applied calculus chapter 4 multiple integrals
TRANSCRIPT
xddyyxfdxdyyxf
dydxyxfdxdyyxf
dycbxayxR
x,yf
b
a
d
c
b
a
d
c
d
c
b
a
d
c
b
a
),(),(
),(),(
then
,:),(
regionr rectangula ain continuous is )( If
Iterated Integral
1
0
2
1
1
1
2
2
30 (ii)
)23( (i)
x
x
ydydx
dydxxyx
Example1
Evaluate the iterated integrals.
14
)1(2)2(226
])1()1(3[])1()1(3[
3)23(
obtain weintegrals, iterated of definition the Using(i)
332
1
3
2
1
2
2
1
2222
2
1
1
1
22
2
1
1
1
2
xdxx
dxxxxx
dxxyyxdydxxyxy
y
Solution :
2
3535
)1515(
1530
obtain weintegrals, iterated of definition the Using(ii)
1
0
53
1
0
42
1
0
2
1
0
2
2
xx
dxxx
dxyydydxxy
xy
x
x
theorem.sFubini’– integrals iteratedan as
calculated becan function continuousany of integrals
double the1943),-(1879 Fubini Gaudio toAccording
dycbxayxR ,:),(
ifregion r rectangula aon theoremsFubini’
b
a
d
c
d
c
b
a
dydxyxfdxdyyxfdAyxf ),(),(),(R
then
Fubini’s Theorem
(-2,1) and (3,1) (0,0), ertices v
h theregion witr triangulaclosed theis ;),( (iii)
sin and 0
,,0 boundedregion theis ;),( (ii)
20,2:),(;4),( (i)
),( Evaluate
2
2
Rxyyxf
xyy
xxRyyxf
yyxyyxRyxyxf
dAyxfR
Example2
5
36
5
2
42
)26(
]})(2[]2)2(2{[
2)4()4(
2
0
543
2
0
432
2
0
32222
2
0
22
2
0
2
2
2
yyy
dyyyy
dyyyyy
dyxyxdxdyyxdAyxyx
yx
y
y
yx
yxR
42
2sin
4
1
)2cos1(4
1
2
sin
2
0
0
0
2
0
sin
0
2
0
sin
0
xx
dxx
dxx
dxy
dydxydAy
xx
x
xy
yR
2
1
2
2
5)49(
2
2
1
0
5
1
0
41
0
222
1
0
3
2
221
0
3
2
22
y
dyy
dyyyy
dyyx
dxdyxydAxy
yx
yx
y
y
yx
yxR
as described is solid The
.0 and 4 , 9
by bounded solid theof volume theFind
22 zzyyx
922 yx
yz 4
0z
R
Example 3
3
3
2
3
3
9
9
2
3
3
9
9
98
24
)4()4(
by given is volume theThus,
2
2
2
2
dxx
dxy
y
dydxydAyV
V
xy
xy
x
x
xy
xyR
3622
)sin(
22
sin36
2
2sin36
)12(cos36cos72
)cos3()sin1(9898
2/
2/
2/
2/
2/
2/
2
2/
2/
2
3
3
2
tt
dtttdt
tdttdxxV
Hence . cos3,sin3let Now tdtdxtx
Example 4
2
0
1
20
2
(ii) sin
(i)
Evaluate
y
x
x
dxdyedydxy
y
as dillustrate is
Rregion The (i)
Solution :
Reversing The Order of
Integration
.integrated becannot sin
But y
y
: nintegratio oforder the reverse ,So dydx dxdy
: becomeregion theThen,
21cos
cossin
)0(sin
sin
sinsin(i)
0
0
0
0
0
0 00
yydy
dyyy
y
dyxy
y
dxdyy
ydydx
y
y
y
y
y
y
yx
x
y
y
yx
x
x
x
y
xy
.integrated becannot But 2
dxex
:n integratio oforder thereverse So, dydx dxdy
: becomeregion theThen,
1
2
1
0
1
0
1
0
1
0
2
0
1
0
2
0
2
0
1
2/
2
2
22
eedue
dxxe
dxye
dydxedxdye
u
u
u
u
x
x
x
x
x
xy
y
x
x
x
xy
y
x
y
y
x
yx
x
scoordinatepolar in the
)( of integral theevaluatemay then we)(
to)(function aconvert can that weSuppose
the
in evaluate easier to isit shape,circular involvingWhen
r,fr, fz
x,yz
s.coordinatepolar
R
dA
dV
A
V
R
dArfV
dArfdV
),(
),(
)( r, fz
)( r, fz
1
0 0
22
2
2
0
4
0
22
)( (ii)
)cos( (i)
scoordinatepolar to
changingby integrals following theEvaluate
2
y
x
dxdyyx
y
dydxyx
Example 5
0r
2r
0
2/
4
4sin
4sinsin
)(cos
)(cos
)cos(
Therefore
2/
0
21
2/
0
4
021
2/
0
4
0
21
2/
0
2
0
2
2
0
4
0
22
2
ddu
dudu
rdrdr
dydxyx
x
Solution :
as described is Rn integratio ofregion The (ii)
8422
1
2
1sin
2
csc
sin2
sin
)sin(
Therefore
2/
4/
2/
4/
22
2/
4/
2
csc
0
22/
4/
csc
0
2
2/
4/
csc
0
2
2
1
0 0
22
2
dd
dr
drdr
rdrdr
r
dxdyyx
y
r
r
y
as described is base its and solid The
3. Example solve toscoordinatepolar the Use
R
3r
R
y
x3 3
Example 6
36cos918)sin918(
sin3
2
)sin4()sin4(
)sin4()4(
bygiven is V volume theThus,
2
0
2
0
2
0
3
0
32
2
0
3
0
2
2
0
3
0
d
dr
r
drdrrrdrdr
dArdAyV
r
r
r
r
RR
A lamina is a flat sheet (or plate) that is so thin as to be considered two-dimensional.
