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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 280 0 "" }{TEXT 
281 23 "Calculus IV with Maple\n" }{TEXT 282 32 "Copyright 2002, Dr. J
ack Wagner\n" }{TEXT -1 30 "j.wagner@intelligentsearch.com" }}}{EXCHG 
{PARA 4 "" 0 "" {TEXT -1 11 "Lesson 12: " }{TEXT 283 48 "The Integral \+
Theorems of Green, Stokes and Gauss" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 9 "Topics:  " }{TEXT 256 136 "Orient
ation of volumes, surfaces and closed curves.  The classical theorems \+
of Green, Stokes and Gauss  are presented and demonstrated. " }{TEXT 
257 28 "\nMaple commands introduced: " }{TEXT 284 12 "animatecurve" }}
}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "Int(P,x)+Int(Q,y);" "6#,&-%$Int
G6$%\"PG%\"xG\"\"\"-F%6$%\"QG%\"yGF)" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "Int(Int(diff(Q,x)-diff(P,y),x),y);" "6#-%$IntG6$-F$6$,&-%%diffG6$%
\"QG%\"xG\"\"\"-F*6$%\"PG%\"yG!\"\"F-F2" }{TEXT -1 0 "" }}{PARA 0 "" 
0 "" {XPPEDIT 18 0 "Int(curlX*n,A);" "6#-%$IntG6$*&%&curlXG\"\"\"%\"nG
F(%\"AG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(X,s);" "6#-%$IntG6$%\"X
G%\"sG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Int(divX,V);
" "6#-%$IntG6$%%divXG%\"VG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(X*n,
A);" "6#-%$IntG6$*&%\"XG\"\"\"%\"nGF(%\"AG" }{TEXT -1 1 " " }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 299 "The orie
ntability of surfaces has previously been discussed in the context of \+
surface integrals.  Because the integral theorems relate integrals ove
r closed sets to integrals over their boundaries we must now be concer
ned with the orientation of the boundaries of closed sets relative to \+
the sets.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 28 "Recall that a closed curve, " }{XPPEDIT 18 0 "gamma(t);" "6#-%&
gammaG6#%\"tG" }{TEXT -1 137 ", is positively oriented if increasing v
alues of t result in counterclockwise movement around the curve.  Alte
rnatively, a closed curve, " }{XPPEDIT 18 0 "gamma(t);" "6#-%&gammaG6#
%\"tG" }{TEXT -1 44 ", which is the boundary of a subset, D, of  " }
{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }{TEXT -1 104 ", is positively o
riented if D is on the left as you move around the curve with increasi
ng values of t. \n" }}}{EXCHG {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 
258 15 "Green's Theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 "Let D be a closed subset of " }{TEXT 259 
1 "R" }{TEXT -1 17 "  with boundary  " }{XPPEDIT 18 0 "gamma;" "6#%&ga
mmaG" }{TEXT -1 30 ".  Let P and Q be functions,  " }{TEXT 260 1 "R" }
{TEXT -1 1 "x" }{TEXT 261 1 "R" }{TEXT -1 2 "->" }{TEXT 262 1 "R" }
{TEXT -1 69 ", defined and continuous on D with continuous derivatives
.  Then,    " }{XPPEDIT 18 0 "Int(P,x)+Int(Q,y);" "6#,&-%$IntG6$%\"PG%
\"xG\"\"\"-F%6$%\"QG%\"yGF)" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(Int
(diff(Q,x)-diff(P,y),x),y);" "6#-%$IntG6$-F$6$,&-%%diffG6$%\"QG%\"xG\"
\"\"-F*6$%\"PG%\"yG!\"\"F-F2" }{TEXT -1 23 " .   Alternatively, if " }
{TEXT 263 1 "X" }{TEXT -1 28 " is a vector field on D and " }{TEXT 
264 1 "n" }{TEXT -1 20 " the unit normal on " }{XPPEDIT 18 0 "gamma;" 
"6#%&gammaG" }{TEXT -1 57 ", we have the geometric, coordinate free fo
rmulation,    " }{XPPEDIT 18 0 "Int(X*n,s);" "6#-%$IntG6$*&%\"XG\"\"\"
%\"nGF(%\"sG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(divX,A);" "6#-%$In
tG6$%%divXG%\"AG" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 265 12 "Example 11.1" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 " Find the area of
 an ellipse of semi-major axis a and semi-minor axis b." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "As in most cases of \+
closed curves, we will use polar coordinates to define the curve " }
{XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 128 " that encloses the \+
elliptic area in the plane.  Notice that this curve is positively orie
nted with respect to the enclosed area." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "restart: with(LinearAl
