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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 296 
"" 0 "" {TEXT -1 68 "Lesson 7:  Applications of Integration 5: Moments
 and Center of Mass" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 26 "Moments a
nd Center of Mass" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 25 "Center of ma
ss of a Wire " }}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "
center of mass" {TEXT -1 26 "Suppose  we  have a wire  " }{XPPEDIT 18 
0 "l " "6#%\"lG" }{TEXT -1 31 "  feet long  whose  density is " }
{XPPEDIT 18 0 "rho(x)" "6#-%$rhoG6#%\"xG" }{TEXT -1 31 "  pounds per f
oot at the point " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 95 " feet fr
om the left hand end of the wire.  What is the total mass of the wire \+
and where is its " }{TEXT 260 14 "center of mass" }{TEXT -1 70 ",    i
.e., the point cm about which the total moment of the wire is 0?" }}}
{PARA 15 "" 0 "Riemann sum" {TEXT 256 4 "Mass" }{TEXT -1 43 "  Chop th
e wire into  n small pieces each  " }{XPPEDIT 18 0 "Delta*x[i]" "6#*&%
&DeltaG\"\"\"&%\"xG6#%\"iGF%" }{TEXT -1 39 " feet long and pick an arb
itrary point " }{XPPEDIT 18 0 "c[i]" "6#&%\"cG6#%\"iG" }{TEXT -1 74 " \+
in each piece.  An approximation to the mass of the ith piece of wire \+
is " }{XPPEDIT 18 0 "rho(c[i])*Delta*x[i]" "6#*(-%$rhoG6#&%\"cG6#%\"iG
\"\"\"%&DeltaGF+&%\"xG6#F*F+" }{TEXT -1 45 ", so an approximation to t
he total mass   is " }{XPPEDIT 18 0 "Sum(rho(c[i])*Delta*x[i],i=1..n)
" "6#-%$SumG6$*(-%$rhoG6#&%\"cG6#%\"iG\"\"\"%&DeltaGF.&%\"xG6#F-F./F-;
F.%\"nG" }{TEXT -1 32 " .   This approximate mass is a " }{TEXT 260 
11 "Riemann sum" }{TEXT -1 33 "    approximating the integral   " }
{XPPEDIT 18 0 "Int(rho(x),x=0..l)" "6#-%$IntG6$-%$rhoG6#%\"xG/F);\"\"!
%\"lG" }{TEXT -1 71 ", and so the mass of the wire is defined as the v
alue of this integral." }}{PARA 15 "" 0 "approximate moment" {TEXT 
257 14 "Center of mass" }{TEXT -1 29 ":    Chopping as above,  the " }
{TEXT 260 18 "approximate moment" }{TEXT -1 53 "    of the ith piece a
bout the center of mass cm is  " }{XPPEDIT 18 0 "(c[i]-cm)*rho(c[i])*D
elta*x[i]" "6#**,&&%\"cG6#%\"iG\"\"\"%#cmG!\"\"F)-%$rhoG6#&F&6#F(F)%&D
eltaGF)&%\"xG6#F(F)" }{TEXT -1 40 " and so the total approximate momen
t is " }{XPPEDIT 18 0 "Sum((c[i]-cm)*rho(c[i])*Delta*x[i],i=1..n)" "6#
-%$SumG6$**,&&%\"cG6#%\"iG\"\"\"%#cmG!\"\"F,-%$rhoG6#&F)6#F+F,%&DeltaG
F,&%\"xG6#F+F,/F+;F,%\"nG" }{TEXT -1 64 ".  This is seen to be a Riema
nn sum approximating the integral  " }{XPPEDIT 18 0 "Int((x-cm)*rho(x)
,x=0..l)" "6#-%$IntG6$*&,&%\"xG\"\"\"%#cmG!\"\"F)-%$rhoG6#F(F)/F(;\"\"
!%\"lG" }{TEXT -1 127 ".   But the center of mass is defined as the po
int about which the total moment is zero so the integral satisfies the
 equation " }{XPPEDIT 18 0 "Int((x-cm)*rho(x),x=0..l)=0" "6#/-%$IntG6$
*&,&%\"xG\"\"\"%#cmG!\"\"F*-%$rhoG6#F)F*/F);\"\"!%\"lGF2" }{TEXT -1 
101 ".   Using properties of integrals, we can solve this equation for
 cm, to get the ratio of integrals  " }{XPPEDIT 18 0 "cm=Int(x*rho(x),
x=0..l)/Int(rho(x),x=0..l)" "6#/%#cmG*&-%$IntG6$*&%\"xG\"\"\"-%$rhoG6#
F*F+/F*;\"\"!%\"lGF+-F'6$-F-6#F*/F*;F1F2!\"\"" }{TEXT -1 146 " .   Not
e the top integral represents the total moment of the wire about its l
eft end (x=0) and the bottom integral is the total mass of the wire." 
