Embed Size (px)
Citation preview
Chapter 6Chapter 6
Unit 5Unit 5Unit 5Unit 5
定积分的几何应用定积分的几何应用定积分的几何应用定积分的几何应用
This section presents various geometric applications
of the definite integral. We will show that area,
volume and length of a curve can be represented as
definite integrals.
That is, “Area is the integral of the length of cross
sections made by lines” and “ Volume is the integral
of the areas of cross sections made by planes” .
1. Introduction1. Introduction
baxfy , above and )(
under region theof area theEvaluating (1)
b
axxfA d)(
a b x
y
o
?A
)(xfy
Method
],,[,],,[],,[ into ,partition to
,
points Selecting )1
12110
1210
nn
nn
xxxxxxba
bxxxxxa
i
n
ii xfA
)(
is areaion approximat The 2)
1
a b x
y
o i ix1x 1ix 1nx
existslimit theif d)(
)(lim
area obtain the to
limit theEvaluate )3
10
b
a
ii
n
i
xxf
xfA
See figureSee figure
a b x
y
o
)(xfy
x dxx
dA
xxf d
.d)()(lim b
axxfxxfA
xxf issction
cross small
of area The
baxU ,on dependent is 1)
If the quantity U satisfies the following conditions:
(2) Method of element(2) Method of element
b
axxfU d)(Then
xxfU )( 3
],[,on defined is where
,
is,that
, , intervalover additivity has quantity The )2(
baxxxU
UU
baU
element of U
.dd and
, ofelement thecalled is ddquantity The
b
a
b
axxfUU
UxxfU
ApplicationsApplications ：：
The method of element has various applications, for
instance, finding the area of a plane region, the
volume of a solid, the length of a plane curve, the
work done by a varying force that moves an object
along a straight line.
2. Area of plane regions2. Area of plane regions
(1) Area of regions in rectangular coordinates(1) Area of regions in rectangular coordinates
0)(under region theof area The 1) xfy
b
axxfAba d)( ,, above and
b
axxfxfA
xfxfbaxxfy
xfy
d)]()([
and ,in for
and )(between region theof area The 2)
12
122
1
x
y
o
)(xfy
a bx xx x
y
o
)(1 xfy
)(2 xfy
a bx x
b
axxfA d)(
b
axxfxfA d)]()([ 12
Example 1Example 1
2
2
and
by boundedregion theof area theFind
xy
xy
Solution 1Solution 12xy
2yx See figure
.3
1
33
2d)(
1
0
321
0
23
x
xxxxA
thus,,1,2 and 0,0 are and of points
cross theand ,
and case, In this
12
21
2
xfxf
xxf
xxf
.3
1
33
2d)(
1
0
321
0
23
x
xxxxA
variableintegral
theas ]1,0[ choose We x
Solution 2Solution 2
2xy
2yx
xxxA d)(d
is area ofelement The2
See figure
Example 2Example 2
23 and 6
by boundedregion theof area theFind
xyxxy
xxxxAx d)6(d],0,2[)1( 231
2xy
xxy 63
then variable,integral
theas ]3,2[ choose We x
SolutionSolution See figure
).9,3(),4,2(),0,0(6
curves twoof points cross The
2
3
xy
xxy
xxxxAx d)6(d],3,0[)2( 322
.12
253
d)6(d)6( 323
0
20
2
3
xxxxxxxxA
Example 3Example 3
4 linestraight theand
2by boundedregion theof area theFind 2
xy
xy
Solution 1Solution 1 See figure
4
2 curves twoof points cross The
2
xy
xy
yy
yA d2
4d2
xy 22
4xy
then
variable,integral theas
]4,2[ Choosing y
).4,8(),2,2(
4
2
24
2.18d
24d y
yyAA
Solution 2Solution 2
xy 22
4xy
then
variable,integral theas
]8,0[ Choosing x
xxxAx d22d],2,0[)1( 1
xxxAx d42d],8,2[)2( 2
18d42d228
2
2
0 xxxxxA
NoticeNotice
Although both approaches to finding the area of the
region in example 3 are valid, the first one is more
convenient.
(2) Area of regions with parameter equations(2) Area of regions with parameter equations
.d)()(
)(under region theof area then the
,)(
)(
equationsby given is function theIf
2
1
21
t
ttttA
xfy
tttty
tx
xfy
Example 4Example 4
SolutionSolution
1 ellipse theof area theFind2
2
2
2
b
y
a
x
The parametric equation of the ellipse is
tby
tax
sin
cos
The part area of the first quadrant
and multiply by 4, we obtain
.
dsin4)cos(dsin4d4 2
20
20
0
ab
ttabtatbxyAa
(3) Area of regions in polar coordinates(3) Area of regions in polar coordinates
Since the polar coordinate is convenient for
describing some curves, we introduce how to
evaluate the areas by applying the polar coordinate.
