高數(shù)下課件-ch12-2二重積分計(jì)算

§12.2二重積分的.討論連續(xù)

fx,y0時(shí)，fx,ydxdy的計(jì)算問D其中Dg1xyg2x)圍成axabx0[ab，作平面xx0．a(chǎn)bg1x0g2x0)]為底zfx0y為曲邊的曲邊梯其面積

g1(x),

x)A(x)

g2(x0

f(x,y) g(x 一般地，x[a,b]，則所得截A(x)

g2(xg1(x

f(x,y)曲頂柱體體積V

A(x)

bg2(x

f(x,y)dy

a

g1(x 即

f(x,y)dx

bg2(x

f(x,y)dy

aD

g1(x axb(1式稱為先yx累次積二次積axb(1)式也可寫f(x,y)dxD

g2(xbg1(xb

f(x,y)o bx若zf(x,y)在區(qū)域Do bx其中D：axb，g1x)yg2x)，且g1x),g2x)f(x,y)dxdy

dxg2(x

f(x,y)

b g1(xbydcox若zf(x,y)在區(qū)域D上連ydcox其中D：cy

(y)x2

(1y),2y)df(x,y)dxdycdD

2(1(

f(x,y)

oxbxoxboxbxoxbxxdcoxyX型區(qū) Y型區(qū)

dDc 例例1計(jì)算二重積分dx ,其中D[3x4,1yD(x解 f(x,y)

(x

在Ddxy21O x(xy21O x

dx

(x4 4D1x1x3 4

1

ln

x2 2計(jì)算曲頂柱體的體積,RR[0xa,0y其其頂是定義在R上的曲面ze (p,q是常數(shù)yaOaxyaOaxaaVeR

dxdy0

0

epxqya e a

eqyep

aq0

1(epa1)(eaa0

1)．3計(jì)算四個(gè)平面xyz1,x0,y0,z0z解xOyxy0，xy1定義在D上的平面為：z1x

x 1V(1xy)dxdyD

dx

(1xy)2 2 [(1x)y

0

dx

1(1x)2dx1． 或V

(1xy)dxdy1D1

dy (1xy)dx 例例4計(jì)算二重積分 dxdy,其中區(qū)域D是由直線xD解

dxdy

dx

yx，及雙曲線yx，及雙曲線xy122y x2y

2 2

x2 122

(x

x)

9．5計(jì)算xydxdyDy2xDyx2yy2解由yx

得交點(diǎn)(11),(42)

y xydxdyD

dy

xy

yx2

y21 121 1

[y(yx

y5]dy4

8xxydxD

0dx

xydy

dxx2

xy6IsinxdxdyDD2y x和直線yx，y2y 由y y由yy由y

得交點(diǎn)(1得交(2得交點(diǎn)(4

ox222ycos 22DIsinD

dxdy

y2

dx 2 2

2 22ycosydy4(2 例 求x2ey2dxdy，其中D是以D(0,1)為頂點(diǎn)的三角形解ey2dy無法用初等函數(shù)

1

x dxdyD

x

y2 e

(1 計(jì)算積分I12yyedxx1422yyyyyy解 exdxyy先改變積分次序I

12

eydyx12 1x(eex)dx12

3e e注例 設(shè)f(x,y)是連續(xù)函數(shù)，則二次積 1dx f(x,y)dy y2 y2(

0

f(x,y)dx

f(x,y)dx (B)0 f(x,y)dxy2 y2(C

0

f(x,y)dx

f(x,y)dxy2y2(D)0 f(x,y)dx0解：0

dx

f(x, 1 1

y 1x2211 yx1y2 y20

f(x,y)dx

f(x,y)dx (C例例計(jì)算I1f(0x f(x)x21ey I

1

y1yxoxey1yxoxxx1xx0

21ey1xx2x1

y2e 0

2yey2

ey2101

e1注 若積分區(qū)域D關(guān)于y軸(或x軸)對稱 f(x,y)關(guān)于x(或是偶函數(shù)(或奇函數(shù) 則f(x,y)dxdy等于對稱半?yún)^(qū)域上積D的兩倍(或?yàn)槔?jì)例計(jì)算二重積I(|x||yx2y2 由于D關(guān)于原點(diǎn)對稱|x||y|關(guān)于x與y均是偶函

則有I

(|x||y|)dxdy4(xy)dx1x2y2 11x11x1

(x

40(

111x

)dx8．32xyyD1yexxe解xx1,2yex3xe1yexxe1O1xDxy|xy1,x1,y1},設(shè)例 yy1,

xey3yex1xeyyex

2

xy1x1,y1

2yex3xeydxdy 1 xey1 xe

2xey3yex1xeyye

2142D

212例例13求兩個(gè)底圓半徑相等 直交圓柱面所圍成的立體體積解x2y2R2，x2z2R2，R2x2R2x2DR2R2x2

dxRR

dx(R2x2)dx

23故所求體

V8V1

163二二.將fx,yd轉(zhuǎn)化為極坐標(biāo)中的二重積分形iiiioxii根據(jù)n

f(x,y)dlimf(i,i)iD在極坐標(biāo)系

0 1()2 1 1[2ii2

(i)2]

1(2ii)ii2i

(2

i

i

i在i內(nèi)取圓周i上一點(diǎn)i,iiiioxii對應(yīng)直iiioxii

iicosi，于

sininlimf(i,i)in0nlimf(icosi,isini)in0

f(x,y)df(cos,sin)d 其中將極坐標(biāo)設(shè)積分D：1(2()，其中1(),2()在[,]上連續(xù) E[

段上點(diǎn)的極徑1()于

f(cos,sin)dd 2()f(cos,sin)dD注 若極點(diǎn)在D的內(nèi)部

1

f(cos,sin)區(qū)域D{(,|02，0(f(cos,sin)dd

(

0 d

f(cos,sin)例 計(jì)算(x2y2)dxdy，其D為由圓x2y22yDx2y24y及直x閉區(qū)域

3y0y 3x0所圍成的平解x2y22x2y24

4x 3yy 3x

16 (x2

y2

3d

2d15( 6 6x2y2dxdyD 22Dxy|1x2y2解D注意：被積函數(shù)也要有對稱性x2y2x2x2y2x2x2y2x2

42

2sin

d 例例16x2y2z2R2x2y2Rx解由對稱R2x2R2x2y2DRx其中D為半圓Rxx

dxYXYXx2y2作極坐標(biāo)變xcos,ysin則積分區(qū)域D{(,)|0，0Rcos2R2R2D 0

dR2R20

0

1(R23

2

R323204R3

2(1sin3)04R3

2)． 例例17z3x2y2z1x2y2z31yx解將此z31yxz3x2z1x2投向xOy投影

x2y2z這正是該立xOy面投影D的邊界線,即積分區(qū)域D．V(3x2y2)dxdy(1x2y2 0(22x22y2)dxdy0D

d

(222)d 例例18計(jì)算雙紐線x2y222a2x2y2解作極坐標(biāo)變xcos，ysin．則雙紐線的極坐標(biāo)方程為22a2cos 且其關(guān)于x軸及AdxD

y軸對

cos22a2sin20

說明：計(jì)算二重積分時(shí)，若被積函數(shù)中含有x2 或yx積分區(qū)域的邊界曲線 含有x2 時(shí)常使用(x2

)dx例 I

x2y2

R(2cos2

2sin2) d 2(cos2

sin2 R

4

2(1cos

1cos

4

(11 解法二I

x2