教程分析高数

D：a£x£

D

D：c£y£ y1(y)£x£

D

x确定D的边界曲线,画出D将D表示成X-型区域(或Y-型区域):

f(x,y)dxdyD

f2(x

f(x,(D

f(x,y)dxdy=

y2(ydy(y f(xdy(y12二.注意被积函数的可积情况及特性，恰三.牢记二重积分化为二次积分 34 x=rcosy=rsin 0£r<+¥,0£f£

M(x,x

=x2+y

M(x, tanf= 0£r<+¥,0£f£5§10.2r(x,(r,qr(x,(r,qq•

y=rf(x,y)ds=

f(x, f(rcosq,rsinq)dsDfx,y)ds在极坐标系下D6用r常数，q常数划分区域Ds=1(r+Dr

2

q =1(2r+Dr =ri+(ri+Dri)

r=ri+r=

=i+DDDsi =ri

A7Dsi=

r=ri+r=

f(x,y)ds=f(rcosq,r f(x,y)dxdy=f(rcosq,r 8

D

r= Dbaff(rcosq,rsinqD=1

rD

=

f(rcosq,r 9a£q£b,0£r£f(rcosq,rsinqD

r=f(qDb f(q

r=fr=f(qDoA0£q£2p,0£r£f(rcosq,rsinqD f(q

转化积分变量;fx,y)dsf(rcosqr 确定θ确定r从极点出发的射线穿过区域D,穿入点对应r的下限，穿出点对应r的上限；。1写出积分fx,y)dxdyDD={(x,y)|1-x解xry=r圆方程为r1直线方程为r

,0£x£==p2011f(rcosq,rsinqD 计算e-x-ydxdy，其中D D原点，半径为a的圆周所围成的闭区域 D：0£r£a，0£q£e-x2-y2

=2pdqae-r2

=p(1-e-a2 ex2dx 求广义积

+¥e-x2S.0S. 令S={(x,y)|0£x£R,0£y£ e-( )dxdyS

Re-x2

e-y2RfiR20e=R20e2. e-xy)0,S夹在以原点为中心,为R 故故e-(x2y2dxdy£e-(22D2S£e-(x2+y2SR1其中Dx,y|x2y2£R2x0,y12D={(x,y)|x2+y2£2R2,x‡0,y‡2 1I=e-x-y 1

p2p

e-

rdr=(1-e-R

=e-x-ydxdy=(1-e-2RD4D p4p4p4p44

)£(Re-x2dx)2£

(1-e-2R2

令Rfi+¥

e-

= +¥e-x2dx=p 计算(x2+y2)dxdy，其D为由D3x2y22yx2y24y及直线x3

y-3x y-3x=0q= x2+y2=4yr=x-3y=0q= x2+y2=2yr=p(x2+y2)dxdy=3p

r2

2 2 5计算二重积分D

dxdy,sin(px2+y2x2+其中积分区域为Dxysin(px2+y2x2+，，D=x2x2+y2x2+

x2+y2D

dxdy= x2+

=42

rdr=- 求球体x2+y2+z2£4a2被圆柱面x2+y2=2ax(a>zyxzyxV= 4a2-x2-y2D其中Dy=2axx2及x轴20£20£r£2acosq,0£q£ 0£r£2acosq,0£q£p2V= 4a2-x2-y2dxdy=

4a2-r2rd