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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 
0 "" {TEXT -1 40 "Plotting the Domain of a Double Integral" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 59 "A double integral of function of two vari
ables has region  " }{TEXT 256 1 "R" }{TEXT -1 675 "  in the plane as \+
its domain.  We use an equivalent iterated integral to find the value \+
of the double integral.  The limit process in a double integral occurs
 simultaneously in both variables.  In an iterated integral the integr
ation (limit) occurs in one variable and then the other, so it is two \+
step process.  Student difficulties with double integrals center aroun
d setting up the limits of the equivalent iterated integral.  One must
 examine the region over which the integration is to take place and de
cide the order of integration.  It may not matter, for example, if the
 region is a rectangle with its sides parallel to the coordinate axes,
 but sometimes it must be  " }{TEXT 257 10 "dA = dx dy" }{TEXT -1 25 "
  and others it must be  " }{TEXT 258 9 "dA =dy dx" }{TEXT -1 421 " . \+
 Sometimes neither order will work and the region must be broken down \+
into parcels where each allows one or the other order to be used.  We \+
are going to start with integrals that are already set up and show how
 to plot their domains.  Then we will reverse the process in the next \+
section.  It helps if we understand the roles played by the components
 of an iterated integral.  If one integrates first with respect to  " 
}{TEXT 259 1 "x" }{TEXT -1 28 "  and then with respect to  " }{TEXT 
260 1 "y" }{TEXT -1 8 " , then " }{TEXT 261 10 "dA = dx dy" }{TEXT -1 
43 "  and we have that the double integral of  " }{TEXT 262 3 "f  " }
{TEXT -1 20 "(with differential  " }{TEXT 263 2 "dA" }{TEXT -1 14 " ) \+
is given by" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 4 "    " }{XPPEDIT 
18 0 "Int(Int(f(x,y),x=(x=curve1)..(x=curve2)),y=(y=point1)..(y=point2
));" "6#-%$IntG6$-F$6$-%\"fG6$%\"xG%\"yG/F+;/F+%'curve1G/F+%'curve2G/F
,;/F,%'point1G/F,%'point2G" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 114 "and the phrase \"curve-to-curve\" is applicable.  If the
 order is reversed then the same double integral is given by" }}}
{EXCHG {PARA 259 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "Int(Int(f(
x,y),y=(y=curve3)..(y=curve4)),x=(x=point3)..(x=point4));" "6#-%$IntG6
$-F$6$-%\"fG6$%\"xG%\"yG/F,;/F,%'curve3G/F,%'curve4G/F+;/F+%'point3G/F
+%'point4G" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 226 "An
 iterated integral works from the inside out, so the first (inside) va
riable must be able to vary from one curve to another, while the secon
d (outside) variable changes from one point to another.  Let \222s con
sider an example." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 264 10 "Example 1:
" }{TEXT -1 10 " Discuss  " }{XPPEDIT 18 0 "Int(Int(y+x^2,y=arctan(x).
.exp(x)),x=-2..1))" "6#-%$IntG6$-F$6$,&%\"yG\"\"\"*$%\"xG\"\"#F*/F);-%
'arctanG6#F,-%$expG6#F,/F,;,$F-!\"\"F*" }{TEXT -1 1 " " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 84 "First, rewrite the integral and insert th
e variables of integration in the limits.  " }{XPPEDIT 18 0 "Int(Int(y
+x^2,y=(y=arctan(x))..(y=exp(x))),x=(x=-2)..(x=1)))" "6#-%$IntG6$-F$6$
,&%\"yG\"\"\"*$%\"xG\"\"#F*/F);/F)-%'arctanG6#F,/F)-%$expG6#F,/F,;/F,,
$F-!\"\"/F,F*" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "
This tells us that for any fixed value of  " }{XPPEDIT 18 0 "x" "6#%\"
xG" }{TEXT -1 4 " ,  " }{XPPEDIT 18 0 "y" "6#%\"yG" }{TEXT -1 29 "  mu
st be able to vary from  " }{XPPEDIT 18 0 "arctan(x)" "6#-%'arctanG6#%
\"xG" }{TEXT -1 6 "  to  " }{XPPEDIT 18 0 "exp(x)" "6#-%$expG6#%\"xG" 
}{TEXT -1 21 "  inside the domain  " }{TEXT 265 4 "R . " }{TEXT -1 70 
"For simplicity, let's define these functions in Maple, after invoking
 " }{TEXT 268 7 "student" }{TEXT -1 5 " and " }{TEXT 269 5 "plots" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "with(stude
nt): with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "h:=x->
exp(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "g:=x->arctan(x);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 218 "We will display four plots i
n parametric form that determine the boundaries of the region.  Make s
ure you can identify each of them.  What roles do the outer limits of \+
our integral play?  They determine the domains of " }{TEXT 266 3 " g \+
" }{TEXT -1 6 " and  " }{TEXT 267 1 "h" }{TEXT -1 46 "  and occur here
 as vertical line segments,   " }{XPPEDIT 18 0 "x=-2" "6#/%\"xG,$\"\"#
!\"\"" }{TEXT -1 8 "   and  " }{XPPEDIT 18 0 "x=1" "6#/%\"xG\"\"\"" }
{TEXT -1 142 " .  As usual, the output can be viewed by placing the cu
rsor on the first Maple command of this worksheet and repeatly hitting
 the enter key.." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "A1:=plo