Suppose the lamina occupies a region D of the xy-plane and its density (in units of mass per area) at a point (x, y) in D is given by ρ(x, y), where ρ is a continuous function on D. This means that
A
myx
lim),(
where Δm and ΔA are the mass and area of a small rectangle that contains (x, y) and the limit is taken as the dimensions of the rectangle approach 0.
Laminas & Density
Definition mass of a planar lamina
of variable density
1 1
0 0
11
2
00
A triangular lamina with vertices 0,0 , 0,1
and 1,0 has density function , .
Find its total mass.
Solution :
,
1 1...
2 24
x
R
x
x y xy
m x y dA xy dydx
m xy dx unit of mass
Example 7
The moment of a point about an axis is the product of its mass and its distance from the axis.
To find the moments of a lamina about the x- and y-axes, we partition D into small rectangles and assume the entire mass of each subrectangle is concentrated at an interior point. Then the moment of Rij about the x-axis is given by
and the moment of Rk about the y-axis is given by
**** ),())(mass( ijijijij yAyxy
**** ),())(mass( ijijijij xAyxx
Moment
m
i
n
j D
ijijijnm
x dAyxyAyxyM1 1
***
,),(),(lim
The moment about the x-axis of the entire lamina is
The moment about the y-axis of the entire lamina is
m
i
n
j D
ijijijnm
y dAyxxAyxxM1 1
***
,),(),(lim
Center of Mass
The center of mass of a lamina is the “balance point.” That is, the place where you could balance the lamina on a “pencil point.” The coordinates (x, y) of the center of mass of a lamina occupying the region D and having density function ρ(x, y) is
where the mass m is given by
D
x
D
ydAyxy
mm
MydAyxx
mm
Mx ),(
1),(
1
D
dAyxm ),(
Moments and Center of Mass of A
Variable Density Planar Lamina
2
Find the mass and center of mass of the lamina
that occupies the region and has the given
density of function .
a) , 0 2, 1 1 ; ,
4 4: , ,0
3 3
) is bounded by , 0, 0, and 1;x
D
D x y x y x y xy
Ans
b D y e y x x
x
322
2 2
,
4 11 1: 1 , ,
4 2 1 9 1
y y
eeAns e
e e
Example 8
Example 9
Find the surface area of the portion of the surface
that lies above the rectangle R in the xy-
plane whose coordinates satisfy
24 xz
40 and 10 yx
Example 10
Find the surface area of the portion of the paraboloid
below the plane 22 yxz 1z
Gregion closed D-3 aon
)( variables threeoffunction a ofn integratio
an is scoordinateCartesian in integrals Triple
x,y,zf
dzdAdzdydxdV
Example 11
Gregion theof volume(ii)The
6),,( wheredV ),,( (i)
evaluate plane,- and plane- ,1
by boundedoctant first in theregion theisG If
G
2
zzyxfzyxf
yzxyyz, xy
obtain weThus .1 and
10by bounded Rregion a is plane-
on theG of projection The1 )(z and
0 )( 6z, )( have wecase In this
2
2
1
yxy
, , xxxy
– y. x,y
x,yzx,y,zf
R
R
yz
z
R
yz
zGG
dAy
dAz
dAzdzzdVdVzyxf
2
1
0
2
1
0
)1(3
3
66),,(
(i)
35
16
75
3
)331()1(
)1(
)1(3
1
0
753
1
0
642
1
0
32
1
0
13
1
0
1
2
2
2
xxxx
dxxxxdxx
dxy
dydxy
x
x
y
xy
x
x
y
xy
♣
RR
yz
zG
dAydAdzdVV )1(
1
0
15
4
103222
1
2
)1(
1
0
531
0
42
1
0
12
1
0
1
2
2
xxxdx
xx
dxy
y
dydxy
x
x
y
xy
x
x
y
xy
(ii)
.
are plane-
on the R projection its andG region theof solid The
.0 and ,25,9by bounded
solid theof volume thefind toscoordinate lcylindrica e Us
2222
xy
zyxzyx
922 yx
2225 yxz
0z
Example 12
3
122
3
61
3
)25(
)25(2
125
25
Thus, 0. z have also We.25 z
obtain we,relation Using
2
0
2
0
9
0
2/3
2
0
9
0
2/1
2
0
3
0
2
2
25
0
22
222
2
1
ddu
dudurdrdr
dArdAdzdVV
-r
r y x
r
r
RR
rz
zG
Solution:
Example 13
are plane- xyon the R projection its andG region theof solid The
9.z plane theand paraboloid by the
bounded solid theof volume thefind toscoordinate lcylindrica Use
22 yx z
2
81
4
81
42
9)9(
)9(
)9(
is volumerequired theThus
2
0
2
0
3
0
422
0
3
0
3
2
0
3
0
2
2
9
2
d
drr
drdrr
rdrdr
dArdAdzdVV
r
r
RR
z
rzG
Solution:
below dillustrate asG region theof solid The
. 3 sphere theinside and
3
1 cone theabove lies that solid theof volume theFind
Example 14
92
9
)1cos(9cos9
sin9sin3
sin
is volumerequired theThus
2
0
2
0
3
2
0
0
2
0 0
2
0 0
3
0
3
2
0 0
3
0
2
3
33
3
d
dd
dddd
ddddVVG
Solution:
“In order to succeed, your desire for success should be greater than your fear of failure. ”