2014级建工第二学期高等数学课件

1由曲顶柱体体积的计算可知f(xy且在D上连续时,若DXD:1(x)y2

y2D axb(b

oay1(x)b

f(x,y)dxdy

dx

f(x,y)d 1(2(y)x(

x2(若D为Y–型区域D: cy

yx1(则

f(x,y)dxdy

dy

2(

f(x,y)xoxd 1(d3当被积函数f(xy)在D上变号时f(x,y)

2f1(x,

f(xf(x,y)f(x,f(x,y)f(x,f2x,y) Df(x,y)dxdyDf1(x,y)dxdDf2(x,y)dxd42说明(1若积分区域既是X–型区域又是Y–2 Df(x,y)dxd

y

2(

f(x,y)d x

2(

(

dy f(x,y) xDxDy

bD(2)若积分域较复杂,可将它分成若 D2型域或Y-型域, 5若是X—型，yx，6例1计算IDxyd,其中Dy＝1x＝2,y＝x所围的闭区域

1y解法1.将D看作X–型区域,则D:1x I 2dxx

2

2xd

yxyd

2xy 11121x31xdx

yx2 1x解法2.将D看作Y–型区域则D

1yI yxyd

2xydy

y2d

2 2

87例2

xyd,其中D是抛物线y2xyx2所围成的闭区域解为计算简便xy积分

y y D:y2xy1y

2x1 yx2x xyd

ydy xyd22

21x2y2

dy2

2[y(y2)2y5]d1y44y32y21y62 8例3.计 sinxdxdy,其中D是直线yx,y0, x 解由被积函数可知x积分不行因此取D为XD:y

y x 0x

sin

dxdy

sin

dx

sinxdxcosx 0说明有些累次积分为了积分方便还需交换积分顺序9例48x8x2 0I0dx f(x,y)dy20

dx0

f(x,y)d8x2解:8x2

0y x x

D2

0y

x2y22D22D0x

2x

y1x将DD1D2视为Y–型区域

12 2

x88If(x,y)dxdy

f(x,1y例5.计算IDx1yy4x2,y3x,x1所围成

)dxdy,其中D11y2解:令f(x,y)xln(y DD1

y4x显然,在D1上,f(xyf(x在D2上,f(x,yf(x

yo

1y2x1y21y21 IDxln(1y21

22

)dxdy例6.计算ò 1+x2-y2D

其中Dy=1,x=-D:1x1,D:1y1,

òy1+x2-y2dxdy=ò1dxò1y1+x2-y2dy

- Dy y1x2y2d1y

D例6

ò1+1+x2-

其中Dy=1,x=-1D:1x1,1

òy1+x2-y2dxdy=

dxò 1+1+x2-1

- 1 (|x|2 30(x1)dx2变换Txx(uv),yy(uv)是uov平面到xoyDf(x,y)dxdyD

f(x(u,v),y(u,

(x,(u,

在极坐标系下r==常数分划区域D

kk(k1,2,,

22k1(rkrk)2k1rk22221[rk(rkrk)]rk2rkrk

krkcosk,krknn0knlimf(rkcosk,rksink)rkn0k Df(x,y)dDf(rcos,rsin)rdrdr

2() 2D

,

(

Do0Do))

f1则可求得D

122() 2思考Dxy轴相切于原点,试问的变化范围是什么?

D

yyoDx答:(1)0

(2)

D(x,y)

a2b2 xacosybsind1Df(x,y) Df(cos,sin)abdd1

0f(cos,sin)abd被积函数中有平方和x2x2x2例7.计算D

其中Dx2y2a2解:在极坐标系下D:0 02

e2d

e2d

1e2

(1ea2a0 a0由于e

)2事实上当DR2时

ex2dx ex2y2dxd

ex2d

ey2d 0利用例7的结果0

dx00

dx

lim(1ea2 例8求球体x2y2z24a2被圆柱面x2y22ax(a0)所截得的(含在柱面内的)立体的体积解:

D:02acos,0 4a24a22 20

4a24a220

x 2x

32a3(

2 例9写出积分 D1式，其中积分区域D{(x,y)|1x10x解

xcosysin直线方程为f(x,D

xx 2d

计算x2y2)dxdyDDx2y22yx2y24y及直线xy 3x

y3y34解y3x0x2y24yx3y0x2y22

24 2 4

2 (

y)dxdy

32

2

2 直角坐标系情形:

yy2Dyy1(x) bxD(x,y)axb,y1(x)yy2

f(x,y)d

bdxy2(x)f(x,y)d xx2(xx2(Dx1(x D(x,y)cyd,x1(y)xx2(y) 则f(xyddd

x2(y)f(x,y)d x(极坐标系情形:D(,),1()2()D则Dfx,y)dD

1(

0((DoD:0(Do

0

(3)

1设f(xC0,1],0f(x)dx1 求I 1dx1f(x)f(y)dy 提示 交换积分顺序后,x,y互

1 yo Id

f(x)f(y)dx

d

f(x)f(y)1x 1x2I

d

f(x)f(y)dy

d

f(x)f(y) 1x1111 1x1111dxf(x)f(y)dyf(x)dx

f(y)d

A2 提示

I2

0

f(,)d

(aa 0I0

aa

f(,aarccosa 2ax备用题1.给定I2axyy

f(x,y)dy(a解：yy

2axx2axa2y2xa2y2原式

0dyy2a2a2a

f(x,y)d

2aaad

f(x,y)dx 2ad a2a2

a

a

f(x,y)d

练习题1fxy)dx