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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 4 "C3M9" }}}{EXCHG {PARA 
257 "" 0 "" {TEXT -1 37 "Double Integrals in Polar Coordinates" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 305 "There are regions in the plane th
at are not easily used as domains of iterated integrals in rectangular
 coordinates.  Sometimes switching to an integral in polar coordinates
 makes a diffcult problem much easier.  We will begin with a simple ex
ample and show how to make the transition.  There are certain " }
{TEXT 256 4 "Do\222s" }{TEXT -1 5 " and " }{TEXT 257 7 "Don\222t\222s
" }{TEXT -1 33 " that we will point out.  First, " }{TEXT 258 6 "ALWAY
S" }{TEXT -1 99 " begin by writing the variables of integration in the
 limits of the rectangular integral, such as \223" }{TEXT 259 1 "x" }
{TEXT -1 8 " =\224and \223" }{TEXT 260 1 "y" }{TEXT -1 14 " =\224.  Se
cond, " }{TEXT 261 6 "ALWAYS" }{TEXT -1 142 " draw a sketch of the dom
ain of the integral after the first step and think about how you would
 draw that same sketch using polar coordinates." }}}{EXCHG {PARA 0 "" 
0 "" {TEXT 262 10 "Example 1:" }{TEXT -1 25 "  Evaluate the integral  \+
" }{XPPEDIT 18 0 "Int(Int(y,y=0..sqrt(4-x^2)),x=1..2)" "6#-%$IntG6$-F$
6$%\"yG/F(;\"\"!-%%sqrtG6#,&\"\"%\"\"\"*$%\"xG\"\"#!\"\"/F3;F1F4" }
{TEXT -1 27 "   using polar coordinates." }}}{EXCHG {PARA 258 "" 0 "" 
{TEXT -1 13 "First Step : " }{XPPEDIT 18 0 "Int(Int(y,y=0..sqrt(4-x^2)
),x=1..2)" "6#-%$IntG6$-F$6$%\"yG/F(;\"\"!-%%sqrtG6#,&\"\"%\"\"\"*$%\"
xG\"\"#!\"\"/F3;F1F4" }{TEXT -1 21 "   is rewritten as   " }{XPPEDIT 
18 0 "Int(Int(y,y=(y=0)..(y=sqrt(4-x^2))),x=(x=1)..(x=2))" "6#-%$IntG6
$-F$6$%\"yG/F(;/F(\"\"!/F(-%%sqrtG6#,&\"\"%\"\"\"*$%\"xG\"\"#!\"\"/F5;
/F5F3/F5F6" }{TEXT -1 1 "\001" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 244 
"Let\222s use Maple to plot the domain.  We will see that we have a po
rtion of the circle of radius 2  in the first quadrant.  A fourth plot
 is used to put the axes in their proper place.  The output results fr
om executing the Maple commands below." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "with(student):  with(plots):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 32 "A:=plot([x,sqrt(4-x^2),x=1..2]):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "B:=plot([1,y,y=0..sqrt(3)]):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "C:=plot([x,0,x=1..2]):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "E:=plot([0,y,y=0..1]):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(A,B,C,E);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "We are providing more details her
e than are really necessary, but maybe someone will benefit from it.  \+
Now we will plot the same region using " }{TEXT 264 9 "polarplot" }
{TEXT -1 92 " .  The output follows from the commands below.  We must \+
use the fact that a vertical line  " }{TEXT 263 5 "x = a" }{TEXT -1 
42 "  in polar coordinates is represented by  " }{XPPEDIT 18 0 "r=a*se
c(theta)" "6#/%\"rG*&%\"aG\"\"\"-%$secG6#%&thetaGF'" }{TEXT -1 10 "  .
