Density and Center of Mass

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Earl Atkinson
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1 Density and Center of Mass Objectives Integrate a density function to find Total Population or Total Mass Integrate or use Suation to find Center of Mass of a Substance with Certain Density Density Density of population is easured in quantity per unit area or per unit volue. For exaple, people per square ile, bacteria per cubic centieter (cc). Density of a substance is easured as ass per unit volue. For exaple, gras per cubic centieter. Finding Total Quantity fro A Density Function Exaple 1 Suppose Cities A, B, and C are as follows: a) City A is a ile square region bounded on the east by a river. The population density, P, is a function of the distance r iles fro the river. That is, P f(r) people/square ile. b) City B is a circular city of radius iles, where the population density, P g(x) people/square ile is a function of the distance, x, fro the city center. c) City C is a very narrow canyon along a highway of length 6k. The population density P h(q) people/k, is a function of the distance q k fro the outh of the canyon. Express each of the populations as a definite integral. 1

2 a) City A b) City B c) City C 1. Since density is constant along a vertical strip, we divide the region into vertical rectangular strips of width r, and height. So the area of a strip is approxiately r square iles.. The population per rectangular strip is approxiately: (area of slice in square iles)(density in people per square ile) r f(r) 3. The total population of City A is f(r)dr people 1. Since the density is constant along a ring around the center of the city, we divide the region into rings of width x, and circuference πx.. The area of each ring is approxiately πx x square iles. 3. The population per ring is approxiately (area of slice in square iles)(density in people per square ile) 4. Total population of City B is πxg(x) x πxg(x)dx people 1. Divide the line into subintervals of length q k.. The population per segent is approxiately (length of segent)(density in people per k) h(q) q

3 3. The total population of City C is 6 h(q)dq people Exaple A rod has length eters. At a distance x eters fro the left endpoint of the rod, the density is given by δ(x) + 6x g/. Find the total ass of the rod. The total ass of the rod is given by + 6xdx x + 3x g Center of Mass Finite Systes Finding the center of ass of a finite syste of n asses i, along the x axis, each located at coordinate x i can be thought of as finding where to put the fulcru of a see-saw to balance n people of weights i located at position x i. Center of Mass x Su of the Moents of the Mass Total Mass of the Syste Exaple 3 Find the center of ass of a syste containing three point asses of 5g, 3g, and 1g located at x 1, x 1, and x respectively. 3

4 x 5( 1) + 3(1) + 1() Continuous Systes Let δ(x) be the ass density (ass per unit length) at a point x. Then Center of Mass x b a xδ(x)dx b a δ(x)dx Exaple 4 A rod of length 3 eters has density δ(x) 1+x gras/eter, and is positioned along the x-axis with left endpoint at. Find the center of ass of the rod. x x(1 + x )dx 1 + x dx x + x3 dx 1 + x dx x + x4 4 3 x + x (1) eters 4

5 Mass ay not always be distributed along a straight line. What if we have a -D or 3-D object, and we want to deterine its center of ass? Our text only deals with objects of constant density here-if you would like to know how this differs when the density is not constant throughout the entire object, see e! For the following, let A x (x) be the area of a slice of the object, perpendicular to the x-axis. Let A y (y) and A z (z) be defined in the sae anner. Note that for a -D figure, you will not have A z (z), and A x (x) and A y (y) are interpreted as the height of a slice. Theore 1 For a region R, of constant density, δ, the center of ass of R is given by the point ( x, ȳ, z), where x, ȳ, and z are defined as follows: x xδax (x)dx Total Mass ȳ yδay (y)dy Total Mass z zδaz (z)dz Total Mass Exaple 5 Find the coordinates of the center of ass of the isosceles triangle, whose vertices are at (, 1), (, 1 ), and (1, ). The triangle has constant density, δ, and ass. Since density is ass per unit area, we ust have δ 1/. It will also be beneficial to know the equations of the lines which define the sides of our triangle. Clearly, x is one of those lines. The other two lines are y 1 x 1 and y 1 1 x. We now find x. x 1 δx [ ( 1 1x) ( 1x 1)] dx 1 x(1 x)dx 1 x x dx 5

6 1 x x dx ( x x3 3 ) 1 ( ) ( 1 6 ) 1 3 It ay be obvious to you that ȳ, but let s show that this is truly the case anyway. We convert our lines into functions of y rather than functions of x, to obtain x y + 1 and x 1 y. It should be evident fro the picture, that we will need two integrals to copute the nuerator of our ȳ. 1/ ȳ δy(1 y)dy + 1/ δy(1 + y)dy [ y 1/ y dy + ] 1/ y + y dy ] [( y y3 3 ) 1/ + ( y + y3 3 ) 1/ ( ( 1/) ( 1/)3 ) + ( (1/) 3 + (1/)3 ) 3 That is, the center of ass for the given object is at (1/3, )! 6

(b)using the left hand end points of the subintervals ( lower sums ) we get the aprroximation

(b)using the left hand end points of the subintervals ( lower sums ) we get the aprroximation (1) Consider the function y = f(x) =e x on the interval [, 1]. (a) Find the area under the graph of this function over this interval using the Fundamental Theorem of Calculus. (b) Subdivide the interval