TICH PHAN KEP Phan 3 Co Hinh Ve

8/12/2019 TICH PHAN KEP Phan 3 Co Hinh Ve

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NG DNG HNH HC CA

TCH PHN KP


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NI DUNG

Tnh din tch min phng

Tnh th tch vt th trong R3

Tnh din tch mt cong


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TNH DIN TCH MIN PHNG

( )

D

S D dxdy

D l min ng v b chn trong R2:

C th dng cch tnh ca tp xc nh

trong GT1 cho nhng bi khng i bin.


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V d

2

1

0

x

x

dx dy 1

3

2,y x y x 1/ Tnh din tch min D gii hn bi:

2y x ( )D

S D dxdy

y x


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2 2 1x y

2/ Tnh din tch min D l phn nm ngoi

ng trn v nm trong ng trn

2 2

2 2

1

2

3

x y

x y x

Ta giao im

i bin: x = rcos, y = rsin

2 2 2

3x y x


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1

3cos2

r

6 6:2

1 cos3

D

r

2 2

2 2

1

23

x y

x y x

1

6

r

2cos

6 3

1

6

( )S D d rdr

3

6 18


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D1= D{x,y)/ y 0} S(D) = 2S(D1)

Min D i xng qua Ox

1

06

:2

1 cos3

D

r

Nu s dng tnh i xng ca D

2cos

6 3

0 1

( )S D d rdr


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BI TON TH TCH

Xt vt th hnh tr c gii hn trn bimt cong z = f2(x, y), mt di l z = f1(x, y),

bao xung quanh l mt tr c ng sinh //

Oz v ng chun l bin ca min D

ng v b chn trong Oxy.

Khi , hnh chiu ca ln Oxy l D.

2 1( ) ( , ) ( , )D

V f x y f x y dxdy


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Cch xc nh hm tnh tch phn v hnh chiu D

Chn hm tng ng vi bin ch xut hin

2 ln trong cc pt gii hn min tnh thtch ().

VD: z ch xut hin 2 ln : z = f1(x, y),

z = f2(x,y), hm tnh tp l

z = |f2(x,y)f1(x,y)|

B1: chn hm tnh tch phn:


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Cch xc nh hm tnh tch phn v hnh chiu D

Gs hm tnh tp l z = f(x,y), D l hnh chiu

ca ln mp Oxyv c xc nh t cc

yu t sau:

1.iu kin xc nh ca hm tnh tp

2.Cc pt khng cha z gii hn min .

3.Hnh chiu giao tuyn ca z = f1(x,y) v

z = f2(x,y) (c th khng s dng)

B2: Xc nh min tnh tp D


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Hnh chiu giao tuyn

1.c tm bng cch khzt cc pt cha z.

2. Cc TH s dng hc giao tuyn.

Tm c t k 1,2

Khng s dng S dng


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D1

D2

S dng xc nh du ca f2f1

f1> f2

f2> f1


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V d

, 0, 0, 1y x y z x z

OxyD hc

0,y y x

1/ Tnh th tch ca vt th gii hn bi:

Cch 1: z xut hin 2 ln nn hm ly tp l

z = 1xv z = 0 (cc hm xc nh trn R2)

cc pt khng cha z

1 0x Hc giao tuyn:1

1

D


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( ) [(1 ) 0]

D

V x dxdy

2

1 1

0

(1 )y

dy x dx

2

1 1

0

4(1 )

15y

dy x dx


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Cch 2: y xut hin 2 ln, chn hm tnh tp l

0,y y x Oxz

D hc

1, 0x z z

Cc pt khng cha y:

0 0x x

Hc giao tuyn:

0x

k xc nh ca

hm tnh tp:

z

x

: , 0, 0, 1y x y z x z


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( ) [ 0]

D

V x dxdz

1 1

0 0

x

dx xdz

1

1/2 3/2

0

4

15x x dx


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1

1z = 1y2

y

z

Cch 3: x xut hin 2 ln, chn hm tnh tp l

2, 1y x x y x z

0, 0y z

Cc pt khng cha x:

21 z y Hc giao tuyn:

0yk xc nh hm:

OyzD hc

: , 0, 0, 1y x y z x z


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2( ) [(1 ) ]

D

V z y dydz

21 1

2

0 0

4(1 )

15

y

dy z y dz


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OxyD hc

OyzD hc

Oxz

D hc


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y x


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y x


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y x


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y x

1x z


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y x

1x z


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y x

1x z


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2 2 2 24 , 0, 2z x y z x y


z xut hin 2 ln nn hm ly tp l:2 24 , 0z x y z

Oxy

D hc

2 2 2x y

Cc pt khng cha z:

