二重積分概念及性質(zhì)

第八 重積z 1x1重積分是定積分的推?和發(fā)展.?樣也是某種確定和式的極限,其基本思想是四步曲:分割、取近似、求和、取極限.定積分的被積函數(shù)是?元函數(shù),其積分區(qū)域是?個(gè)確定區(qū)間.??重、三重積分的被積函數(shù)是?元、三元函數(shù),其積分域是?個(gè)平?有界2

double二重積分的性質(zhì)小結(jié)思考題第八第八 重積曲頂柱體的體積?曲頂柱體的體積的計(jì)算問題41.1.1.曲頂柱體以xOy?上的閉區(qū)域D為底側(cè)?以曲?z=f(x,y(fx,y)0且在D上連續(xù)). z=f(x, Dx

5分 特點(diǎn)平=底?積×高回 曲邊梯形?積是如何思想是以直代以不變代變?nèi)绾蝿?chuàng)造條件使平與曲這 6O之和近似表?O

z=f(x,

????D D

V=limf(x,h lfi

i7 ,Dsn(Dsi表?第i個(gè)?域的?積相應(yīng)地此曲頂取近 在每個(gè)?域內(nèi)任取?

i=1,2,3,

8求 ni V i取極限令n個(gè)?域的直徑中的最?值(nnV=limf(xi,hi)Dslfi0i9在點(diǎn)x,y)處的面密度為mx,y假定mx,y) nDMi?m(xi,hi)Dsnnn

M=limm(xi,hilfi01.1.1.定義設(shè)f(xy)是有界閉區(qū)域D上的有界函數(shù)將閉區(qū)域D任意分成n ,Dsn其中Dsi表?第i個(gè)?區(qū)域也表?它的?積 (i=1,2, 并作和f(xi,hi)Dsi n④如果當(dāng)nf(xy)在閉區(qū)域D上的二重積分記為fxy)ds,i=1nDf(Df(x,積被積分積分區(qū)函變域數(shù)

lfi被 積 表 達(dá) z=fx,y)0在底D上的?重積分 V=f(x,y)dsD它的?密度mxy)在薄片D上的?重積分 M=m(x,y)dsD ff(x,D注注重積分中ds0,定積分中dx可正可負(fù)yDOxf(x,y)ds=f(x,yDOx 設(shè)f(xy)是有界閉區(qū)域D上的連續(xù)函數(shù)或是分片連續(xù)函數(shù)時(shí),則f(x,D存在注今后的討論中注當(dāng)fxy0時(shí)?重積分是fxy0時(shí)?重積分是柱體體積的負(fù)值?在其它的部分區(qū)域上是負(fù)的那末f(xy)在D上代數(shù)和 設(shè)D為圓域x2+y2￡ ?重積分D= z

R2-xR2-x2-y2xRR2R2-x2-y23R2-x2-y2ds=2πR33D性質(zhì)1(線性性質(zhì))設(shè)a、b為常數(shù)[af(x,y)–bg(x,D=af(x,y)ds–bg(x, x2+y2Dx2+y2D -2πa3

)ds其中D為x2y2￡b>a>=πa 性質(zhì)2區(qū)域可加性)將區(qū)域D分為兩個(gè)?域D1,f(x,y)ds=f(x,

(D=D1+D2 +f(x,性質(zhì)3(幾何應(yīng)用)

D1s=1ds=ds

1外?公共點(diǎn)注ds既可看成是以D為底以1注D柱體體積?可看成是D的?積性質(zhì)4(正性 f(x,y)?0,(x,y)?D,f(x,y)ds?D (A(C)等于

= x2y2d= 1￡x2+y1￡x2+y2(B(D性質(zhì)5(單調(diào)性設(shè)f(x,y)￡g(x, (x,y)?D,f(x,y)ds￡g(x,

性質(zhì)6(絕對(duì)可積性若f(xy)可積則|f(xy)|f(x,D

￡D

f(x,y)設(shè)I1

x2+y2ds, =cos(x2+y2)ds考研數(shù)學(xué)(三,四)(4分2I3=cos(xD+y)ds,其中D={x,y) 考研數(shù)學(xué)(三,四)(4分2 D((A)I3>I2>I1.(C)I2>I1>I3x?[0, [0, cos2x2+y2由于x2y22￡x2x2+y2

(B)I1>I2>I3(D)I3>I1>I2xx2+所以

￡cos(x2+y2)￡cos(x2+y2比較I1=(x+y)2ds與 =(x+性質(zhì)5(性質(zhì)5(單調(diào)性2的??,其中D:(x-2)2+(y-1)2￡1,則 (A)I1=I2(C)I1￡I2yx+y=?1 ??? ??