}}{EXCHG {PARA 288 "" 0 "" {TEXT -1 143 "Exercise:   Find the center o
f mass of a  wire  1 foot long  whose density at a point x inches from
 the left end is  10 + x + sin(x)  lbs/inch." }}}}{SECT 0 {PARA 4 "" 
0 "" {TEXT -1 40 " Center of mass of a solid of revolution" }}{PARA 4 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "If  " }
{XPPEDIT 18 0 "y *`=` *f(x) >= 0" "6#1\"\"!*(%\"yG\"\"\"%\"=GF'-%\"fG6
#%\"xGF'" }{TEXT -1 6 " for  " }{XPPEDIT 18 0 "a <= x *`<=` b" "6#1%\"
aG*(%\"xG\"\"\"%#<=GF'%\"bGF'" }{TEXT -1 20 ", then let S be the " }
{TEXT 260 22 "solid of revolution   " }{TEXT -1 179 " obtained by rota
ting the region under the graph of f around the x axis. We know how to
 express the volume of S as an integral: Just integrate from a to b th
e  crossectional area " }{XPPEDIT 18 0 "Pi*f(x)^2" "6#*&%#PiG\"\"\"*$-
%\"fG6#%\"xG\"\"#F%" }{TEXT -1 24 " of the solid S to get  " }
{XPPEDIT 18 0 "Volume = Int(Pi*f(x)^2,x=a..b);" "6#/%'VolumeG-%$IntG6$
*&%#PiG\"\"\"*$-%\"fG6#%\"xG\"\"#F*/F/;%\"aG%\"bG" }}}{EXCHG {PARA 0 "
" 0 "center of mass" {TEXT -1 27 "Now how would we  find the " }{TEXT 
260 14 "center of mass" }{TEXT -1 203 "    of the solid, assuming it's
 made of a homogeneous material? Well, it's  clear that the center of \+
mass will be somewhere along the x-axis between a and b.    Let CM be \+
the center of mass.  Partition  " }{XPPEDIT 18 0 "[a,b]" "6#7$%\"aG%\"
bG" }{TEXT -1 21 " into n subintervals " }{XPPEDIT 18 0 "[x[i],x[i+1] \+
]" "6#7$&%\"xG6#%\"iG&F%6#,&F'\"\"\"F+F+" }{TEXT -1 36 " and using pla
nes  perpendicular to " }{XPPEDIT 18 0 "[a,b]" "6#7$%\"aG%\"bG" }
{TEXT -1 73 " approximate the solid S with the n disks where the ith o
ne has volume   " }{XPPEDIT 18 0 "Delta V[i] = Pi*f(x[i])^2*Delta x[i]
" "6#/*&%&DeltaG\"\"\"&%\"VG6#%\"iGF&**%#PiGF&*$-%\"fG6#&%\"xG6#F*\"\"
#F&F%F&&F26#F*F&" }}{PARA 0 "" 0 "" {TEXT -1 62 "Now the signed moment
 of the ith disk about the point CM is   " }{XPPEDIT 18 0 "M[i] = (x-C
M)* Delta V[i]" "6#/&%\"MG6#%\"iG*(,&%\"xG\"\"\"%#CMG!\"\"F+%&DeltaGF+
&%\"VG6#F'F+" }{TEXT -1 99 "  and the sum of these moments will be app
roximately 0, since CM is the center of mass.  If we let " }{XPPEDIT 
18 0 "Delta x[i]" "6#*&%&DeltaG\"\"\"&%\"xG6#%\"iGF%" }{TEXT -1 83 " g
o to zero  this  approximate equation becomes an equation for the cent
er of mass:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "CMequation :=  int((
x-CM)*Pi*f(x)^2,x=a..b) =0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Us
eing properties of integrals, we can solve this equation for CM." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "sol := solve(CMequation,\{CM\} )   \+
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "Notice that the center of m
ass of the solid of revolution is the same as the center of mass of a \+
wire whose density at " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 34 " is
 the area of the cross-section." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
126 "We can define a word cenmass  which takes a function  f, an inter
val [a,b], and locates the center of the solid of revolution." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "cenmass := proc(f,a,b)  \n int(x*f(
x)^2,x=a..b)/int(f(x)^2,x=a..b) end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 92 "For example, the center of  the solid obtained by rotating the \+
region R under the graph of  " }{XPPEDIT 18 0 "y=cos(x)" "6#/%\"yG-%$c
osG6#%\"xG" }{TEXT -1 21 " for x between 0 and " }{XPPEDIT 18 0 "pi/2
" "6#*&%#piG\"\"\"\"\"#!\"\"" }{TEXT -1 3 " is" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "cenmass(cos,0,Pi/2);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 73 "Now we can define a word to draw the solid and locate t
he center of mass." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 374 "drawit := pr
oc(f,a,b) \n local cm, solid;\n### WARNING: the definition of the type
 `symbol` has changed'; see help page for details\n  cm := plots[point
plot3d]([evalf(cenmass(f,a,b)),0,0],color=red,symbol=box,thickness=3):
\n  solid := plots[tubeplot]([x,0,0],x=a..b,radius=f(x),numpoints=20,t
ubepoints=30, style=wireframe); \nplots[display]([cm,solid],scaling=co
nstrained); end: " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Test this ou
t." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " drawit(cos+2,0,7);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "numFrames := 40;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "b := 10;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 69 "We can animate the motion of the center of mass as
 the solid changes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 89 "plots[display]( [seq(drawit(2+cos,0,(i/numFrames)*
b), i=1..numFrames)],\ninsequence=true);" }}}{EXCHG {PARA 288 "" 0 "" 
{TEXT 262 9 "Exercise:" }{TEXT -1 63 "  Find the center of mass of a h
omogeneous hemispherical solid." }}{PARA 288 "" 0 "" {TEXT 263 9 "Exer
cise:" }{TEXT -1 227 "  A homogeneous solid is in the shape of a parab
olic solid of revolution obtained by rotating the graph of  y = x^2 , \+
x in [0,a]  around the the y axis, for some positive number a.  If the
 center of mass is at  y= 10, what's a?" }}}}}}{MARK "0 0 0" 5 }
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