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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 32 "A Cylinder in a Uniform Field -
-" }}{PARA 256 "" 0 "" {TEXT -1 41 " An Application of the Divergence \+
Theorem" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT
-1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT -1 21 "School of Mathematic
s" }}{PARA 259 "" 0 "" {TEXT -1 12 "Georgia Tech" }}{PARA 260 "" 0 ""
{TEXT -1 21 "herod@math.gatech.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 27 " The
Divergence Theorem" }{TEXT -1 150 " states that if T is a solid regio
n bounded by a closed surface S which is piecewise smooth and F is a f
unction which has continuous derivatives, then" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " \+
" }{TEXT 262 0 "" }{TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT -1 19 " Dou
bleIntegral( F*" }{XPPEDIT 18 0 "eta" "6#%$etaG" }{TEXT -1 2 " d" }
{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 20 " ) = TripleIntegral(
" }{TEXT 257 0 "" }{TEXT -1 12 " div(F) dV )" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "where the double integral is ov
er S (" }{XPPEDIT 18 0 "eta" "6#%$etaG" }{TEXT -1 63 " is the outward \+
unit normal) and the triple integral is over T." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 454 " In this example, we compute the flux through a cylinder p
laced in a uniform, constant force field. Without using any computing \+
devices except the above Divergence Theorem, we know the flux through \+
the cylinder for such a field must be zero. It is, perhaps, of computa
tional curosity to verify this result by computing the flux over the s
urface of the cylinder through a computation of the first integral abo
ve. This will be broken into three parts." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "with(linalg): with
(plots):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
103 " Before computation of the first integral, we define the fiel
d, draw the cylinder, and verify that " }{TEXT 261 5 "Maple" }{TEXT
-1 28 " can compute the divergence." }}{PARA 0 "" 0 "" {TEXT -1 1 " "
}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "The Field" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "F:=(x,y,z)->[-1,3,2];" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 88 "fieldplot3d(F(x,y,z),x=-1..1,y=-1..1,z=-1..1,gri
d=[3,3,3],\n axes=NORMAL,color=BLUE);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA
3 "" 0 "" {TEXT -1 12 "The Cylinder" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 "Side:=(s,t)->[cos(t),sin(t),s];" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 34 "Top:=(r,t)->[r*cos(t),r*sin(t),1];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Bot:=(r,t)->[r*cos(t),r*sin(
t),0];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "P1:=plot3d(Side(
s,t),s=0..1,t=0..2*Pi,color=RED,axes=NORMAL):\nP2:=plot3d(Top(r,t),r=0
..1,t=0..2*Pi,color=magenta,axes=NORMAL):\nP3:=plot3d(Bot(r,t),r=0..1,
t=0..2*Pi,color=magenta,axes=NORMAL):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 20 "display([P1,P2,P3]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3
"" 0 "" {TEXT -1 24 "Computing the Divergence" }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 26 "diverge(F(x,y,z),[x,y,z]);" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 76 " As a result of the computation indicated in \+
the last line, we know that" }}{PARA 0 "" 0 "" {TEXT -1 40 " \+
" }{TEXT -1 2 " " }{TEXT 259 0 "" }
{TEXT -1 34 " TripleIntegral ( div(F) dV ) = 0." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "Our next goal is to make
the computation of the integral that comes first in the statement of \+
the Divergence Theorem. Because the computation of the integral that c
ame second was zero and because of the validity of the Divergence Theo
rem, we expect to get that the value of the first integral is zero. "
}}{PARA 0 "" 0 "" {TEXT -1 139 " We break the computation into thr
ee parts; the sides, the top and the bottom. We compute the integral a
long the sides and leave as an " }{TEXT 260 24 "Exercise for the Stude
nt" }{TEXT -1 69 " the computation of the flux over the top and bottom
of the cylinder." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 59 "The computation of the flux over the sides of the \+
cylinder." }}{PARA 0 "" 0 "" {TEXT -1 82 " Think a minute on what \+
the normal to the sides should be. Now we verify this." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "assume(s,real); assume(t,real);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Side:=(s,t)->[cos(t),sin(t)
,s];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "dSds:=diff(Side(s,t
),s);\ndSdt:=diff(Side(s,t),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 24 "N:=crossprod(dSdt,dSds);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We c
ompute the dotproduct of the field with this normal." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 27 "FdotN:=dotprod(F(x,y,z),N);" }}}{PARA 0 "
" 0 "" {TEXT -1 30 "We compute the first integral." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 42 "IntSide:=int(int(FdotN,s=0..1),t=0..2*Pi);"
}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 203 "You \+
compute IntTop and IntBottom -- the integral of the flux over the top \+
and the bottom of the cylinder. If you don't get that the sum of these
two is zero, you may need to think about your orientation." }}{EXCHG
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