  If   " }{XPPEDIT 18 0 "a=0" "6#/%\"aG\"\"!" }{TEXT -1 16 "   then we
 use  " }{XPPEDIT 18 0 " theta=Pi/2" "6#/%&thetaG*&%#PiG\"\"\"\"\"#!\"
\"" }{TEXT -1 3 " .\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "A1
:=polarplot([2,t,t=0..Pi/3]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 36 "B1:=polarplot([sec(t),t,t=0..Pi/3]):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "C1:=polarplot([r,0,r=1..2]):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 51 "angleline:=polarplot([r,Pi/3,r=0..2.2],color=b
lue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display(A1,B1,C1,E
,angleline);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "We included the p
lot " }{TEXT 265 9 "angleline" }{TEXT -1 94 " to show the last angle n
eeded to plot the figure.  Now we are ready to set up our integral.  \+
" }{TEXT 266 5 "NEVER" }{TEXT -1 21 " simply substitute   " }{XPPEDIT 
18 0 "x=r*cos(theta)" "6#/%\"xG*&%\"rG\"\"\"-%$cosG6#%&thetaGF'" }
{TEXT -1 9 "   and   " }{XPPEDIT 18 0 "y=r*sin(theta)" "6#/%\"yG*&%\"r
G\"\"\"-%$sinG6#%&thetaGF'" }{TEXT -1 49 "   into the limits of the re
ctangular integral.  " }{TEXT 267 6 "ALWAYS" }{TEXT -1 46 " make those
 substitutions into the integrand  " }{XPPEDIT 18 0 "f(x,y)" "6#-%\"fG
6$%\"xG%\"yG" }{TEXT -1 29 "  and multiply the result by " }{TEXT 268 
2 " r" }{TEXT -1 12 "  to form   " }{XPPEDIT 18 0 "f(r*cos(theta),r*si
n(theta))*r*dr;" "6#*(-%\"fG6$*&%\"rG\"\"\"-%$cosG6#%&thetaGF)*&F(F)-%
$sinG6#F-F)F)F(F)%#drGF)" }{TEXT -1 1 " " }{TEXT 269 1 "d" }{XPPEDIT 
18 0 "theta" "6#%&thetaG" }{TEXT -1 115 "  .  Look at the figure and a
sk yourself,  \223How do I draw this figure in polar coordinates?\224 \+
 \223Which functions of  " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT 
-1 7 "  does " }{TEXT 270 1 "r" }{TEXT -1 46 " range between?\224  And
 \223For which values does  " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }
{TEXT -1 77 "   first touch the figure and last touch the figure?\224 \+
 The process looks like" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 3 "   " 
}{XPPEDIT 18 0 "Int(Int(f(x,y),y=(y=g(x))..(y=h(x))),x=(x=a)..(x=b))" 
"6#-%$IntG6$-F$6$-%\"fG6$%\"xG%\"yG/F,;/F,-%\"gG6#F+/F,-%\"hG6#F+/F+;/
F+%\"aG/F+%\"bG" }{TEXT -1 15 "    becomes    " }{XPPEDIT 18 0 "Int(In
t(f(r*cos(theta),r*sin(theta))*r,r=(r=u(theta))..(r=v(theta))),theta=(
theta=theta[1])..(theta=theta[2]))" "6#-%$IntG6$-F$6$*&-%\"fG6$*&%\"rG
\"\"\"-%$cosG6#%&thetaGF.*&F-F.-%$sinG6#F2F.F.F-F./F-;/F--%\"uG6#F2/F-
-%\"vG6#F2/F2;/F2&F26#F./F2&F26#\"\"#" }{TEXT -1 2 "  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 230 "Think of a line emanating from the origi
n and extending out at about 30 degrees.  It will first touch the figu
re on the vertical line and exit the figure at the circle.  And, this \+
will remain true as you place that line along the " }{TEXT 271 2 "x-" 
}{TEXT -1 115 "axis and rotate it up to the point where the line and a
rc intersect.  Using Maple to substitute into the integrand:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "f:=(x,y)->y;\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "grand:=simplify(f(r*cos(t),r*sin(t)
),symbolic);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The polar integ
ral becomes" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 
0 "Int(Int(r*sin(theta)*r,r=(r=sec(theta))..(r=2)),theta=(theta=0)..(t
heta=Pi/3))=Int(Int(r^2*sin(theta),r=(rsec(theta))..(2)),theta=(0)..Pi
/3)" "6#/-%$IntG6$-F%6$*(%\"rG\"\"\"-%$sinG6#%&thetaGF+F*F+/F*;/F*-%$s
ecG6#F//F*\"\"#/F/;/F/\"\"!/F/*&%#PiGF+\"\"$!\"\"-F%6$-F%6$*&F*F7-F-6#
F/F+/F*;-%%rsecG6#F/F7/F/;F;*&F>F+F?F@" }{TEXT -1 4 "    " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "answer1:=Doubleint(grand*r,r=sec(t)
..2,t=0..Pi/3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "answer2:
=value(answer1);" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 272 11 "Example 2: " }{TEXT -1 24 " Find the integral of
   " }{XPPEDIT 18 0 "f(x,y)=exp(x^2+y^2)" "6#/-%\"fG6$%\"xG%\"yG-%$exp