0? and ,
on continous is where, and
between under region of area theFind
r
For exampleFor example
SolutionSolution See figure
xo
d
d
.d)]([2
1 2
A
)(r
d)]([2
1d
is area ofelement The
2A
The following examples apply this technique.
Example 5Example 5
2cos
by boundedregion theof area theFind22 a
The total area = the area
of the part in the first
quadrant and multiply
by 4, that is,
.d2cos2
14 4 2
0
21
4 aaAAA
xy
2cos22 a
1A
SolutionSolution
Example 6Example 6
0cos1
by boundedregion theof area theFind
aar
d)cos1(2
1d 22 aA
By the symmetry, we have
d
SolutionSolution
.2
32sin
4
1sin2
2
3
d)coscos21(
d)cos1(2
12
2
0
2
0
22
0
22
aa
a
aA
3. Computing volumes3. Computing volumes
This section offers further practice in setting up a
definite integral for the volume of a solid. We will
find that “volume is the integral of cross-sectional
area”
(1) The volume of a solid of revolution(1) The volume of a solid of revolution
The solid formed by revolving a region in a
plane about a line in that plane is called a solid of solid of
revolutionrevolution. See the following figures
circular cylinder circular cone circular truncated cone
Question?Question?
.revolution of solid resulting the
of volume theFind . axis about the revolved
is , above and under region The
x
baxfy
SolutionSolution See figure
x dxx x
y
o
)(xfy
xxfV
bax
d)]([d
is volumeof
ion approximat local the
then variable,integral the
as ],[ Choosing
2
xxfVb
ad)]([
thereforeis solid theof volumeThe
2
The following examples we use the formula to
find the volume of such a solid.
Example 7Example 7
figure. See
cone.circular resulting theof volume theFind
axis. about the revolved is length base and
height with triangleorthogonal theofregion The
xrh
y
r
h
P
xo
SolutionSolution
xh
ry
OP
is ofequation The
xxh
rV
hx
dd
issection cross typical theof area the
then variable,integral theas ],0[ Choose
2
.33
d
is conecircular theof volume theThus,
2
0
3
2
22
0
hrx
h
rxx
h
rV
hh
y
r
h
P
xo
Example 8Example 8
SolutionSolution
.revolution of solid resulting theof
volume theFind axis. about the revolved is
0by boundedregion The 3
2
3
2
3
2
x
aayx
a ao
y
x
],[,
,3
3
2
3
22
3
2
3
2
3
2
aaxxay
xay
.105
32d 3
3
3
2
3
2
axxaVa
a
d
cyyV
y
y
dyc,yyx
d)]([
is revolution of solid resulting
theof volumethe
Then . axis about the
revolved is axis and
,by
bonundedregion theIf
2
Similarly,Similarly,
x
y
o
)( yx c
d
The following examples we use the formula to
find the volume of such a solid.
Example 9Example 9
SolutionSolution
ly.respectiveaxis,
and axis about the 0 and cos1
,sinby boundedregion therevolvingby
obtained revolution of solid theof volume theFind
y
xytay
ttax
a2a
)(xy
2
0
22
22
0
d)cos1()cos1(
d)(
is axisabout revolving volumeThe 1
ttata
xxyV
xa
x
32
2
0
323
5
d)coscos3cos31(
a
tttta
figure. See axis.-about
and region plane revolvingby revolution
of solid ebetween th volume theof difference
theis axis-about revolving of volumeThe 2
yOBC
OABC
y
o
y
xa2A
BCa2 )(2 yxx )(1 yxx yyxV
a
y d)(22
02 yyx
ad)(22
01
2
22 dsin)sin( ttatta
.6dsin)sin(
dsin)sin(
332
0
23
0
22
atttta
ttatta
Example 10Example 10
SolutionSolution
.3 line about the
revolved is 0 and 4by bounded
region that thesolid theof volume theFind2
x
yxy
See figure
yQMPMV d][d22
,d412
d])43()43([ 22
yy
yyy
3
dyP
QM
is volume theofion approximat
local then thevariable,
integral theas ]4,0[ Choosing y
.64d4124
0 yyV
(2) The volume of a solid with known area of cros(2) The volume of a solid with known area of cros
s sections made by planess sections made by planes
Suppose that we wish to compute the volume of a
solid S, and we know the area A(x) of each cross
section made by planes in a fixed direction (see
Fig.)
xo a bx dxx
In order to evaluate the total volume we first approxim
ate the volume of the region bordered by two parallel p
lanes a distance dx apart
solid. the
of volume theion toapproximat local typical
a is d)(dThen .point at the axis
thelar toperpendicu plane by the made solid
theofsection cross theof area thebe Let
xxAVxx
xA
.d)( of Volume b
axxAS
In other words, volume is the definite integral of
cross-sectional area .
