微积分a2第5次习题课答案

r2x微积分A(2)第五次习题课参r2x1.（1）若

D

(xy)2d,

D

(xy)2d,D:0y ,则( )

I1I2.(B)I1I2

I1I2.(D)I2之大小相等关系不定而与r有关D

0（2）设0R1，则二重积分I x2y2

x2yedBe1（A） x2y2x0,

x2ye e1

x2y2

x2yed（C）e1

x2y2x0,

x2yed。e1 （3）f(xC[0,1],且0f(x)dxA,求0dxxf(xfy)dy x

1(3积分顺序(1) f(x,y)dy f(x,y)dy

(2)I1 3

f(rcos,rsin I2 22

f(rcos,rsin x

1(3

32解

f(x,y)dy

f(x,y)dy

f(x,y y arccos（2）I1

rdr3f(rcos,rsin

I2

2

f(rcos,rsin3计算积分00

[xy]d[xy]d0dd2d 132331 x2 ，zx,xx2解：V

x)dxdy

2

1cos[1r(1cos)]rdrx2D12 x2D12

(xa

y2)x2b

y2xarcosybrsin,J

D'{(r,):02,0r a2cos2b2sin2 Ddxdy Da2a2cos2b2sin2ab0d 1ab(a2b222求积分|xyx2y2|2D

D{(x,y)|x2y2Dxydxdy,D为圆域x2y2a2D xydxdy42D

rsinrcos

r3dr2|2 r3dr2|24sin2d4sin2d|a

4 (xy)dxdyD(xyx1)2y1)22yxD (xy)dxdy4 2(rcosrsin)rdr 解法二：由(x1)2y1)22得r2(sincos (xy)dxdy 4 2(sincos)(rcosrsin 3 4

(cossin)r32(sincos 001,|x||y|f(x,D

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

,D{(x,y)||x||y|2}2D化f(xy)dxdyDxy0x1,0xy1}为极坐标下的累次积分，并DDxy0x1,0xy1},用极坐标变换后, f(x,y)dxdyd rf(rcos,rsin)dr2dcossinrf(rcos,rsin 0

arccos1212 2rdr2f(rcos,rsin)d4

rdr

2rf(rcos,rsinr2 r2

f(rcos,rsin)d

rf(rcos,rsin 2 f(x,y)

2D0f(xy)(D

2f)dxdy0设函数f(t)连续，f(t)0，求积 x2y2

af(x)bf(y)dxdyf(x)f(f(xy为连续函数,f(xy)fyx).证明 0dx0f(x,y)dy0dx0f(1x,1x1uy1v,0v1,0uv,J 0dx0f(1x,1y)dy0dv0f(u, 0dv0f(v,u)du0dx0f(x,yyxf(x, D{(x,y)xD

a,x0,y0}解：由3xucos4vysin4vD'{(u,v|0ua,0v2J4usin3vcos3vDf(x,D f(usin4v,ucos3v)4usin3vcos3 42

usin3vcos3vf(usin3v,ucos3 a040du2usin3vcos3vf(usin3v,ucos30D(xy)sin(xy)dxdy,D{(x,y)|0xy,0xyDyexydxdy,D{(x,y)|xy1,x0,yD(1)x1(uvy1(uvD'{(u,v|0u,0vD J(u,v)12(xysin(xy)dxdyusinv1dudv1udusinvdv1 2 (2)xuvyuD'{(u,v|0uv,0v1},J(u,v)1 1于是exydxdyevdudv1dvvevdu (e D试作适当变换,把f(xy)dxdy,Dx,y|xDx1(uvy1(uv

21u1,1v1;f(xy)dxdy1

12f(u)du1dv

fxy 2 f(x在[abDab[cdF(x,y)f(x(x,yD。证明：F(x,y)在D上可积。证：将[ab]、[cdax0x1x2xn1xn和cy0y1y2ym1ymd则D分成了nm个小矩形Dij(i1,2,n,j1,2,m)记if(x在小区间[xi1xiij(FF在Dijij(F)i ij(F)ijixiyj(dc)ixii,j i,jn

f(x)在[ab上可积，可知i

00)limn(F

lim(d

n

0i,j

0

ixi F(xy在Df(xy在有界闭区域D上的非负连续函数,且在D上不恒为零,则f(xy)dD证：由题设存在P0(x0,y0)D使得f(P0)0.令f(P0),则由连续函数的局部保号性知:0使得f(P) ,PD(DU(P,)D).又因f(x,y)0且连续，所 f(x,y)d f(x,y)d f(x,

D Df(x,y)d

2证：由条件，该平面曲线Ll

/2(t)/2 为有限值。对0L分成nl

1段：L，L，LLP[ 使P与其一端点的弧长为 ，以P为中心作边长为的正方形，则 Lii(i1,2,nLi，记i，则为一多边形，设

Wn2

]

1)2(l)L的面积WLWl)由的任意性有WL0Lf(x在[abbb[f(x)dx]2(ba)f2bb f(x在[ab上连续,f2x在[ab[ab上可积,2 b bbf2(x)f2(2[af