G6#,&*$F'\"\"#\"\"\"*$F(F.F/" }{TEXT -1 69 "  over the annular region \+
in the first quadrant between the circles  " }{XPPEDIT 18 0 "x^2+y^2=1
" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(F(" }{TEXT -1 7 "  and  " }
{XPPEDIT 18 0 "x^2+y^2=4" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"\"%" }
{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "It is easiest t
o graph this using polarplot ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "with(student): with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "P1:=polarplot([2,t,t=0..Pi/2]):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 31 "P2:=polarplot([1,t,t=0..Pi/2]):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "display(P1,P2);\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 104 "It is easy to see that this integral cannot be
 done as one iterated integral in rectangular coordinates." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f:=(x,y)->exp(x^2+y^2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "grand:=simplify(f(r*cos(t),r*sin(t)
),symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Polint:=Dou
bleint(grand*r,r=1..2,t=0..Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Polint:=value(Polint);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 79 "In addition to problems with the region, there is no know
n antiderivative for  " }{XPPEDIT 18 0 "exp(x^2)" "6#-%$expG6#*$%\"xG
\"\"#" }{TEXT -1 32 "  .  The presence of the extra  " }{TEXT 273 1 "r
" }{TEXT -1 78 "  in the polar integrals makes such problems a straigh
tforward substitution,  " }{XPPEDIT 18 0 "u=r^2" "6#/%\"uG*$%\"rG\"\"#
" }{TEXT -1 4 "  .\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 274 13 "C3M9 Pro
blems" }{TEXT -1 68 "   Use Maple and polar coordinates to evaluate th
e given integrals.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "1.  " }
{XPPEDIT 18 0 "Int(Int(exp(-x^2-y^2),y=0..sqrt(4-x^2)),x=-2..2)" "6#-%
$IntG6$-F$6$-%$expG6#,&*$%\"xG\"\"#!\"\"*$%\"yGF.F//F1;\"\"!-%%sqrtG6#
,&\"\"%\"\"\"*$F-F.F//F-;,$F.F/F." }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 30 "2.  The double integral over  " }{XPPEDIT 18 0 "S
" "6#%\"SG" }{TEXT -1 5 "  of " }{XPPEDIT 18 0 "x^2/(x^2+y^2)" "6#*&%
\"xG\"\"#,&*$F$F%\"\"\"*$%\"yGF%F(!\"\"" }{TEXT -1 10 " , where  " }
{TEXT 275 2 "S " }{TEXT -1 32 " is the annular region between  " }
{XPPEDIT 18 0 "x^2+y^2=1" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(F(" }
{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "x^2+y^2=3" "6#/,&*$%\"xG\"\"#\"\"
\"*$%\"yGF'F(\"\"$" }{TEXT -1 7 "  with " }{XPPEDIT 18 0 "y>=0" "6#1\"
\"!%\"yG" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "3.  \+
The double integral over  " }{TEXT 276 1 "R" }{TEXT -1 6 "  of  " }
{XPPEDIT 18 0 "sqrt(x^2+y^2)" "6#-%%sqrtG6#,&*$%\"xG\"\"#\"\"\"*$%\"yG
F)F*" }{TEXT -1 9 "  where  " }{TEXT 277 2 "R " }{TEXT -1 55 " is the \+
triangle with vertices (0,0), (3,0), and (3,3)." }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 4 "4.  " }{TEXT 278 11 "Challenge! " }{TEXT -1 2 "  " }
{XPPEDIT 18 0 "Int(Int(x,x=-sqrt(2*y-y^2)..sqrt(2*y-y^2)),y=0..2)" "6#
-%$IntG6$-F$6$%\"xG/F(;,$-%%sqrtG6#,&*&\"\"#\"\"\"%\"yGF2F2*$F3F1!\"\"
F5-F-6#,&*&F1F2F3F2F2*$F3F1F5/F3;\"\"!F1" }{TEXT -1 2 "  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 10 "5.   Let  " }{TEXT 279 1 "k" }{TEXT -1 
115 "  be an instructor assigned constant as described in the preface.
  Find the exact value of the double integral of  " }{XPPEDIT 18 0 "ex
p(k*(x^2+y^2)" "6#-%$expG6#*&%\"kG\"\"\",&*$%\"xG\"\"#F(*$%\"yGF,F(F(
" }{TEXT -1 100 "  over the \"slice of pie\" shaped region in the firs
t quadrant bounded by the curves with equations  " }{XPPEDIT 18 0 "x^2
+y^2=1" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(F(" }{TEXT -1 3 " , " }
{XPPEDIT 18 0 "y=0" "6#/%\"yG\"\"!" }{TEXT -1 7 " , and " }{XPPEDIT 
18 0 "y=x" "6#/%\"yG%\"xG" }{TEXT -1 2 " ." }}}}{MARK "0 0 0" 0 }
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