x x x y y x x x y x x x x x x  xydydx xydx x xdx x x xdx x x xdx x x x                                            Moment  of mass of an object taken at a point (  x ,  y ), is the quantity of distance   multiplies  its mass .      For a lamina with region  R , the moment taken about the y  -axis  is given by:   ( , )  y R  M x x y dA          For a lamina with region  R , the moment taken about the x  -axis  is given by:   ( , )  x R  M y x y dA       Application of Double Integrals: (b) Calculating the Moment of a Lamina  4 Chapter 2: Multiple Integrals Example 1 A triangular lamina with vertices (0,0),(0,1) and (1,0) has density ( , )  x y xy      . Find its moment of mass about  x -axis. Solution Moment of mass about  x -axis: ( , )  x R  M y x y dA       1 10 011 30 01 301 3 201 4 3 2015 4 230 ( )3( 1)3( 3 3 1)33 331 33 5 4 21 1 3 113 5 4 2160  x x  y xy dydx xydx x xdx x x x xdx x x x x   dx x x x x                                     A point in a lamina where it is in a state of equilibrium  is called center of mass , which can be obtained by dividing the moment of mass about  y -axis and  x -axis by the mass.   Application of Double Integrals: (c) Calculating the Center of Mass of a Lamina
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