c4l2 Double Integrals Over Nonrectangular Region

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.


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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

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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

c4l2 Double Integrals Over Nonrectangular Region

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