2 20 4 x y Hc hiu giao tuyn:

22

Hnh chiu giao tuyn khng s dng


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2 2( ) [(4 ) 0]D

V x y dxdy

2 2

2

0 0

(4 )d r rdr

6


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2 2

4z x y

2 2 2x y

0z


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2 2 2 24 , 2 2z x y z x y


Hm tnh tp:2 2

2 24 , 12

x yz x y z :

OxyD hc

2 2

2 2

2 2

4 12

2

x yx y

x y

(hc giao tuyn)

2 22 2( ) (4 ) 1

2D

x yV x y dxdy


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2 22 2( ) (4 ) 1

2D

x yV x y dxdy

2 21 6 3 3

2 D

x y dxdy

2 2

2

0 0

3(2 ) 32 d r rdr


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2 24z x y

2 2

1

2

x yz

Hnh chiu: x2+ y2 2


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2 22 1, 1z x y x y


v cc mt ta .

Cc mt ta bao gm: x = 0, y = 0, z = 0

Hm tp:2 22 1, 0z x y z

:Oxy

D hc 1, 0, 0x y x y (Khng c gt ca 2 mt cong tnh tp)

2 2( ) 1 2D

V x y dxdy

2 2


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2 22 1z x y

1x y

D


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5/ T h th t h t th h bi


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2 2 2 2 24, 2 , 0x y z x y y z 5/ Tnh th tch ca vt th cho bi:

Hm tp : 2 24 , 0z x y z

:Oxy

D hc 2 2 2 24, 2x y x y y

2 2 2( ) 4

D

V x y dxdy

2 2sin

2

0 0

( ) 2 4V d r rdr

s dng tnh i xng ca D:


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2 24z x y


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TNH DIN TCH MT CONG


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TNH DIN TCH MT CONG

OxyD hc S

Mt cong S c phng trnh: z = f(x, y),

Din tch ca S tnh bi cng thc

2 21 ( ) ( )x yD

S f f dxdy

C h t h di t h t


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Gi s S c pt tng qut F(x,y,z)=0

1.Chn cch vit tp mt cong S( tng ng

vi bin xut hin t nht trong pt cc mtchn v pt ca S)

2.Tnh phn vi phn mt cho hm ly tp.

3.Tm hnh chiu D(ging nh tnh th tch)

Cch tnh din tch mt cong

V D


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V D

2 2 2x y y

:Oxy

D hc

2 2 2 24, 2x y x y y

2 24z x y 1/ Tnh din tch ca

b chn trong mt tr

2

D

Pt mt cong: 2 24z x y

2 2 2 2,

4 4

x y

x yz z

x y x y

2 2( ) ( )S d d


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2 21 ( ) ( )x yD

S z z dxdy

2 2

2

4D

dxdyx y

2 2sin

20 0

22

4

rdrd

r

4 8

D

2


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2 24z x y

2 2 2x y y

222/ Tnh din tch ca phn mt tr:


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22z x2/ Tnh din tch ca phn mt tr:

b chn bi cc mt 2 0, 2 0,x y y x

Phng trnht mt cong:2

2

xz

:Oxy

D hc

2 0, 2 0, 2 2x y y x x 2 2

2 2x


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2 21 ( ) ( )x yD

S f f dxdy

21D

x dxdy

2 2 22

0 2

1 13x

x

dx x dy

2 2

2

2

x

z

22z x


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2z x

D

3/ Tnh din tch ca phn mt nn:


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2 2z x y

3/ d t c ca p t

b chn bi mt cu:

2 2 2 2x y z

:Oxy

D hc 2 2 1x y

2 21 ( ) ( )x yD

S f f dxdy 2D

dxdy

2 ( ) 2S D

(S(D) l din tch hnh trn c R = 1)


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4/ Tnh din tch ca phn mt cu:


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2 2 2 4x y z

4/ Tnh din tch ca phn mt cu:

b chn bi cc mt:

, 3 , 0x z z x x Phn mt cu gm 2 na S1v S2:

2 2

1,2 4y x z Hnh chiu ca S1v S2ln Oxz ging nhau

v xc nh bi:

S = S1+ S2

2 24 0,:

, 3 , 0

x zD

z x z x x

2 24 0x z


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4 0,:

, 3 , 0

x zD

z x z x x

z

x

4

2 2


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2 21 2

2 2

1 ( ) ( )

2

4

D

z

D

xS S dxdz

dxdz

x z

y y

4 2

26 0

2

4

rdrd

r

12

1 26

S S S

2 24y x z


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TICH PHAN KEP Phan 3 Co Hinh Ve

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