t([x,g(x),x=-2..1],color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 39 "A2:=plot([x,h(x),x=-2..1],color=green):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 43 "A3:=plot([-2,y,y=g(-2)..h(-2)],color=blue):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "A4:=plot([1,y,y=g(1)..h(1)
],color=magenta):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "displa
y(A1,A2,A3,A4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "Note how if v
ertical line is moved from left to right on this plot it will always i
ntersect the graph of  " }{TEXT 270 1 "g" }{TEXT -1 35 "  on the botto
m and the graph of   " }{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 182 "  o
n the top, i.e., curve-to-curve. What would happen if we moved horizon
tal line from the bottom to the top of the plot?   On the left it woul
d intersect either the vertical line   " }{XPPEDIT 18 0 "x=-2" "6#/%\"
xG,$\"\"#!\"\"" }{TEXT -1 20 "  or the graph of   " }{XPPEDIT 18 0 "h
" "6#%\"hG" }{TEXT -1 66 "  ,  while on the right it would cut across \+
either the graph of   " }{TEXT 271 1 "g" }{TEXT -1 24 "  or the vertic
al line  " }{XPPEDIT 18 0 "x=1" "6#/%\"xG\"\"\"" }{TEXT -1 187 " .  Th
is violates the principle of \223curve-to-curve, point-to-point\224, w
hich means that the reverse order of integration will not work for thi
s region.  You may be surprised to note that   " }{XPPEDIT 18 0 "f(x,y
)=y+x^2" "6#/-%\"fG6$%\"xG%\"yG,&F(\"\"\"*$F'\"\"#F*" }{TEXT -1 61 "  \+
 does not play a role in the domain of the integral unless " }{TEXT 
272 2 " f" }{TEXT -1 35 "  is not defined at some point in  " }{TEXT 
273 1 "R" }{TEXT -1 46 " .  In Maple the integral would be defined by:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "a:=Doubleint(y+x^2,y=ar
ctan(x)..exp(x),x=-2..1);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
9 "value(a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalf(value
(a));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "int(int(y+x^2,y=ar
tan(x)..exp(x)),x=-2..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
57 "evalf(value(int(int(y+x^2,y=artan(x)..exp(x)),x=-2..1)));" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Think about the syntax we used to \+
plot the bottom and top curves above." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 274 10 "Example 2:" }{TEXT -1 11 "  Discuss \001" }{XPPEDIT 18 
0 "Int(Int(y+x^2,x=y^2..2-y),y=0..1)" "6#-%$IntG6$-F$6$,&%\"yG\"\"\"*$
%\"xG\"\"#F*/F,;*$F)F-,&F-F*F)!\"\"/F);\"\"!F*" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 23 "As before, start with \001" }{XPPEDIT 18 0 "Int(Int(
y+x^2,x=(x=y^2)..(x=2-y)),y=(y=0)..(y=1))" "6#-%$IntG6$-F$6$,&%\"yG\"
\"\"*$%\"xG\"\"#F*/F,;/F,*$F)F-/F,,&F-F*F)!\"\"/F);/F)\"\"!/F)F*" }
{TEXT -1 43 "  and define the inner limits as functions." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "r:=y->y^2;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 11 "s:=y->2-y;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 31 "Now plot just these two curves:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "display(plot([r(y),y,y=0..1]),plot([s(y),y,y=0..1]));
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Since
 we integrated first with respect to " }{TEXT 275 1 "x" }{TEXT -1 157 
" , we must be able to draw horizontal lines that always begin on the \+
left curve and end on the right curve.  Sliding horizontal line up the
 page, the first  " }{TEXT 276 1 "y" }{TEXT -1 50 "  value for which t
he line touches the region is  " }{XPPEDIT 18 0 "y=0" "6#/%\"yG\"\"!" 
}{TEXT -1 25 "  and the last value is  " }{XPPEDIT 18 0 "y=1" "6#/%\"y
G\"\"\"" }{TEXT -1 4 "  .\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 277 14 "C
3M7 Problems:" }{TEXT -1 51 "  Plot the domains of the double integral
s listed.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 " 1.  " }{XPPEDIT 
18 0 "Int(Int(f(x,y),y=x^2..2x),x=0..1)" "6#-%$IntG6$-F$6$-%\"fG6$%\"x
G%\"yG/F,;*$F+\"\"#*&F0\"\"\"F+F2/F+;\"\"!F2" }{TEXT -1 21 "          \+
       2.  " }{XPPEDIT 18 0 "Int(Int(f(x,y),x=-y..y^2),y=0..2)" "6#-%$
IntG6$-F$6$-%\"fG6$%\"xG%\"yG/F+;,$F,!\"\"*$F,\"\"#/F,;\"\"!F2" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 " 3.  " }{XPPEDIT 18 0 "Int(Int(f(x,
y),x=y..2-y),y=0..1)" "6#-%$IntG6$-F$6$-%\"fG6$%\"xG%\"yG/F+;F,,&\"\"#
\"\"\"F,!\"\"/F,;\"\"!F1" }{TEXT -1 18 "              4.  " }{XPPEDIT 
18 0 "Int(Int(f(x,y),y=x^2..2+x),x=-1..2))" "6#-%$IntG6$-F$6$-%\"fG6$%
\"xG%\"yG/F,;*$F+\"\"#,&F0\"\"\"F+F2/F+;,$F2!\"\"F0" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 10 " 5.  Let  " }{TEXT 278 1 "k" }{TEXT -1 67 "  be
 an instructor assigned constant as described in the preface.  " }
{XPPEDIT 18 0 "Int(Int(f(x,y),y=-x^2..100+x^2),x=-k..k+5)" "6#-%$IntG6
$-F$6$-%\"fG6$%\"xG%\"yG/F,;,$*$F+\"\"#!\"\",&\"$+\"\"\"\"*$F+F1F5/F+;
,$%\"kGF2,&F:F5\"\"&F5" }}}}{MARK "39" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 
1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }