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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "MATH 243-015" }}{PARA 0 "
" 0 "" {TEXT -1 32 "Analytic Geometry and Calculus C" }}{PARA 0 "" 0 "
" {TEXT -1 19 "Prof. D. A. Edwards" }}{PARA 0 "" 0 "" {TEXT -1 14 "Cop
yright 2001" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "\nHere we are going to
 use some dot products and lengths, so we need to assume something abo
ut t.  Here it will be more convenient to " }{TEXT 0 6 "assume" }
{TEXT -1 11 " that t is " }{TEXT 0 8 "positive" }{TEXT -1 19 ", rather
 than real." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 55 "restart;\nwith(plots):\nwith(linalg):\nassume(t,positive):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 256 "" 0 
"" {TEXT -1 49 "Section 16.8: Cylindrical and Spherical Integrals" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 207 "A
s in the two-dimensional case, Maple has no predefined commands for do
ing cylindrical and spherical integrals.  Therefore, we must make the \+
conversions ourselves.  Suppose we want to integrate the function " }
{XPPEDIT 18 0 "x^2+2*(y+3*z);" "6#,&*$%\"xG\"\"#\"\"\"*&F&F',&%\"yGF'*
&\"\"$F'%\"zGF'F'F'F'" }{TEXT -1 82 " over the sphere of radius R.  Fi
rst we do the integral in Cartersian coordinates:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 337 "fc3:=x^2+2*(y+3*z);\nacx3:=int(int(int(fc3,z=-sqrt
(R^2-x^2-y^2)..sqrt(R^2-x^2-y^2)),y=-sqrt(R^2-x^2)..sqrt(R^2-x^2)),x=-
R..R);\nacy3:=int(int(int(fc3,z=-sqrt(R^2-x^2-y^2)..sqrt(R^2-x^2-y^2))
,x=-sqrt(R^2-y^2)..sqrt(R^2-y^2)),y=-R..R);\nacz3:=int(int(int(fc3,x=-
sqrt(R^2-z^2-y^2)..sqrt(R^2-z^2-y^2)),y=-sqrt(R^2-z^2)..sqrt(R^2-z^2))
,z=-R..R);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 181 "Here we get the answer, but it does take a while.
\n\nNext we do the integral in cylindrical coordinates.  First we conv
ert fc3 into polar coordinates using the standard substitutions:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "fp3:=eval(fc3,[x=r*cos(theta),y=r*s
in(theta)]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "For the ranges, w
e see that now " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 16 " go
es from 0 to " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT 
-1 44 " and r goes from 0 to R.  Since z went from " }{XPPEDIT 18 0 "-
sqrt(R^2-x^2-y^2)" "6#,$-%%sqrtG6#,(*$%\"RG\"\"#\"\"\"*$%\"xGF*!\"\"*$
%\"yGF*F.F." }{TEXT -1 4 " to " }{XPPEDIT 18 0 "sqrt(R^2-x^2-y^2)" "6#
-%%sqrtG6#,(*$%\"RG\"\"#\"\"\"*$%\"xGF)!\"\"*$%\"yGF)F-" }{TEXT -1 57 
", we see that the new range (from the definition of r) is" }{XPPEDIT 
18 0 " -sqrt(R^2-r^2" "6#,$-%%sqrtG6#,&*$%\"RG\"\"#\"\"\"*$%\"rGF*!\"
\"F." }{TEXT -1 4 " to " }{XPPEDIT 18 0 "sqrt(R^2-r^2)" "6#-%%sqrtG6#,
&*$%\"RG\"\"#\"\"\"*$%\"rGF)!\"\"" }{TEXT -1 132 ".  Of course, this m
eans we must do the z integration first.  Also, we must make sure to a
dd the extra r from the differential term:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 79 "ap3:=int(int(int(fp3*r,z=-sqrt(R^2-r^2)..sqrt(R^2-r^2
)),r=0..R),theta=0..2*Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 225 "Ag
ain, we have a problem, basically since we have introduced complicated
 square roots once we do the first integration.  Therefore, next we do
 the integral in spherical coordinates.  First we make the necessary s
ubstitutions:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "fs3:=eval(fc3,[x=r
ho*cos(theta)*sin(phi),y=rho*sin(theta)*sin(phi),z=rho*cos(phi)]);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Now the ranges of integration ar
e rather straightforward.  We must remember to insert the proper expre
ssion " }{XPPEDIT 18 0 "rho^2*sin(phi)" "6#*&%$rhoG\"\"#-%$sinG6#%$phi
G\"\"\"" }{TEXT -1 23 " from the differential:" }{MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "as3:=int(int(int(fs3*rho^2*s
in(phi),rho=0..R),phi=0..Pi),theta=0..2*Pi);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 162 "To make sure we have the right answer, we substitute R
=4 into each expression.  (It takes a while because Maple has to calcu
late the integral in ap3 numerically.)" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 42 "evalf(eval([acx3,acy3,acz3,ap3,as3],R=4));" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 10 "