(B)I1>I2(D)?法比較(x,y)?D,x+y>(x+y)2￡(x+y)3x設(shè)fx,y)￡gx,y)x,y?則設(shè)fx,y)￡gx,y)x,y?則fx,y)ds￡gx, 在D上的最?、最?值σ為D的?積,ms￡f(x,y)ds￡D

設(shè)fx,y0x,y?D, mds￡f(x,y)ds￡M 再?性質(zhì)1和性質(zhì)3,證畢估值性質(zhì)m估值性質(zhì)ms￡fx,y)ds￡MD例不作計(jì)算,Iexy)ds的值D

￡1,(0<b<a). 解區(qū)域D的?積sm

0￡x2+y2￡eaea1=e0￡ex2+y2 s￡e( )ds￡sD abπ￡e(x+y)ds￡ ms￡f(x,y)dsms￡f(x,y)ds￡Dfxy)dsf(x,h)s(s為D的?積D?何意義設(shè)fx,y0,x,y?D則曲頂柱體體積等于以D為底以f(x,h為?的平頂柱體體積.證顯然s0.將性質(zhì)7中不等式各除以s,m￡1f(x,y)ds￡sm￡1f(x,y)ds￡ss即是說確定的數(shù)值1sD

f(x,

f(xy)的最?值M與最?值m之間的在D上?少存在?點(diǎn)(x,h使得函數(shù)在該點(diǎn)的值與這個(gè)確定的數(shù)值相等,即1f(x,y)ds=f(x,hsf(xy)是有界閉區(qū)域D:x2y2￡r2連續(xù)函數(shù)極限

f(x,y)ds是 rfi0πr不存在

x2+y2￡r2ff (D)f提?利?積分中值定理求1求1rfi02f(x,x2+y2f(x,y)ds=fD解利?積分中值定理即得

其中點(diǎn)(x,hx2y2￡r2內(nèi)的?點(diǎn)顯然,當(dāng)rfi0時(shí)點(diǎn)(x,hfi(0,0).由函數(shù)的連續(xù)性知2 2

f(x,y)ds=limf(x,h)=frfirfi0

x2+y2性質(zhì)9設(shè)區(qū)域D關(guān)于x軸對(duì)稱,D1為D限中的部分,f(x,y)關(guān)于坐標(biāo)y為偶函數(shù)(即f(x,-y)=f(x, 則f(x,y)ds=2f(x, 設(shè)區(qū)域D關(guān)于x軸對(duì)稱f(xy)關(guān)于坐標(biāo)y為奇函數(shù)(即fx,y)fx,y)), f(x,y)ds= f(xy)關(guān)于坐標(biāo)yf(xy)關(guān)于坐標(biāo)y第?象限中的部分,如果函數(shù)f(xy)關(guān)于坐標(biāo)為偶函數(shù)(即f(-x,y)=f(x,y)), f(x,y)ds=2f(x, f(xy)關(guān)于坐標(biāo)x(即f(xyfx,yf(x,y)ds=D例計(jì)算二重積分A[sin(xy2sinx2y)]dsD其中Dxy)1￡x￡1,-1￡y￡解積分區(qū)域D關(guān)于x軸,y軸都對(duì)稱sinxy2和sinx2y)分別關(guān)于x和y是奇函數(shù),A=[sin(xy2)+sin(x2D=sin(xy2)ds+sin(x2 =0+0=研究?考題,選擇,3則xycosxsiny)dxdy等于(A).D2cosxsiny(C)4(xy+cosxsiny)dxdy.

(D)記I

(xy(xy+cosxsinD

D

則I=I1I2I1=

I2

D

xydxdy+D0=0+00

D3+

D

=cosxsinD

=cosxsinydxdy+cosxsinD1+

D3+2cosxsincosx2cosxsin=2cosxsinydxdy+0所以II1I22cosxsin2009年考研題,選擇題,4 如圖,正?形x,y|x|￡1,|y|￡1}被對(duì)?線劃分為四個(gè)區(qū)域Dk(k=1,2,3,4), 則max{I

A

(A) (B)

(C) (D)

- OD4 與被積函數(shù)的奇偶性 -D2、D4兩區(qū)域關(guān)于x軸對(duì)稱?fx,yycosxfx,即被積函數(shù)是關(guān)于y的奇函數(shù),I2I42009年考研題,選擇題,4 如圖,正?形xy|x|￡1,|y|￡1}劃分為四個(gè)區(qū)域Dk(k=1,2,3,4), 則max{I}= A

OD4 OD4(A)I (B) (C) (D)I-1 -1D1、D3兩區(qū)域關(guān)于y軸對(duì)稱 ?f(-x,y)=ycos(-x)=f(x, -即被積函數(shù)是關(guān)于x的偶函數(shù),I1=

ycosxdxdy>{(x,I1=

y?x,0￡x￡ycosxdxdy