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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 41 "Maple Illustrations of It
erated Integrals" }}{PARA 19 "" 0 "" {TEXT -1 13 "J. A. Ziegler" }}
{PARA 256 "" 0 "" {TEXT -1 37 "Southern Polytechnic State University" 
}}{PARA 257 "" 0 "" {TEXT -1 17 "jziegler@spsu.edu" }}{PARA 258 "" 0 "
" {TEXT 265 44 "http://www2.SPSU.edu/math/ziegler/index.html" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 349 "Abstract:  A method is given for using M
aple to illustrate the usual geometric interpretation of the evaluatio
n of iterated double integrals in rectangular and sphereical coordinat
es.  This is suitable for student use, individually, or in a computer \+
laboratory setting.  An elaboration of this method yields illustration
s suitable for classroom use." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "A Volume in Rectangular Coordinat
es" }}{EXCHG {PARA 4 "" 0 "" {TEXT -1 23 "Sweeping Out the Volume" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 153 "A typical exercise in evaluating double integrals by using ite
rated integrals is to find the volume enclosed between the graph of th
e function f(x,y)  = " }{XPPEDIT 18 0 "4-(x/2)^2-(y/3)^2;" "6#,(\"\"%
\"\"\"*$*&%\"xGF%\"\"#!\"\"\"\"#F**$*&%\"yGF%\"\"$F*\"\"#F*" }{TEXT 
-1 68 " and the x,y-plane, over the region R enclosed by the curves x \+
= 0, " }}{PARA 0 "" 0 "" {TEXT -1 19 "y = x + 3, and y = " }{XPPEDIT 
18 0 "x^2+1;" "6#,&*$%\"xG\"\"#\"\"\"\"\"\"F'" }{TEXT -1 51 ".  Here R
 is \"y-simple.\"  The \"lower curve\" is y = " }{XPPEDIT 18 0 "x^2+1;
" "6#,&*$%\"xG\"\"#\"\"\"\"\"\"F'" }{TEXT -1 38 " and the \"upper curv
e\" is y = x + 3.  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "The volum
e between the surface z = f(x,y) and the x,y-plane over the region R i
s given by the integral " }{XPPEDIT 18 0 "Int(Int(z,x = R .. ``),y = `
` .. ``);" "6#-%$IntG6$-F$6$%\"zG/%\"xG;%\"RG%!G/%\"yG;F-F-" }{TEXT 
-1 51 ".  Its value may be found via the iterated inegral " }{XPPEDIT 
18 0 "int(int(4-(x/2)^2-(y/3)^2,y = x^2+1 .. x+3),x = 0 .. 2);" "6#-%$
intG6$-F$6$,(\"\"%\"\"\"*$*&%\"xGF*\"\"#!\"\"\"\"#F/*$*&%\"yGF*\"\"$F/
\"\"#F//F3;,&*$F-\"\"#F*\"\"\"F*,&F-F*\"\"$F*/F-;\"\"!\"\"#" }{TEXT 
-1 107 ".  Maple can be used in the following way to help us visualize
 what is meant by \"sweeping out the volume\". " }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 36 "restart:with(plots):with(plottools):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We begin by defining the function
, f, and two functions whose graphs are the upper and lower curves, re
spectively." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "f:=(x,y)->4-
(x/2)^2-(y/3)^2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Uc:=x->
x+3:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Lc:=x->x^2+1:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Next, we create the graphs of f o
ver R and of the curves which bound R.  We collect the latter as R." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "s:=plot3d(f(x,y),x=0..2,y=
Lc(x)..Uc(x),grid=[10,5],style=wireframe,color=black):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "L1:=line([0,1,0],[0,3,0],color=blue
,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "L2:=space
curve(\{[x,Uc(x),0],[x,Lc(x),0]\},x=0..2,color=blue,thickness=2):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "R:=\{L1,L2\}:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "display(s,R,view=[0..2,0..5,0..4],o
rientation=[-23,56],axes=normal);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
161 "To help us to see the solid, we create the graphs of the lines jo
ining the vertices of R with the corresponding points on the graph of \+
f.  We collect these as C." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
55 "C1:=line([0,1,0],[0,1,f(0,1)],color=green,thickness=3):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "C2:=line([0,3,0],[0,3,f(0,3)
],color=green,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
55 "C3:=line([2,5,0],[2,5,f(2,5)],color=green,thickness=3):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "C:=\{C1,C2,C3\}:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "display(s,R,C,view=[0..2,0..5,0..4]
,orientation=[-23,56],axes=normal);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 227 "Now we define, as an \"image-valued\" function, that portion o
f the plane which \"sweeps out\" the volume as described by the \"oute
r\" integration.  The area of this portion of the plane is, of course,
 given by the \"inner\" integral." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 77 "A:=x->plot3d([x,y,z],y=Lc(x)..Uc(x),z=0..f(x,y),color
=red,style=patchnogrid):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 193 "We m
ake a sequence of pictures showing the succesive positions of this \"c
ross section\" together with the other graphs as background.  It is im
portant that the background appear in each \"frame.\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "SV:=seq(display(A(0),A(i/20+0.001),
s,C,R),i=0..40):" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
152 "Finally, we display the result.  To run the movie, we click on th
e image and press the \"play\" button of the \"controls\" which appear
 below the toll bar. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "di
splay(SV,insequence=true,axes=normal,view=[0..2,0..5,0..4],orientation
=[-23,56]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 4 "" 0 "" {TEXT -1 20 "Generating the Solid" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 515 "A somewhat more elaborate procedure may be used t
o \"generate the solid.\" This makes a good classroom demonstration, b
ut is unnecessarily complicated for an introductory Maple laboratory. \+
This procedure is a good illustration of the use of \"image-valued\" M
aple functions.  (These are an invention of my own, although very like
ly many others working with Maple have invented them, too.)  The 0.001
 in the functional arguments prevent the plotting commands from produc
ing what would be essentially \"empty plot\" errors." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 87 "surf:=t->plot3d(f(x,y),x=0..t,y=Lc(x)..Uc
(x),grid=[10,5],color=blue,style=patchnogrid):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 84 "Lside:=t->plot3d([x,Lc(x),z],x=0..t,z=0..f(x,L
c(x)),color=yellow,style=patchnogrid):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 84 "Uside:=t->plot3d([x,Uc(x),z],x=0..t,z=0..f(x,Uc(x)),c
olor=yellow,style=patchnogrid):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 73 "LLine:=t->line([t,Lc(t),0],[t,Lc(t),f(t,Lc(t))],color=black,th
ickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "ULine:=t->li
ne([t,Uc(t),0],[t,Uc(t),f(t,Uc(t))],color=black,thickness=3):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "Bline:=t->line([t,Lc(t),0],[
t,Uc(t),0],color=black,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 159 "SV2:=seq(display(A(0),A(i/20+0.001),surf(i/20+0.001)
,Lside(i/20+0.001),Uside(i/20+0.001),LLine(i/20+0.001),ULine(i/20+0.00
1),Bline(i/20+0.001),s,C,R),i=0..40):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 84 "display(SV2,insequence=true,axes=normal,view=[0..2,0.
.5,0..4],orientation=[-23,56]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "A Volume in Spherica
l Coordinates" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "restart:wit
h(plots):with(plottools):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 9 "Exer
cise:" }{TEXT -1 48 " Draw the volume enclosed below by the cone z = \+
" }{XPPEDIT 18 0 "sqrt(x^2+y^2);" "6#-%%sqrtG6#,&*$%\"xG\"\"#\"\"\"*$%
\"yG\"\"#F*" }{TEXT -1 33 " and above by the hemisphere z = " }
{XPPEDIT 18 0 "sqrt(9-x^2-y^2);" "6#-%%sqrtG6#,(\"\"*\"\"\"*$%\"xG\"\"
#!\"\"*$%\"yG\"\"#F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}{TEXT 260 9 "Solution:" }{TEXT -1 158 " Note that when y = 0 and x > \+
0, the first equation reduces to z = x.  Likewise, when x = 0 and y > \+
0, z = y.  Thinking in spherical coordinates, we see that " }{XPPEDIT 
18 0 "phi = Pi/4;" "6#/%$phiG*&%#PiG\"\"\"\"\"%!\"\"" }{TEXT -1 116 " \+
at the wall of the cone.  So the portion of the hemisphere we need may
 be drawn in the following way.  By choosing " }{TEXT 257 18 "style = \+
wireframe," }{TEXT -1 71 " we will be able to see into the solid.  It \+
is convenient to use t for " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }
{TEXT -1 11 " and p for " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 
1 "," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "S:=plot3d(3,t=0..2*
Pi,p=0..Pi/4,coords=spherical,style=wireframe,color=black):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "It is quite easy to draw the cone
 as a parametric surface in spherical coordinates.    Along the wall o
f the cone, " }{XPPEDIT 18 0 "phi = Pi/4;" "6#/%$phiG*&%#PiG\"\"\"\"\"
%!\"\"" }{TEXT -1 10 "  , while " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }
{TEXT -1 22 " runs from 0 to 3 and " }{XPPEDIT 18 0 "theta;" "6#%&thet
aG" }{TEXT -1 11 " from 0 to " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"
%#PiGF%" }{TEXT -1 3 ".  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
88 "c:=plot3d([rho,t,Pi/4],rho=0..3,t=0..2*Pi,coords=spherical,style=w
ireframe,color=black):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Now dis
play the result." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "display
(S,c,scaling=constrained,orientation=[-25,75]);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 121 "Even when fully magnified, there are just too many \+
grid lines.  This can be improved by inserting the optional command   \+
" }{TEXT 258 1 " " }{TEXT -1 72 "grid=[25,5]   in the definition of S \+
and grid=[5,25] to that of c.  Thus" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 90 "S:=plot3d(3,t=0..2*Pi,p=0..Pi/4,coords=spherical,styl
e=wireframe,color=black,grid=[25,5]):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 100 "c:=plot3d([rho,t,Pi/4],rho=0..3,t=0..2*Pi,coords=sph
erical,style=wireframe,color=black,grid=[5,25]):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 54 "display(S,c,scaling=constrained,orientation=[
-25,75]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "That is much better.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 69 "Finding the v