gebra):with(VectorCalculus):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "P
arametrization of the ellipse" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 35 "e := <a*cos(theta), b*sin(theta)>;\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Now we define a carefully
 selected vector field. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 
"X := VectorField(<x/2, y/2>, 'cartesian'[x, y]);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 23 "Divergence(X, [x, y]);\n" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 57 "Therefore, from the alternative form of Green's t
heorem, " }{XPPEDIT 18 0 "Int(X*n,s)" "6#-%$IntG6$*&%\"XG\"\"\"%\"nGF(
%\"sG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(``,A);" "6#-%$IntG6$%!G%
\"AG" }{TEXT -1 4 " = A" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "
de := diff(e, theta);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "
nds := <de[2], -de[1]>;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Now \+
reparametrize X in polar coordinates. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "X := evalVF(X, e);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "dA := X.nds;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 24 "int(dA, theta=0..2*Pi);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
43 " In the same vein, setting P=-y and Q=x,   " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{XPPEDIT 18 0 "Int(P,y)-Int(Q,x);" "6#,&-%$IntG6$%\"PG%
\"yG\"\"\"-F%6$%\"QG%\"xG!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(
Int(diff(P,x)-diff(Q,y),x),y);" "6#-%$IntG6$-F$6$,&-%%diffG6$%\"PG%\"x
G\"\"\"-F*6$%\"QG%\"yG!\"\"F-F2" }{TEXT -1 4 " = 2" }{TEXT 273 1 "A" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 
"This formula is useful for working with parameterized curves, but wor
king with curves  parameterized by trigonometric functions can be tric
ky." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
266 12 "Example 11.2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 52 "Find the area enclosed by the y axis and the cu
rve, " }{XPPEDIT 18 0 "gamma(theta) = [cos(x)^2, sin(x)^3];" "6#/-%&ga
mmaG6#%&thetaG7$*$-%$cosG6#%\"xG\"\"#*$-%$sinG6#F-\"\"$" }{TEXT -1 3 "
,  " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 6 " = 0.." }
{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 " " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots): with(L
inearAlgebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "C := <cos(theta)^2, sin(theta)^3>;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 36 "plot([C[1], C[2], theta=0..2*Pi]);\011\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "X := C[1]:Y := C[2]:" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "dX := diff(C[1], theta); d
Y := diff(C[2], theta);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
31 "int(X*dY-Y*dX, theta=0..2*Pi);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 32 "         So we try the top half." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "plot([C[1], C[2], theta=0..Pi]);\n" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 29 "int(X*dY-Y*dX, theta=0..Pi);\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "Now what is the problem?  The pro
blem is that the curve we are looking at is really two curves superimp
osed on one another, each oriented oppositely.  The plots from 0 to " 
}{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 12 "  and
 from  " }{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 
4 " to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 200 " are identical \+
but in opposite directions.  To see this use the animatecurve function
 in the plots package.  This function shows the actual drawing of the \+
curve for increasing values of the parameter." }}}{EXCHG {PARA 0 "" 0 
"" {HYPERLNK 17 "animatecurve" 2 "animatecurve" "" }{MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Right click on the plot and sel
ect " }{TEXT 274 4 "play" }{TEXT -1 10 " from the " }{TEXT 275 8 "anim
ate " }{TEXT -1 9 "sub-menu." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 54 "animatecurve([C[1], C[2], theta=0..Pi/2], color=red);\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "animatecurve([C[1], C[2], th
