Bai9 Tich Phan Kep

8/12/2019 Bai9 Tich Phan Kep

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TCH PHN BI

Chng 2:

Phn 1 : TCH PHN KP


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BI TON TH TCH

Xt vt th hnh tr c gii hn trn bi

mt cong z = f(x, y) > 0, mt di l Oxy, baoxung quanh l mt tr c ng sinh // Oz v

ng chun l bin ca min D ng v b

chn trong Oxy. Tm th tch .D


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z = f(x, y)z

xy

D


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Xp x bng cc hnh tr con


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Dij

Th tch xp x ca hnh tr con * *

( ) ( , )ij ij ij ij V S D f x y

,( ) ij

i j V V


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NH NGHA TCH PHN KP

Cho hm s z = f(x, y) xc nh trong minD ng v b chn.

D


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Dk

Phn hoch D thnh cc min con D1, D2, , Dn

Sk l din tchca min conDk.

d (D k ) = ng knh Dk = khong cch

ln nht gia 2 im trong Dk.

1,max{ ( )}

k k nd d D ng knh phn hoch


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)( k k D S S

Mk

f (Mk)

1( )

n

n k k k

S f M S Tng tch phn ca f

Mk c chn ty trong Dk

D


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1( )

n

n k k k

S f M S

f kh tch nu:0

lim nd

S

vi phn hoch ty ca D

Tch phn kp ca f trn D l gii hnnu c ca Sn

0( , ) lim n

d Df x y ds S


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Phn hoch D theo cc ng // ox, oy

Dij


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Khi f kh tch, vic tnh tch phn khng phthuc vo phn hoch. Do c th phnhoch D theo cc ng song song Ox, Oy.

Dk l hnh ch nht vi cc cnh x, y

Thay cch vit tp kp

( , ) ( , )D D

f x y dxdy f x y ds

Sk = x. y


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Nhn dng hm kh tch

ng cong (C) : y = y(x)trn ti M(x0,y0) (C) nu y(x) lin tc ti x0.

(C) trn tng khc nu (C) c chia thnhhu hn cc on trn.

Nu f(x,y) lin tc trn min D ng, b chnv c bin trn tng khc th f kh tch trnD.


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( ) 11/D

S D dxdy

Tnh cht hm kh tch

2 / . ( , ) . ( , )

( ) D D

D D D

c f x y dxdy c f x y dxdy

f g dxdy fdxdy gdxdy

Cho D l min ng v b chn

1 2 1 2

1 2 1 23 / ,

D D D D

D D D D D

fdxdy fdxdy fdxdy

va khong dam nhau(toi a chdnh bien)

(Din tch D)


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nh l gi tr trung bnh

0

1( ) ( , )

( ) Df M f x y dxdy

S D

D l min lin thng nu 2 im ty trong D cth ni nhau bi 1ng cong lin tc trong D.

Cho f lin tc trn tp ng, b chn, linthng D. Khi tn ti M0(x0, y0) D sao cho

1( , )

( ) D

f x y dxdy S D

gi l gi tr trung

bnh ca f trn D.


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CCH TNH TCH PHN KP

2 ( )y y x

1( )y y x

1 2( ):

( )y x y aD

x x b

y D

a b

2

1

( )

( )

( ,( ), )y b

a y D

x

x

f x y dy f x y dxdy dx

2

1

( )

( )

( , )y x

y x

f x y dy b

a

dx

Cch vit:


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D

c

d2 ( ) x x y

1( ) x x y

1 2( ):

( ) x y x c D

y y d

x

2

1

( )

( )

( ,( ), )

x d

c x D

y

y

f x y dx f x y dxdy dy

2

1

( )

( )

( , ) x y d

x y c

f x d y y dx

Cch vit:


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V D

D

I xydxdy

0

1

0

x

xydy I dx 1

0

2

02

x

x dx y

1/ Tnh

vi D l tam gic OAB,O(0, 0), A(1, 0), B(1, 1)

1

1

O

A

B

1 3

0

1

2 8

x dx

CCH 10 1

: x

D 0 y x


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DI xydxdy

1

1

O

A

B

1 2

0

1 1

2 8

y y dy

11

0 y

xydx dy

1 1

0

2

2y

y dy x

CCH 20 1

:y

D 1y x


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( )D

I x y dxdy 2/ Tnh

vi D: x2 + y

2 1, y 0

1-1

211

1 0( )

x

x y dy I dx

212

0

1

1 2

x

dx y

xy

21y x

1 1:

x D

1 2

2

1

1 21

2 3

x x x dx

20 1y x


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( )D

I x y dxdy

1-1

2

2

1 1

01

( )y

y

I dy x y dx 1

2

0

2 1y y dy

21y x

0 1:

y D

23

2 21 1y x y


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( 1)D

I x dxdy 3/ Tnh

vi D gii hn bi cc ng y = x, y = x2

y = x2

2

0 1:

x D

x y x

2

1

0( 1)

x

x

I dx x dy

12

0

( 1)( ) x x x dx

13

0

1( )

4 x x dx


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( 1)D

I x dxdy 4/ Tnhvi D gii hn bi cc ng y2 + 8x = 16,y2 24x = 48

y2 24x = 48 y2 + 8x = 16

2 2

2 2: 48 8

24 24

y y x D

y


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5/ Tnh din tch min D gii hn bi cc ng 2(2 ) , 2 y x x y x x

Honh giao im 2(2 ) 2

0, 20

x x x x x x x

2

0 2:

2 (2 )

x D

x x y x x

( )D

S D dxdy 2

2 (2 )

0 2

x x

x x

dx dy


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6/ Tnh 2

4

y

D

xedxdy

y

min D gii hn bi cc ng: y = 0,y= 4 x2, x 0,

22 4 2

0 04

x y

xeI dx dy y Kh lynguynhm

4 4 2

0 0

4

y y xeI dy dx

y

4

i th t

2

24y x


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4 4 2

0 04

y y xeI dy dx y

4 2

02

y edy

8 14 4

e

4 42 2

004 2

y y

e x dy y


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7/ Tnh D

x y dxdy

min D gii hn bi cc ng: y = 0, y= 2 x2

22

2

1

D1

D2

2 2


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6/ Tnh D

x y dxdy

min D gii hn bi cc ng: y = 0, y= 2 x2

2

1

D2

D1

2 2

1

2

( )

( )

D

D

I y x dxdy

x y dxdy


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7/ V min ly tch phn v i th t ly tptrong cc VD sau

1 2

01 / ( , )

y

y

I dy f x y dx

4 4

02 / ( , )

y

y

I dy f x y dx

2 2

1 2

3 / ( , )y

y

I dy f x y dx


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1 2

01/ ( , )

y

y

I dy f x y dx

x y


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1 2

01/ ( , )

y

y

I dy f x y dx

x y

2 x y


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1 2

01/ ( , )

y

y

I dy f x y dx

x y

2 x y

2 x y y

0 1y


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1 2

01/ ( , )

y

y

I dy f x y dx

x y

2 x y

2 x y y 20

0 1

y

x

x

0 2

1 2

y

x

x

0 1y

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