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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 21 "The Theorem of Pappus" }}{PARA
257 "" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT -1 14 "jh
erod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 144 " There is a simple relatio
n between centroids and volumes of revolution which was first pointed \+
out by Pappus of Alexandria (circa 300 A.D.)." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 17 "Theorem of Pappus" }
{TEXT -1 22 ": When a plane region " }{XPPEDIT 18 0 "Omega" "6#%&Omega
G" }{TEXT -1 137 " is revolved about an axis which lies in its plane b
ut does not cross it, the volume of the resulting solid of revolution \+
is the area of " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 76 " mu
ltiplied by the circumference of the circle described by the centroid \+
of " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 1 "," }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 " In stating thi
s theorem, we assumed a complete revolution. If " }{XPPEDIT 18 0 "Omeg
a" "6#%&OmegaG" }{TEXT -1 111 " is only partially revolved about the g
iven axis, then the volume of the resulting solid is simply the area o
f " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 75 " multiplied by t
he length of the circular arc described by the centroid of " }
{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "
" {TEXT -1 31 " Here are two applications." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Application 1" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 249 "We have \+
a sphere of radius a. We remove the center cylindrical core of radius \+
a/2. What is the volume remaining? The method we use to work the probl
em is to find the centroid of the area left after removing the core an
d apply the Theorem of Pappus. " }}{PARA 0 "" 0 "" {TEXT -1 123 " \+
First we draw a picture of a cross section of what remains after takin
g out the core. To draw the picture, take a = 1." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "a:=1;\nplot
(\{sqrt(a^2-(a/2)^2),sqrt(a^2-x^2),\n -sqrt(a^2-(a/2)^2),-sqrt(a^2
-x^2)\},\nx=-a/2..a/2,y=-a..a,scaling=constrained,color=BLACK);\na:='a
':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 243 " To apply the Theore
m of Pappus, we find the y coordinate of the centroid. Is it not clear
that the x coordinate of the centroid is zero? (If this is not clear,
compute it!) Remember the formula for the y coordinate of the center \+
of mass: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
80 " y coordinate of the center of
mass = " }{XPPEDIT 18 0 "int(int(y,y),x)/int(int(1,y),x)" "6#*&
-%$intG6$-F%6$%\"yGF)%\"xG\"\"\"-F%6$-F%6$F+F)F*!\"\"" }{TEXT -1 0 ""
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "assume(a>0):\nint(int(y,y
=sqrt(a^2-(a/2)^2)..sqrt(a^2-x^2)),x=-a/2..a/2)/\nint(int(1,y=sqrt(a^2
-(a/2)^2)..sqrt(a^2-x^2)),x=-a/2..a/2);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 16 "ybar:=normal(%);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 89 "area:=int(int(1,y=sqrt(a^2-(a/2)^2)..sqrt(a^2-x^2)),x
=-a/2..a/2);\nvolume:=2*Pi*ybar*area;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(subs(a=1,volume));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 14 "Application 2 " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }
{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 110 " be a square with si
de length 2 and center at [0,2]. Find the volume of the solid of revol
ution upon rotating " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1
22 " about the line y = x." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 38 "First we draw a picture of the region " }
{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 22 " and of the line y = \+
x" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 109 "plot(\{[[1,1],[1,3],[-1,3],[-1,1],[1,1]],[[-2,-2],[2
,2]]\},\nx=-2..2,y=-2..4,scaling=constrained,color=BLACK);\n\n" }}}
{PARA 0 "" 0 "" {TEXT -1 40 "The centroid of the solid of the region \+
" }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 12 " is [0 , 2]." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "y2:=2;" }}}{PARA 0 "" 0 ""
{TEXT -1 418 " The problem before us is to find the distance the c
entroid travels as it rotates about this line y = x. We could use resu
lts from earlier calculus studies for finding the distance from a poin
t to a line. Instead, why not let Maple do the calculation using its d
elightful little geometry package? The following will compute the dist
ance from the point P = [0 ,2] to a line containing A = [-2 ,-2] and B
= [ 2 , 2]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(geometr
y):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "point(P,[0,y2]): poi
nt(A,[-2,-2]): point(B,[2,2]):line(L,[A,B]):" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 34 "r:=distance(P,L); SquareArea:=2*2;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "NewVolume:=2*Pi*r*SquareArea;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 10 "Assignment" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 157 "1. Calculate the \+
volume of the doughnut shaped region called a torus, obtained when the
circle of radius 1 centered at [3 ,0] is revolved about the y - axis.
" }}{PARA 0 "" 0 "" {TEXT -1 90 "2. Consider the area under the curve \+
of sin(x) and above the x axes between x = 0 and x = " }{XPPEDIT 18 0
"Pi;" "6#%#PiG" }{TEXT -1 93 ". What is the volume of the region obtai
ned when this region is rotated about the line y = x." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 25 "plot([x,sin(x)],x=0..Pi);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
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