Example 11Example 11
SolutionSolution See figure
cylinder.circular
cut the plane that thesolid theof volume theFind
. is circle base theand plane ebetween th angle
cross theand cylinder,circular theof radiuswith
circle base theoforgion he through tpass planeA
R
R
R
x
yo 222 is circle
base theofequation The
Ryx x
,tan)(2
1)( is area its
triangleorthogonalan issection cross theSince
22 xRxA
.tan3
2
dtan)(2
1 solid theof Volume
3
22
R
xxRR
R
Example 12Example 12
SolutionSolution See figure
.hight of
triangleisosceles are axis- thelar toperpendicu
sections cross theand radius of circle a
is base hein which t solid theof volume theFind
h
x
R
x
y
o Rx
22)(
issection cross theof area typical theAnd
xRhyhxA
222 is circle
theofequation The
Ryx
.2
1d 222 hRxxRhV
R
R
4. The arc length of a plane curve4. The arc length of a plane curve
Definition 1Definition 1
xo
y
0MA
nMB 1M
2M 1nMSee figure
n
iii
nn
i
MMl
ABn
BMM
MMMA
AB
11
1
10
lim
as oflength thedefine wesections,
into ,,
,,,by
curve thePartition
(1) Length of a curve in rectangular coordinates(1) Length of a curve in rectangular coordinates
curve? the
oflength theFind ].,[on abledifferenti and
continous is where, by
given is curve theofequation that theSuppose
ba
xfbxaxfy
SolutionSolution See figure
xo
y
a bx xx d
dy
then ],,[d,
variable.
integral theas choose We
baxxx
x
.d1 is curve theoflength theSo 2 xysb
a
xyyxs d1)d()d(d
:issecant small typical theoflength the
222
Example 13Example 13
SolutionSolution
baxxy , ,3
2 curve theoflength theFind 2
3
,d1d)(1d
,
221
21
xxxxs
xy
a b
See figure
].)1()1[(3
2d1
islength theHence,
23
23
abxxsb
a
Example 14Example 14
SolutionSolution
nxxny nx
0 dsin
curve theoflength theFind
0
,sin1
sin n
x
nn
xny
tnt
xn
xxys
ntx
nb
a
dsin1
dsin1d1
0
0
2
.4d2
cos2
sin
d2
cos2
sin22
cos2
sin
0
0
22
nttt
n
ttttt
n
(2) Length of a curve with parametric equations(2) Length of a curve with parametric equations
by given is curve theofequation theIf
curve? theoflength theFind .,on
derivable and continous are )( and )( where
),( ,)(
)(
tt
tty
tx
SolutionSolution
ttt
tttyxs
d)()(
)d)](()([)d()d(d
22
22222
.d)()(
is curve theoflength the
22 ttts
Example 15Example 15
SolutionSolution
0
curve theoflength total theFind
3
2
3
2
3
2
aayx
)20( ,sin
cos
is curve thisofequation parametric The
3
3
t
tay
tax
1 4length total theObviously, ss
The length in the first quadrant
.6dcossin34d4 2
0
2
0
22 atttatyx
(3) Length of a curve in polar coordinates(3) Length of a curve in polar coordinates
curve? theoflength theFind
.,on derivable and continous is )( where
)( )(
bygiven is curve theofequation theIf
r
rr
SolutionSolution
,d)()()d()d(d
)( ,sin)(
cos)(
2222
rryxs
ry
rx
.d)()( 22
rrs
Example 16Example 16
SolutionSolution
30 ,0
3sin
curve theoflength total theFind3
aar
,3
cos3
sin3
1
3cos
3sin3
22
aar
.2
3d
3sin
d3
cos3
sin3
sin
d)()(
3
0
2
3
0
242
62
22
aa
aa
rrs
Example 17Example 17
SolutionSolution
.20 0,
Archimedes of spiral theoflength theFind
aar
See figure
d)()(
,
22
rrs
ar
.)412ln(4122
22 a
2
0
2
2
0
222
d1
d
a
aa
This section presented various geometric applications
of the definite integral. We showed that area, volume
and length of the curve can be represented as definite
integrals. Next section we will present various
physical applications of the definite integral.