eta=Pi/2..Pi], color=red);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "T
his makes very explicit the fact that the plot from 0 to " }{XPPEDIT 
18 0 "Pi;" "6#%#PiG" }{TEXT -1 144 "  traverses the curve twice; once \+
in each direction.  The integral over this interval is therefore 0.  W
ith this knowledge in hand we try again." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "int(X*dY-Y*dX, theta=0..Pi/2);\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 37 "What happens when you integrate from " }{XPPEDIT 
18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 4 " to " }{XPPEDIT 
18 0 "Pi;" "6#%#PiG" }{TEXT -1 47 " ?   The critical role of orientati
on is clear!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int(X*dY-Y*
dX, theta=Pi/2..Pi);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 4 "" 0 "" {TEXT 267 15 "Stokes' Theorem" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 16 "Stokes theorem  " }{XPPEDIT 18 0 "Int(curlX*n,A);" "6#-
%$IntG6$*&%&curlXG\"\"\"%\"nGF(%\"AG" }{TEXT -1 4 " =  " }{XPPEDIT 18 
0 "Int(X,s)" "6#-%$IntG6$%\"XG%\"sG" }{TEXT -1 113 "  , is a generaliz
ation of Green's theorem to non-planar surfaces. To see this, consider
 the projection operator " }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT -1 
25 " onto the x-y plane. Let " }{XPPEDIT 18 0 "tau;" "6#%$tauG" }
{TEXT -1 31 " be the unit tangent vector to " }{XPPEDIT 18 0 "gamma(t)
;" "6#-%&gammaG6#%\"tG" }{TEXT -1 58 ", the projection of the boundary
 of the surface.  Then, = " }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT 
-1 1 " " }{XPPEDIT 18 0 "Int(X,s);" "6#-%$IntG6$%\"XG%\"sG" }{TEXT -1 
3 " = " }{XPPEDIT 18 0 "Int([P, Q]*tau,s);" "6#-%$IntG6$*&7$%\"PG%\"QG
\"\"\"%$tauGF*%\"sG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int([P, Q]*diff
(gamma,s),s = a .. b);" "6#-%$IntG6$*&7$%\"PG%\"QG\"\"\"-%%diffG6$%&ga
mmaG%\"sGF*/F/;%\"aG%\"bG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int([P*di
ff(x(t),t)+Q*diff(y(t),t)],t);" "6#-%$IntG6$7#,&*&%\"PG\"\"\"-%%diffG6
$-%\"xG6#%\"tGF1F*F**&%\"QGF*-F,6$-%\"yG6#F1F1F*F*F1" }{TEXT -1 2 " =
" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Int(P,x)+Int(Q,y);" "6#,&-%$IntG6$%
\"PG%\"xG\"\"\"-F%6$%\"QG%\"yGF)" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 13 "         Let " }{XPPEDIT 18 0 "alpha,beta,gamma;" "6
%%&alphaG%%betaG%&gammaG" }{TEXT -1 24 "  be the angles between " }
{TEXT 276 1 "n" }{TEXT -1 52 " and the x, y, and z axes respectively. \+
 Note that  " }{TEXT 277 2 "dA" }{XPPEDIT 18 0 "cos(gamma);" "6#-%$cos
G6#%&gammaG" }{TEXT -1 3 " = " }{TEXT 278 4 "dxdy" }{TEXT -1 6 "  and \+
" }}{PARA 0 "" 0 "" {TEXT 279 2 "n=" }{XPPEDIT 18 0 "[cos(alpha), cos(
beta), cos(gamma)];" "6#7%-%$cosG6#%&alphaG-F%6#%%betaG-F%6#%&gammaG" 
}{TEXT -1 55 " . Considering only the projection onto the x-y plane, \+
" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "Int(curlX*n,A);" "6#-%$IntG
6$*&%&curlXG\"\"\"%\"nGF(%\"AG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(
Int((diff(Q,x)-diff(P,y))*cos(gamma)/cos(gamma),x),y);" "6#-%$IntG6$-F
$6$*(,&-%%diffG6$%\"QG%\"xG\"\"\"-F+6$%\"PG%\"yG!\"\"F/-%$cosG6#%&gamm
aGF/-F66#F8F4F.F3" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Int(Int(diff(Q,x)
-diff(P,y),x),y);" "6#-%$IntG6$-F$6$,&-%%diffG6$%\"QG%\"xG\"\"\"-F*6$%
\"PG%\"yG!\"\"F-F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "Stokes the
orem is therefore the result of summing the results of Green's theorem
 over the projections onto each of the coordinate planes." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 12 "Example \+
11.3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 49 "Use Stokes' theorem to find the line integral of " }{XPPEDIT 
18 0 "(x+y)^3*dx-(x-y)^3*dy+z^3*dz;" "6#,(*&,&%\"xG\"\"\"%\"yGF'\"\"$%
#dxGF'F'*&,&F&F'F(!\"\"F)%#dyGF'F-*&%\"zGF)%#dzGF'F'" }{TEXT -1 51 "  \+
around the intersection of the elliptic cylinder " }{XPPEDIT 18 0 "[3*
cos(theta), 2*sin(theta), z];" "6#7%*&\"\"$\"\"\"-%$cosG6#%&thetaGF&*&
