Post on 03-Jun-2018
8/12/2019 c4l2 Double Integrals Over Nonrectangular Region
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DOUBLE INTEGRALS OVERNONRECTANGULAR REGION
C4L2
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Learning Outcomes
At the end of the lesson the studentshould be able to:
1. Find the iterated integral with non-constant of integration.
2. Define a Tye 1 and Tye 2 region.
!. "se the roerty of Tye 1 and Tye 2.
8/12/2019 c4l2 Double Integrals Over Nonrectangular Region
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#terated #ntegrals with $on-constant Limits of #ntegration Definition
dydxyxfdydxyxf
dxdyyxfdxdyyxf
yh
yh
d
c
yh
yh
d
c
xg
xg
b
a
xg
xg
b
a
=
=
),(),(
),(),(
)(
)(
)(
)(
)(
)(
)(
)(
2
1
2
1
2
1
2
1
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%&amle 1
Evaluate the iterated integral:
54
401
5179
2
1
1
2
1
0
9
0
3
0
2
2/
2
5
1
.4.39.2.1
:
)(.4.3
.26.1
2
22
2
Answer
dydxyxdydxxy
dydxydydxyx
x
x
x
x
yy
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Double integral over nonre!tangular region
Tye 1 'egion
A tye 1 region is bounded on the left and right by(ertical lines &)a and &)b and is bounded below andabo(e by continuous cur(es y) and y)where
.)()( 21 bxaforxgxg
)(1 xg )(2 xg
a b
)(2 xgy=
)(1 xgy=
The arrow enters from y = g1 to y = g2 ,
and the value of y increases so the
double integral is to be integrated first in
terms of y
dxdyyxfdAyxf
xg
xg
b
aR
),(),(
)(
)(
2
1
=
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Tye 2 'egion
A tye 2 region is bounded below and abo(e by
hori*ontal lines y)c and y)d and is bounded on theleft and right by continuous cur(es
dycfor
yhyhwhereyhxandyhx
== )()()()( 2121
d
c
As the arrow enters from the
curves x=h1 to x=h2 the
double integral is to be
integrated first in terms of x
dydxyxfdAyxf
yh
yh
d
cR
),(),(
)(
)(
2
1
=
)(1 yhx=)(
2
yhx=
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Theore"
a !f " is a ty#e 1 region which f$x,y is continuous,
then
b !f " is a ty#e 2 region which f$x,y is continuous,
then
dxdyyxfdAyxf
xg
xg
b
aR
),(),(
)(
)(
2
1
=
dydxyxfdAyxf
yh
yh
d
cR
),(),(
)(
)(
2
1 =
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Ste# to $ollo%:
%&etch the region " and indicate the
boundaries'
Classify as Ty#e 1 or Ty#e 2 by drawing a
vertical arrow or a hori(ontal arrow'
)ind the interval for x and y'
*rite the formula with the corres#onding
limits
+valuate using the iterated integral'
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E&a"#le ': (ill in the "iing li"it o$integration b) uing the $igure belo%*
2xy=
xy=
2
Figure 1
2xy=
Figure 2
dxdyyxfdAyxfb
dydxyxfdAyxfa
R
R
),(),()
),(),()
?
?
?
?
?
?
?
?
=
=
$1,1
$2,4
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E&a"#le +
+valuate the double integral in two ways using
iterated integral viewing " as the region' -se thea##ro#riate Ty#e of region'
8and,/16byboundedregiontheis
2
===
xxyxyR
dAxR
X = 8
y = xy =16/x
(4,4)$.,2
x
y
$.,.
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+olution
Type 1 Region (Using er!ica" arro# asguide)
Type 2 Region (Using $ori%on!a" arro# asguide, di&ide !$e 'igure in!o 2 par!s)
576),( 2
/16
8
4
== dxdyxdAyxfx
xR
576),( 288
4
2
8
/16
4
2
=+= dydxxdydxxdAyxfyyR
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E&a"#le ,
%(aluatexyxyyR
dAxyR
====
and0,2,1byen!osedregiontheis
"2
y = x
y=2
y=1R
10/312
0
2
1
2 == dydxxydAxyy
R
s Type 2 Region
$2,22
1$1,1
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Refer to Figure 1 below:
1'
y=4
x
y
1
4
y= 4x
/ecide the
ty#e of region'
dAyxfR
),(
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See Figure 2 below
2'
1
2
$1,2
X= y/2 ?),( dAyxfR
/ecide the ty#e
of region
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0ore +xam#les
/raw the reuired figure and
evaluate.
15
{ }
{ }
{ }.,21),(,.3
0,10),(,1
2.2
.,20),(,.1
3/
2
23
yxyyyxRdAe
xyxyxRdAx
y
xyxxyxRdAyx
yx
R
R
R
=
=+
=
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Reversing the r!er of"ntegr#tion +valuate the double integral by reversing the
order of integration
dydxe
dydxxdxdye
x
y
y
y
x
3
2
24
0
2
1
2/
2
0
4
4
1
0
.3
)(os.2.1
3ote !t would be easier to understand if the figure is drawn