\nExercise\n" }{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 39 "As an
 exercise, integrate the function " }{XPPEDIT 18 0 "x*y*sin(y*z);" "6#
*(%\"xG\"\"\"%\"yGF%-%$sinG6#*&F&F%%\"zGF%F%" }{TEXT -1 139 " over the
 right circular cylinder of radius 3 whose base is in the xy-plane, wh
ose axis is the positive z-axis, and whose height is also 3." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "hca
rt:=x*y*sin(y*z);\nhpol:=eval(hcart,[x=r*cos(theta),y=r*sin(theta)]);
\nint(int(int(hpol,z=0..3),theta=0..2*Pi),r=0..3);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 
256 "" 0 "" {TEXT -1 27 "Section 17.1: Vector Fields" }}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 120 "We define a vector field just as we did a vect
or function of one variable.  We begin with the following vector field
 g3:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "g3:=vector([y^3,x^2+exp(-y)
]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "To plot a vector field, we
 use the command " }{TEXT 0 9 "fieldplot" }{TEXT -1 114 ".  The first \+
argument is the vector field we wish to plot; the others are the range
s of the independent variables:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 55 "field1:=fieldplot(g3,x=-1..1,y=-1..1):\ndisplay(field1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "We can also plot vector fields in \+
3-D.  Consider the following vector field: " }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "g4:=vector([x^2+y,y^2-z,x-z]);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 45 "To show the field in 3-D, we use the command " }
{TEXT 0 11 "fieldplot3d" }{TEXT -1 30 ", whose syntax is the same as \+
" }{TEXT 0 9 "fieldplot" }{TEXT -1 1 ":" }{MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "field2:=fieldplot3d(g4,x=-2..2,y=-2
..2,z=0..7):\ndisplay(field2,axes=normal,orientation=[8,70]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 256 "" 0 "" {TEXT -1 28 "Section \+
17.2: Line Integrals" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Suppose we
 want to integrate the function " }{XPPEDIT 18 0 "x^2+2*y" "6#,&*$%\"x
G\"\"#\"\"\"*&F&F'%\"yGF'F'" }{TEXT -1 26 " along the curve given by \+
" }{TEXT 256 1 "r" }{TEXT -1 5 "(t)=(" }{XPPEDIT 18 0 "t^(1/2),(t+2/3)
^(3/2);" "6$)%\"tG*&\"\"\"F&\"\"#!\"\"),&F$F&*&F'F&\"\"$F(F&*&F,F&F'F(
" }{TEXT -1 60 ") from t=1 to t=25.  First we define the relevant func
tions:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "f1:=x^2+2*y;\nr:=vector([
t^(1/2),(t+2/3)^(3/2)]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Next \+
we write f1 in terms of t and calculate the velocity vector " }{TEXT 
257 1 "v" }{TEXT -1 4 "(t):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "fint
:=eval(f1,[x=r[1],y=r[2]]);\nv:=map(diff,r,t);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 54 "To get an idea of what we're doing, we plot the curv
e " }{TEXT 258 1 "r" }{TEXT -1 35 "(t) in the xy-plane and the curve (
" }{TEXT 259 1 "r" }{TEXT -1 6 "(t),f(" }{TEXT 260 1 "r" }{TEXT -1 
177 "(t))) in space.  Thus, we are trying to find the area of the surf
ace outlined below.  You should know how to generate p1 and p2, though
 the generation of p3 we will cover later." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 265 "p1:=spacecurve([r[1],r[2],fint],t=1..25,color=black,
thickness=3):\np2:=spacecurve([r[1],r[2],0],t=1..25,color=black,thickn
ess=3):\np3:=plot3d([r[1],r[2],z],z=0..fint,t=1..25,style=patchcontour
):\ndisplay(p1,p2,p3,axes=normal,labels=['x','y','z'],orientation=[-45
,64]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "To find the value of th
e integral, we must first calculate the speed:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "sp:=simplify(norm(v,2));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 59 "Then we multiply fint by sp and integrate from t=1 to t=2
5:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "int(fint*sp,t=1..25);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "This answer is of little utility u
nless we assign a number to it:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "e
valf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Now let's find the ce
ntroid of a spring decribed by the curve given by (" }{XPPEDIT 18 0 "t
,t^2,t^3" "6%%\"tG*$F#\"\"#*$F#\"\"$" }{TEXT -1 43 ") from t=0 to t=5.
  A graph is shown below:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "r2:=v
ector([t,t^2,4*t^(3/2)/3]);\nspacecurve(r2,t=0..5,thickness=3,color=bl
ack,axes=normal,orientation=[-22,43]);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 113 "First we find the mass of the spring, using 1 for the de
nsity.  To do this, we just integrate the length of v(t):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "sp2:=simplify(norm(map(diff,r2,t),2
));\nmas:=int(sp2,t=0..5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "The
n we find each of the three moments by integrating x, y, and z times t
he speed:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "myz:=int(r2[1]*sp2,t=0
..5);\nmxz:=int(r2[2]*sp2,t=0..5);\nmxy:=int(r2[3]*sp2,t=0..5);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "To get the coordinates of the cent
roid, we just divide the appropriate moment by the mass:" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 38 "cen:=evalm(vector([myz,mxz,mxy])/mas);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Let's calculate the line integral
 of the vector field g3 from the previous section about the ellipse " 
}{XPPEDIT 18 0 "x^2+4*y^2=1" "6#/,&*$%\"xG\"\"#\"\"\"*&\"\"%F(*$%\"yGF
'F(F(F(" }{TEXT -1 27 ", which is parametrized by " }{TEXT 261 1 "r" }
{TEXT -1 33 "(t)=(cos(t),sin(t)/2) for t=0 to " }{XPPEDIT 18 0 "2*Pi" 
"6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 ":" }{MPLTEXT 1 0 31 "\nr3:=vecto
r([cos(t),sin(t)/2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Therefor
e, we are looking at how the velocity field varies along this curve:" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "curve1:=plot([r3[1],r3[2],t=0..2*
Pi],thickness=3,color=red):\ndisplay(field1,curve1);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 159 "To calculate the circulation, we take the dot \+
product of the velocity and the vector field to find the integrand, an
d then write this expression in terms of t:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "vel:=map(diff,r3,t);\ndotprod(g3,vel);\ninteg:=eval(%
,[x=r3[1],y=r3[2]]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Then we i
ntegrate the expression from t=0 to t=2*Pi:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "circ:=int(integ,t=0..2*Pi);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 100 "We can also evaluate line integrals in 3-D.  Consider \+
the following helix, where t ranges from 0 to " }{XPPEDIT 18 0 "2*Pi" 
"6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 ":" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 30 "r4:=vector([cos(t),sin(t),t]);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 98 "We wish to calculate the line integral of the field g4 fr
om the previous section along the helix: " }{MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "curve2:=spacecurve(r4,t=0..
2*Pi,thickness=3,color=black):\ndisplay(field2,curve2,axes=normal,orie
ntation=[8,70]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "The line int
egral is again given by the integral of the dot product of the field a
nd the velocity vector:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "vel4:=m
ap(diff,r4,t);\ndotprod(g4,vel4);\nint4:=eval(%,[x=r4[1],y=r4[2],z=r4[
3]]);\nwork:=int(int4,t=0..2*Pi);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 9 "Exercises" }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "1.  Integrate
 the function (numerically if necessary) " }{XPPEDIT 18 0 "cos(x)+y^2+
sin(z);" "6#,(-%$cosG6#%\"xG\"\"\"*$%\"yG\"\"#F(-%$sinG6#%\"zGF(" }
{TEXT -1 54 " over the helix (cos(t),sin(t),t) for t between 0 and " }
{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 "." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "fun
c:=cos(x)+y^2+sin(z);\nrfun:=vector([cos(t),sin(t),t]);\nfhel:=eval(fu
nc,[x=rfun[1],y=rfun[2],z=rfun[3]]);\nvf:=map(diff,rfun,t);\nspf:=simp
lify(norm(vf,2));\nint(fhel*spf,t=0..2*Pi);\nevalf(%);" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "2.  Integ
rate the vector field (" }{XPPEDIT 18 0 "x, y, sin(y*z)" "6%%\"xG%\"yG
-%$sinG6#*&F$\"\"\"%\"zGF)" }{TEXT -1 42 ") over the helix in the prev
ious exercise." }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Solution" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "We use the definitions from above.
\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "pot:=vector([x,y,sin(y*z)]);
\nint2:=eval(dotprod(pot,vf),[x=rfun[1],y=rfun[2],z=rfun[3]]);\nevalf(
int(%,t=0..2*Pi));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 257 "" 0 "
" {TEXT -1 1 "\n" }}}}{MARK "6" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }