Wed 26 Jun
Double Over
15 . 1
Integrals Rectangles
Calculus: Review
If(u) da
< X 2 <Xn b
Divide [a b) n-subintervals
=
, into S t .
.
a
=
Xo .
. . .
f(u) & Xi =
Xi -
Xi -
norm
IIPI) =
max Paxi 4 where I sin
&
limf(xi)Axi = (a) du
b i 0
=
a
Thu 27 Jun
Double Integrals
~
deal with rectangle
(i Y ; ) Rectangle Rij
~
.
>
flu y)
-
D-
,
defined by the
rectangle
Y; -
su ·
R =
G(x y))a , = X = b and c =
y =
d)
Yi - 1
-
Divide [a b] , into n-subintervals
·
a =
Mo<d , < . . .
<an = b Dui =
di-ci-
[ - ,
Divide [c d] , into m-subintervals
d is : is ; b
d
< <
(yj Dyi Ye
= =
yj
=
c
yo y
. . .
· -
.
,
Area ↑
Aij =
Aui
Dy ;
·
dij: itloyik =
diagonal length
.
f(a y),
·
11011 = max Gadij , zien , 122m ↑
mi
[E
I im
↑ lIP11 - 0
f(xi , yi) .
DAij
j= 1 i= 1
if this limit and f inite it R
exists called double integral of function
(my) rectangle
·
is :
is the over
donated by Iff(a y ,
d
· if
flag) >0
,
//fley)dA =
volume under the surface t =
flug) and above the region R
·
if flue ,
y) =
1
, /flay(dA :
SdA =
area of e
Integrating Double Integrals
· I. If(m y) , du
de
1
integrate with respect to m
,
treating y
as a constant
limit to
2. plug a
,
we end
by an
integral with respect to
y
& or
vice-verca
,Example Calculate
(/flay) for
flay) 100-bary D 02
-Ey *
:
1 : = :
, ,
. 100-baly dudy c .
% 21. "100- bary
. ,
dyde
=
/: /1004 2ay) dy
-
=
/100y- Surya) ! do
S:
=
200-1by dy
=
10 1,00 -
3x2)-1-100 -
302) da
=
(200y -
sys] :
=
1: 200 da
(2004) :
=
=
(192) -
1-208)
= I
400 400
, Thu 27 Jun
15 2.
Double Integrals Over Genval Regions
Double Integrals
·
SJ f(x y) , dA -R
!
R
-
↑
we define flay) =
flucy) ,
( :
y)ER
0
,
(x ,
y) + (0/ )
+
: +(a ,
y) da =
/flayl a s
Fubini's Theorem
·
Let fluy) be continues on a
Region R
1. If R is defined
by acusb ,
g . (u) <y (gall) Ig ., go continues on [a , b]1 then :
·
/ + im y) da = fla y) dydn
,
&
- -vertical segmentations
- -do this if
y
is a function of we
a is
2 .
IfD is defined by cyed ,
holy) E
Ehaly) /h ., ha continues on
Saibj) ,
then :
·
(Jf(u y)dA ,
=
Ihl ,
flay) dudy
-
- -horizontal segmentations
- - do this if a is a function of
y
a is
Example Find the volume of solid that lies under paraboloid ur and above the D the bounded
my-plane
>
the region by
:
the z = + in
y
the line za and the parabola x
y
=
y
=
Y 2x x2
y
=
2x =
·
x2 0
Cur
-
2x =
x(x -
2) = p
Vertical Integration +
yadyde
x = 2 :
Solary
24
y
= x = +
jys]ardo
· (0(2x + 843) -
(n + juP)
-I-5u-n + d
-u
- -in t -
=x5 +
=n4]
-
Double Over
15 . 1
Integrals Rectangles
Calculus: Review
If(u) da
< X 2 <Xn b
Divide [a b) n-subintervals
=
, into S t .
.
a
=
Xo .
. . .
f(u) & Xi =
Xi -
Xi -
norm
IIPI) =
max Paxi 4 where I sin
&
limf(xi)Axi = (a) du
b i 0
=
a
Thu 27 Jun
Double Integrals
~
deal with rectangle
(i Y ; ) Rectangle Rij
~
.
>
flu y)
-
D-
,
defined by the
rectangle
Y; -
su ·
R =
G(x y))a , = X = b and c =
y =
d)
Yi - 1
-
Divide [a b] , into n-subintervals
·
a =
Mo<d , < . . .
<an = b Dui =
di-ci-
[ - ,
Divide [c d] , into m-subintervals
d is : is ; b
d
< <
(yj Dyi Ye
= =
yj
=
c
yo y
. . .
· -
.
,
Area ↑
Aij =
Aui
Dy ;
·
dij: itloyik =
diagonal length
.
f(a y),
·
11011 = max Gadij , zien , 122m ↑
mi
[E
I im
↑ lIP11 - 0
f(xi , yi) .
DAij
j= 1 i= 1
if this limit and f inite it R
exists called double integral of function
(my) rectangle
·
is :
is the over
donated by Iff(a y ,
d
· if
flag) >0
,
//fley)dA =
volume under the surface t =
flug) and above the region R
·
if flue ,
y) =
1
, /flay(dA :
SdA =
area of e
Integrating Double Integrals
· I. If(m y) , du
de
1
integrate with respect to m
,
treating y
as a constant
limit to
2. plug a
,
we end
by an
integral with respect to
y
& or
vice-verca
,Example Calculate
(/flay) for
flay) 100-bary D 02
-Ey *
:
1 : = :
, ,
. 100-baly dudy c .
% 21. "100- bary
. ,
dyde
=
/: /1004 2ay) dy
-
=
/100y- Surya) ! do
S:
=
200-1by dy
=
10 1,00 -
3x2)-1-100 -
302) da
=
(200y -
sys] :
=
1: 200 da
(2004) :
=
=
(192) -
1-208)
= I
400 400
, Thu 27 Jun
15 2.
Double Integrals Over Genval Regions
Double Integrals
·
SJ f(x y) , dA -R
!
R
-
↑
we define flay) =
flucy) ,
( :
y)ER
0
,
(x ,
y) + (0/ )
+
: +(a ,
y) da =
/flayl a s
Fubini's Theorem
·
Let fluy) be continues on a
Region R
1. If R is defined
by acusb ,
g . (u) <y (gall) Ig ., go continues on [a , b]1 then :
·
/ + im y) da = fla y) dydn
,
&
- -vertical segmentations
- -do this if
y
is a function of we
a is
2 .
IfD is defined by cyed ,
holy) E
Ehaly) /h ., ha continues on
Saibj) ,
then :
·
(Jf(u y)dA ,
=
Ihl ,
flay) dudy
-
- -horizontal segmentations
- - do this if a is a function of
y
a is
Example Find the volume of solid that lies under paraboloid ur and above the D the bounded
my-plane
>
the region by
:
the z = + in
y
the line za and the parabola x
y
=
y
=
Y 2x x2
y
=
2x =
·
x2 0
Cur
-
2x =
x(x -
2) = p
Vertical Integration +
yadyde
x = 2 :
Solary
24
y
= x = +
jys]ardo
· (0(2x + 843) -
(n + juP)
-I-5u-n + d
-u
- -in t -
=x5 +
=n4]
-