olume using iterated integrals in spherical coordinates." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 262 48 "Visualization of a possible order of int
egration" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Let us add a represen
tative radius line to our picture." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 72 "R:=spacecurve([r,0,Pi/6],r=0..3,coords=spherical,colo
r=red,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "disp
lay(S,c,R,scaling=constrained,orientation=[-25,75]);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 37 "It is clear that, inside our volume, " }
{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 39 " may run from 0 to 3, in
dependently of " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " an
d " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 8 ", while " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 16 " runs from 0 to " }{XPPEDIT 18 0 
"pi/4;" "6#*&%#piG\"\"\"\"\"%!\"\"" }{TEXT -1 20 " , independently of \+
" }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 93 ".  Having realize
d this, we can now write the necessary integrals.  Integrating in the \+
order " }{XPPEDIT 18 0 "rho,phi,theta;" "6%%$rhoG%$phiG%&thetaG" }
{TEXT -1 53 ", and remembering that in spherical coordinates dV = " }
{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }{XPPEDIT 18 0 "sin(phi);" "6
#-%$sinG6#%$phiG" }{XPPEDIT 18 0 "d*rho;" "6#*&%\"dG\"\"\"%$rhoGF%" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "d*phi;" "6#*&%\"dG\"\"\"%$phiGF%" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "d*theta;" "6#*&%\"dG\"\"\"%&thetaGF%" }
{TEXT -1 10 ", we have " }{XPPEDIT 18 0 "Int(Int(Int(rho^2*sin(phi),rh
o = 0 .. 3),phi = 0 .. Pi/4),theta = 0 .. 2*Pi);" "6#-%$IntG6$-F$6$-F$
6$*&%$rhoG\"\"#-%$sinG6#%$phiG\"\"\"/F+;\"\"!\"\"$/F0;F4*&%#PiGF1\"\"%
!\"\"/%&thetaG;F4*&\"\"#F1F9F1" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 68 "V:=Int(Int(Int(rho^2*sin(phi),rho=0..3),phi=0.
.Pi/4),theta=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "va
lue(V);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 39 "Illustrating the \+
process of integration" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The fir
st integration, of " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 27 " f
rom 0 to 3 while holding " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 
5 " and " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 84 " fixed, p
roduces the line segment we have already seen.  The second integration
, of " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 11 " from 0 to " }
{XPPEDIT 18 0 "Pi/4;" "6#*&%#PiG\"\"\"\"\"%!\"\"" }{TEXT -1 15 " while
 holding " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 220 " fixed \+
produces a wedge-shaped surface.  We may show this surface by using th
e following parametric surface, which we define as a function of s in \+
a way that will be useful in a moment.  Again, we abbreviate \"phi\" b
y p. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "s1:=s->plot3d([rh
o,2*Pi*s/60,p],rho=0..3,p=0..Pi/4,coords=spherical,color=red,style=pat
chcontour):  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "display(S,
c,s1(0),scaling=constrained,orientation=[-25,75]);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 122 "Now we make use of our \"image-valued\" function
, s1, to make an animation showing the surface \"sweeping out\" the vo
lume as " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 16 " runs fro
m 0 to " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 80 
".  Change the magnification to its maximum setting before running the
 animation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "pic:=seq(dis
play(S,c,s1(t)),t=0..59):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
70 "display(pic,insequence=true,scaling=constrained,orientation=[-25,7
5]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 53 "A more el
aborate illustration of the last integration" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 201 "By redefining the top and bottom surfaces as \"image-v
alued\" functions, we may make an animation that really generates the \+
volume.  The arguments \"s + 0.001\" in the functions sa and ca are us
ed to keep " }{TEXT 259 6 "plot3d" }{TEXT -1 157 " from generating an \+
error when s = 0.  The line, L, has been added to improve visibility. \+
 Again, set the maximum magnification before running the animation." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "sa:=s->plot3d(3,t=0..2*Pi*
s/60,p=0..Pi/4,coords=spherical,style=patchnogrid,color=green):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "ca:=s->plot3d([rho,t,Pi/4],r
ho=0..3,t=0..2*Pi*s/60,coords=spherical,style=patchnogrid,color=blue):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "L:=line([0,0,0],[0,0,3]
,color=navy,thickness=4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
70 "pic1:=seq(display(L,S,c,s1(0),s1(s),sa(s+0.001),ca(s+0.001)),s=0..
60):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(pic1,insequ
ence=true,scaling=constrained,orientation=[-25,75]);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "2 0 0" 35 }{VIEWOPTS 1 1 0 
1 1 1803 }