W8.3 - February 27, 2008 Physics for...

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Unformatted text preview: February 27, 2008 Physics for Scientists&Engineers 1 1 Physics for Scientists & Physics for Scientists & Engineers 1 Engineers 1 Lecture 29 February 27, 2008 Physics for Scientists&Engineers 1 2 Volume Integrals Volume Integrals in Cartesian Coordinates in Cartesian Coordinates It is by far easiest to express the volume element dV in Cartesian coordinates Then dV is simply the product of the three individual coordinate elements The three-dimensional volume integral written in Cartesian coordinates becomes f ( ! r ) dV = f ( ! r ) dx x min x max ! " # $ $ % & y min y max ! dy " # $ $ % & dz z min z max ! V ! dV = dxdydz February 27, 2008 Physics for Scientists&Engineers 1 3 Volume Integrals Volume Integrals in in Cylindrical Coordinates Cylindrical Coordinates Because we are using the angle as one of the coordinates in cylindrical coordinates, our volume element is not cube-shaped any more For a given differential angle d , the size of the volume element depends on how far away from the z-axis the volume element is located The differential volume is The volume integral is then dV = r ! dr ! d " dz f ( ! r ) dV = f ( ! r ) r ! dr ! r ! min r ! max " # $ % % & ( ( ) min ) max " d ) # $ % % & ( ( dz z min z max " V " February 27, 2008 Physics for Scientists&Engineers 1 4 Volume Integrals Volume Integrals in in Spherical Coordinates Spherical Coordinates In spherical coordinates we use two angular variables The size of the volume element for a given value of the differential coordinates depends on the distance r to the origin as well as the angle relative to the = 0 axis (equivalent to the z-axis in Cartesian or cylindrical coordinates) The differential volume element in spherical coordinates is The volume integral is dV = r 2 dr sin ! d ! d " f ( ! r ) dV = f ( ! r )sin ! d ! ! min ! max " # $ % % & ( ( ) min ) max " d ) # $ % % & ( ( r 2 dr r min r max " V " February 27, 2008 Physics for Scientists&Engineers 1 5 Center of Gravity for Half-Sphere Center of Gravity for Half-Sphere Now lets apply our knowledge to a simple object A half-sphere with constant density and radius R Clearly the center of gravity will lie on the z-axis February 27, 2008 Physics for Scientists&Engineers 1 6 Symmetry Planes Symmetry Planes Examine integrals Suppose an object is symmetric with respect to exchange In this case, the density is an even function of x and the boundaries are symmetric with respect to x = 0 => integration yields X = 0. In the case that an object is symmetric, the center of mass must lie within the symmetry plane. 1 1 1 ( ) ; ( ) ; ( ) V V V X x r dxdydz Y y r dxdydz Z z r dxdydz M M M ! ! !...
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This note was uploaded on 04/06/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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W8.3 - February 27, 2008 Physics for...

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