第12講二重積分的計(jì)算

第十二講二重積分的計(jì)算二重積分的計(jì)化為二次積處理絕對(duì)利用對(duì)稱性和輪換不變 練習(xí)十三/ dx

計(jì)算二次積111y

yyx211 x1y1y 34 112

1y|341y| PreviousNextx2y2x2y2Dyyo其中D0yyyo

dxdyy22x.x2x2y2D

4 d4

d 2 48cosd23

x2y2PreviousNext練習(xí)十三/y求exydxdy其中D由直線x0,y0xy圍成D解：xy1原

1xy1xyo1 2dcos

ecos

PreviousNext 2ecos2

(cossin

2令u 1cossin1則du

(cossin)2原式2

1eudu0

1(e2PreviousNext例設(shè)f(x)在點(diǎn)x0處連續(xù),且f(0) f(0)f(01,

f(x2y2)d1x2y2R21解：原式 R0

R R6 dR

f

2)R2f(2 Rlim

(型)PreviousNextR0

2f(R2 6R

R0

f(R2)3R4

0型0

f(R2)2R

f(R2 3R0

4R

6R0 R2

f(R2) 6

PreviousNext22

42

x2y2dy2

22x4x2x2y2x2y2o22

x2y2 2 2x解:原式

2

3 (1cos328 PreviousNext計(jì)算6

6y6yo6 解法一:交換積分次 原式

cos

dx

dy

16cos 16x 6x解法二:利用分部積分 cos原式6[6 cos

cos yy

dx|0x0

6y0

dy0y

6cosydy PreviousNext練習(xí)十三/f(x)連續(xù)D:|x|1,|y|1 1D

f(xy)d

f(xy)d

2dx

f(x (令ux 2

x12f(u)(du)1

121

x12f1

PreviousNext u f(u)du12dxf(u)du21 u u u1(u1)f(u)du

0(1u)f(u)1111

1ux22o1ux22oux212PreviousNext|y|yx2D

d,D:|x|1,0y解D1

|yx2|dyyx2

yx2dx2x2

x2

dy

y2yD12yD1o1 xPreviousNext計(jì)算|x2y24|d其中Dx2y2D解x2y24|D (4x2y2

y2)d

(x24x2y29

y2 2d2(42)d

4)324132PreviousNext例：計(jì)算(xy)2d其中D:|x||y|D解：原式x2dy2d 由對(duì)稱性,得2xyd 由輪換不變性得x2dy2 式2x2d8x2 D/

81x2dx1xdy

PreviousNext練習(xí)十三/設(shè)f(x)為連續(xù)函數(shù)求[xyf(x2y2xD其中D由曲線yx3x1y圍成yyyoD2解：添上輔助線yx3分積分區(qū)域D為D1與D2. PreviousNext得xyf(x2y2dxdy0,xyf(x2y2dxdy xyfx2y2dxdyD又xdxdy ydxdy xdxdy y y

xdx

dy

26

PreviousNext練習(xí)十三/設(shè)D由xy1x0,y圍成計(jì)算二重積分D

x3y3d解：由輪換不變性3D

x3y3dxdyD

y3x33 x3y3dxdyD

PreviousNext11

f(x)dxyy1o1

f(x)f(y)dy. 解原式(輪換)

0dyyf(y)fx 2[0 f(x)f(y)dy0dy

f(y)f1f(x)f( (D:0x,y2 2 f(x)dx

f(y)dy2

PreviousNext2exy(2xy1)d1 其中Dxy|0y1y1x1證fxyexy(2xy

exy(2xyy2y則yexy(2x

xyx 函數(shù)f(x,y)在D內(nèi)無臨界點(diǎn) z12x (1x

PreviousNextminz1z1(1)3,maxz1z1(1)邊界y1

x

(1x2z3ex2

(2x

x1)minz2z2(1) maxz2z2(0)y1

x

(0x3zexx2(2x2x1)3

z3在(0,1)內(nèi)無駐點(diǎn)minz3z3(0)0,maxz3z3(1)PreviousNext在D上3f(x,y3f(x,yd 精度不DD1{(x,y)|0y1,y1xD2{(x,y)|0y1,0x1 線段x z4y (0y xminz4z4(0) xmax

PreviousNextf(1,0)

f(0,0) 在D上,3f(x,y0,3 D2D1

f(x,y)d0在D2上1f(x,y

1f(x,y)d1 2f(x,y)df(x,y)df(x,y)d PreviousNext練習(xí)十三/二lim

t

dx t0t2

1

y4分析：原式lim 2 t

D1 (利用積分中值定理txyottxyott0t

1

4(0, t 1104

PreviousNext練習(xí)十三/二若連續(xù)函fx,y)滿3f(x,y)5(x2y2)2 u2v2

f(u,v)dudv,則x2y2

f(x,y)dxdy 分析：令k

x2y2

f(x,3則fx,y5x2y22

PreviousNext在區(qū)域x2y2上積分有k f(x,y)dxdy

3[5(x2y2)

kx2y2 x2y2 5d

k1解答1PreviousNext練習(xí)十三/b設(shè)f(x)在[ab]上連續(xù)且恒正利用二重積分證bb f(x)cosb2b

[ f(x)sinb2bb

f證：左式 f(x)cos f(y)cos f(x)sin f(y)sin a f(x)f(y)(coskxcoskysinkxsinPreviousNextf(x)f(y)cos(kxky)D(D:ax ayf(x)f(y)D af(x)dxaf(y)bb

PreviousNext練習(xí)十三/設(shè)f(x)是[0,1]上單調(diào)減少的連續(xù)函數(shù)且恒正1xf2

f211利用二重積分證明 0 .111110xf f11110

2(x)dx11011

f(x)dx

f2(x)dx

0xf(x)101

2(x)dx0

f(y)dy

f2(x)dx101

yf(y)PreviousNextxf2(x)f(y)dyf2(x)f(y) (D:0x 0