\"\"#F&-%$sinG6#F*F&%\"zG" }{TEXT -1 16 "  and the plane " }{XPPEDIT 
18 0 "x+y+2*z = 2;" "6#/,(%\"xG\"\"\"%\"yGF&*&\"\"#F&%\"zGF&F&F)" }
{TEXT -1 4 " .  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 65 "In approaching any problem of this sort, a picture
 is invaluable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "restart:
 with(plots): with(LinearAlgebra):with(VectorCalculus):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "S := <3*cos(theta), 2*sin(theta), z
>;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "P1 := plot3d(S, theta
=0..2*Pi, z=-6..6, color=blue, style=wireframe, grid=[40,40]):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "P := <x, y, 1-x/2-y/2>;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "P2 := plot3d(P, x=-4..4, y=-
4..4, color=pink, style=patchnogrid):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "F := VectorField(<(x+y)^3, -(x-y)^3, z^3>, 'cartesian
'[x, y, z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "P3 := field
plot3d(F, x=-4..4, y=-4..4, z=-6..6, color=black, arrows=SLIM, grid=[5
, 5, 5]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "display(P1, P2
, P3, axes=framed, labels=[x, y, z]);\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 44 "To apply Stokes' theorem, we will integrate " }{TEXT 285 
6 "curl F" }{TEXT -1 62 " over the elliptic area cut out by the cylind
er on the plane. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "curlF \+
:= Del &x F;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Find the unit n
ormal to the plane." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dP[x
] := diff(P, x);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dP[y]
 := diff(P, y);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "N := d
P[x] &x dP[y];\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Normalizing..
.note n is consistent with the orientation of  the x-y plane" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "n := Normalize(N, 2, inplace
);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "The dot product " }{TEXT 
287 12 "(curl F) . n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "cur
lFn := evalVF(curlF, <x, y, z>).n;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 45 "Find dA for the cross section of the cylinder" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 39 "C := <3*r*cos(theta), 2*r*sin(theta)>;\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J := Jacobian(C, [r, th
eta]);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "How does this compare
 to the Jacobian for the circular transformation " }{XPPEDIT 18 0 "<r*
cos(theta), r*sin(theta)>" "6#-%$<,>G6$*&%\"rG\"\"\"-%$cosG6#%&thetaGF
(6#*&F'F(-%$sinG6#F,F(" }{TEXT -1 1 "?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "dA := simplify(Determinant(J));\n" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 15 "Reparameterize " }{TEXT 286 6 "curlFn" }{TEXT -1 
1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "curlFn := subs(\{x=
S[1], y=S[2], z=S[3]\}, curlFn);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 20 "Put it all together!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
44 "Int(Int(curlFn*dA, r=0..1), theta=0..2*Pi);\n" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 12 "Example 11.4" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Evaluate the li
ne integral of  " }{XPPEDIT 18 0 "F = y^2*dx+x^2*dy+z^2*dz;" "6#/%\"FG
,(*&%\"yG\"\"#%#dxG\"\"\"F**&%\"xGF(%#dyGF*F**&%\"zGF(%#dzGF*F*" }
{TEXT -1 74 "  around the curve formed by the intersection of the elli
ptic paraboloid, " }{XPPEDIT 18 0 "z = 2*x^2+3*y^2;" "6#/%\"zG,&*&\"\"
#\"\"\"*$%\"xGF'F(F(*&\"\"$F(*$%\"yGF'F(F(" }{TEXT -1 17 " , and the p
lane " }{XPPEDIT 18 0 "z = 3*y+5*x+2;" "6#/%\"zG,(*&\"\"$\"\"\"%\"yGF(
F(*&\"\"&F(%\"xGF(F(\"\"#F(" }{TEXT -1 144 " .\nIt is not always neces
sary to reparameterize in terms of alternative parameterizations.  Som
etimes Cartesian coordinates do just as nicely. \n" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 63 "restart: with(plots): with(LinearAlgebra):w
ith(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "S :
= <x, y, 2*x^2+3*y^2>;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 
"P := <x, y, 3*y+5*x+2>;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 89 "Now we need a picture of the intersection
 and a computation of the limits of integration." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 65 "P1 := plot3d(S, x=-2..4, y=-Pi..Pi, color=blu
e, style=wireframe):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "P2 \+
:= plot3d(P, x=-2..4, y=-Pi..Pi, color=pink, style=patchnogrid):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq := S[3]=P[3];\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "In
tersection of Plane and Surface" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 21 "e1 := solve(eq, y); \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "eq1 := 33-24*x^2+60*x=0;  \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "x limits of integration " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "X := fsolve(eq1, x);       \+
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "e2 := solve(eq, x);\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eq2 := 41-24*y^2+24*y; \+
   \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 24 "y limits of integration " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "Y := fsolve(eq2, y)\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "The equation of the curve
 of intersection comes in two parts." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "E1 := subs(y=e1[1], P);\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "E2 := subs(y=e1[2], P);\n" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 92 "P3 := spacecurve(\{evalm(E1), evalm(E2)\}, x=-2.
.4, color=black, thickness=3, numpoints=1000):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 83 "display(\{P1, P2, P3\}, labels=[x, y, z], axes
=framed, view=[-2..4, -2..4, 0..25]);\n\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 28 "Now we compute the integral." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 57 "F := VectorField(<y^2, x^2, z^2>, 'cartesian'[x, y,
 z]);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "curlF := Del &x \+
F;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dP[x] := diff(P, x)
;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dP[y] := diff(P, y);
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "N := dP[x] &x dP[y];
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "n := Normalize(N, 2, in
place);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "This is our unit nor
mal to the surface enclosed by the curve." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "J := Jacobian(S, [x, y, z]);\n" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 26 "J := DeleteColumn(J, [3]);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 51 "dA := simplify(sqrt(Determinant(Transpose
(J).J)));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "curlFn := ev
alVF(curlF, <x, y, z>).n;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
49 "Int(Int(curlFn*dA, x=X[1]..X[2]), y=Y[1]..Y[2]);\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(%);\n" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 198 " If you try this with some trigonometric parameteri
zation you will find that it is not only more complicated, but Maple w
ill not produce a numerical answer even after a much longer processing
 time. " }}{PARA 0 "" 0 "" {TEXT -1 2 "  " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT 270 18 "Divergence Theorem" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "G
reen's theorem also generalizes to volumes. Let V be a closed subset o
f " }{XPPEDIT 18 0 "R^3;" "6#*$%\"RG\"\"$" }{TEXT -1 88 "  with a boun
dary consisting of surfaces oriented by outward pointing normals.  The
n,   " }{XPPEDIT 18 0 "Int(divX,V) = Int(X*n,S);" "6#/-%$IntG6$%%divXG
%\"VG-F%6$*&%\"XG\"\"\"%\"nGF-%\"SG" }{TEXT -1 103 "  The idea is to s
lice the volume into thin slices.  On each slice, Green's theorem hold
s in the form, " }{XPPEDIT 18 0 "Int(X*n,s) = Int(divX,A);" "6#/-%$Int
G6$*&%\"XG\"\"\"%\"nGF)%\"sG-F%6$%%divXG%\"AG" }{TEXT -1 83 " .  By su
mming over the slices and taking limits we obtain the divergence theor
em. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
271 12 "Example 11.5" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 96 "Find the flux, through the surface of a sphere \+
of radius R , of the vector field F=[x, y, z] .  " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots): with(LinearAlgebra): w
ith(VectorCalculus):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 57 "The sphere is parameterized with geographic coo
rdinates. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "V := <r*cos(p
hi)*cos(theta), r*cos(phi)*sin(theta), r*sin(phi)>;\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "F := VectorField(<x, y, z>, 'cartes
ian'[x, y, z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Here is a pict
ure of what we are talking about." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "S := subs(r=2, V);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 86 "P1 := fieldplot3d(F, x=-2..2, y=0..2, z=0..2, color=r
ed, grid=[4, 4, 4], arrows=SLIM):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "P2 := plot3d(S, theta=0..Pi, phi=0..Pi/2, color=blue,
 style=wireframe):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "displ
ay(P1, P2, scaling=constrained, axes=framed);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 35 "J := Jacobian(V, [r, phi, theta]);\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "dV := simplify(sqrt(Determin
ant(Transpose(J).J))) assuming real;\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "divF := Divergence(F, [x, y, z]); \n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 37 "Integrate over the positive quadrant." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Int(Int(Int(divF*dV, r=0..R
), phi=0..Pi/2), theta=0..Pi);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "simplify(4*value(%), symbolic);\n" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
272 8 "Practice" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "1. Given the e
llipsoid,  " }{XPPEDIT 18 0 "98*x^2+4*y^2+36*z^2 = 36;" "6#/,(*&\"#)*
\"\"\"*$%\"xG\"\"#F'F'*&\"\"%F'*$%\"yGF*F'F'*&\"#OF'*$%\"zGF*F'F'F0" }
{TEXT -1 69 " , find the flux through the surface of the vector fields
 defined by:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "     a)    " }
{XPPEDIT 18 0 "[x/sqrt(x^2+y^2+z^2), y/sqrt(x^2+y^2+z^2), z/sqrt(x^2+y
^2+z^2)];" "6#7%*&%\"xG\"\"\"-%%sqrtG6#,(*$F%\"\"#F&*$%\"yGF,F&*$%\"zG
F,F&!\"\"*&F.F&-F(6#,(*$F%F,F&*$F.F,F&*$F0F,F&F1*&F0F&-F(6#,(*$F%F,F&*
$F.F,F&*$F0F,F&F1" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "     \+
b)   " }{XPPEDIT 18 0 "[x^2/sqrt(x^2+y^2+z^2), y^2/sqrt(x^2+y^2+z^2), \+
z^2/sqrt(x^2+y^2+z^2)];" "6#7%*&%\"xG\"\"#-%%sqrtG6#,(*$F%F&\"\"\"*$%
\"yGF&F,*$%\"zGF&F,!\"\"*&F.F&-F(6#,(*$F%F&F,*$F.F&F,*$F0F&F,F1*&F0F&-
F(6#,(*$F%F&F,*$F.F&F,*$F0F&F,F1" }{TEXT -1 2 "  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 13 "2. Integrate " }{XPPEDIT 18 0 "3*x*y^3*dx+4*y^2
*x*dy;" "6#,&**\"\"$\"\"\"%\"xGF&%\"yGF%%#dxGF&F&**\"\"%F&*$F(\"\"#F&F
'F&%#dyGF&F&" }{TEXT -1 78 "  counterclockwise around the square forme
d by the lines,  y=1, y=3, x=1, x=3." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "3. Consider the curve f
ormed by the intersection of the surface " }{XPPEDIT 18 0 "z = ln(x+y)
;" "6#/%\"zG-%#lnG6#,&%\"xG\"\"\"%\"yGF*" }{TEXT -1 84 "  and the cyli
nder of radius 1 centered at the point (2, 2). Find the integral of   \+
" }{XPPEDIT 18 0 "5*x*y^2*dx+3*x*y^2*dy;" "6#,&**\"\"&\"\"\"%\"xGF&%\"
yG\"\"#%#dxGF&F&**\"\"$F&F'F&F(F)%#dyGF&F&" }{TEXT -1 20 " around this
 curve. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 13 "5. Integrate " }{XPPEDIT 18 0 "x*dx-z^2*dy+y^2*dz;
" "6#,(*&%\"xG\"\"\"%#dxGF&F&*&%\"zG\"\"#%#dyGF&!\"\"*&%\"yGF*%#dzGF&F
&" }{TEXT -1 123 "  around the curve formed by the intersection of \nt
he cylinder of radius 2 whose long axis is the x axis, and the surface
, " }{XPPEDIT 18 0 "z+4 = 2*x^2;" "6#/,&%\"zG\"\"\"\"\"%F&*&\"\"#F&*$%
\"xGF)F&" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }
{TEXT -1 3 " .\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }