最新工作文档高等数学下复旦大学出版习题十答案详解优秀名师资料.docx
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最新工作文档高等数学下复旦大学出版习题十答案详解优秀名师资料
工作文档高等数学下_复旦大学出版_习题十答案详解
习题十
21.根据二重积分性质,比较ln()dxy,,[ln()]dxy,,与的大小,其中:
,,,DD
(1)D表示以(0,1),(1,0),(1,1)为顶点的三角形;
(2)D表示矩形区域.{(,)|35,02}xyxy,,,,
解:
(1)区域D如图10-1所示,由于区域D夹在直线x+y=1与x+y=2之间,显然有
图10-1
12,,,xy
从而0ln()1,,,xy
2故有ln()[ln()]xyxy,,,
2ln()d[ln()]dxyxy,,,,,所以,,,,DD
(2)区域D如图10-2所示.显然,当时,有.(,)xyD,xy,,3
图10-2从而ln(x+y)>1
2故有ln()[ln()]xyxy,,,
2ln()d[ln()]dxyxy,,,,,所以,,,,DD
2.根据二重积分性质,估计下列积分的值:
IxyDxyxy,,,,,,,4d,{(,)|02,02},
(1);,,D
22IxyDxyxy,,,,,,sinsind,{(,)|0,π,0π}
(2);,,D
2222(3)IxyDxyxy,,,,,,(49)d,{(,)|4},.,,D
02,,x解:
(1)因为当时,有,(,)xyD,02,,y
因而.04,,xy
从而2422,,,xy
2d4d22d,,,,,,xy故,,,,,,DDD
2d4d22d,,,,,,xy即,,,,,,DDD
d,,,而(σ为区域D的面积),由σ=4,,D
84d82,,,xy,得.,,D
22
(2)因为,从而0sin1,0sin1,,,,xy
220sinsin1,,xy
220dsinsind1d,,,,,xy故,,,,,,DDD
220sinsindd,,,xy,,,即,,,,DD
2,,π而
2220sinsind,,xy,π所以,,D
22(3)因为当时,所以(,)xyD,04,,,xy
22229494()925,,,,,,,xyxy
229d(49)d25d,,,,,,,xy故,,,,,,DDD
229(49)d25,,,,,,,xy即,,D
2,,,,π24π而
2236π,,,,(49)d100xy,π所以,,D
3.根据二重积分的几何意义,确定下列积分的值:
22222()d,{(,)|};axyDxyxya,,,,,,
(1),,D
222222
(2)axyDxyxya,,,,,d,{(,)|}.,,,D
22解:
(1)在几何上表示以D为底,以z轴为轴,以(0,0,a)为顶()d,axy,,,,,D
1223点的圆锥的体积,所以axya,,,,()dπ,,D3
222
(2)在几何上表示以原点(0,0,0)为圆心,以a为半径的上半球axy,,d,,,D
22223的体积,故axya,,,,dπ.,,D3
12224.设f(x,y)为连续函数,求.fxyDxyxxyyr,,,,,,lim(,)d,{(,)|()()}002,,Dr,0rπ
解:
因为f(x,y)为连续函数,由二重积分的中值定理得,使得,,(,),,,D
2fxyfrf(,)d(,),,,,,,,,,,π(,),,D
r,0又由于D是以(x,y)为圆心,r为半径的圆盘,所以当时,(,)(,),,,,xy0000
112lim(,)dlimfxyrff,,,,,,,,π(,)lim(,)22,,Drrr,,,000rrππ于是:
,ffxylim(,)(,),,00,,,xy(,)(,)00
fxy(,)d,5.画出积分区域,把化为累次积分:
,D
(1);Dxyxyyxy,,,,,,{(,)|1,1,0}
2
(2)Dxyyxxy,,,,{(,)|2,}
2(3)Dxyyyxx,,,,{(,)|,2,2}x
解:
(1)区域D如图10-3所示,D亦可表示为.yxyy,,,,,,11,01
11,y所以fxyyfxyx(,)dd(,)d,,,,,,Dy01,
2
(2)区域D如图10-4所示,直线y=x-2与抛物线x=y的交点为(1,-1),(4,2),区域D
2可表示为.yxyy,,,,,,2,12
图10-3图10-4
22y,所以fxyyfxyx(,)dd(,)d,,2,,,,Dy,1
2(3)区域D如图10-5所示,直线y=2x与曲线的交点(1,2),与x=2的交点为(2,y,x
224),曲线y,与x=2的交点为(2,1),区域D可表示为,,,,yxx2,12.xx
图10-5
22xfxyxfxyy(,)dd(,)d,,所以.2,,,,D1x
6.画出积分区域,改变累次积分的积分次序:
elnx22yd(,)dxfxyy
(1);
(2);d(,)dyfxyx2,,,,100y
πsinx132,y(3);(4)d(,)dxfxyy;d(,)dyfxyxx,,,,,0sin0y2
1233yy,d(,)dd(,)dyfxyyyfxyx,(5).,,,,0010
2解:
(1)相应二重保健的积分区域为D:
如图10-6所示.02,2.,,,,yyxy
图10-6
xD亦可表示为:
04,.,,,,xyx2
224yx所以d(,)dd(,)d.yfxyxxfxyy,x2,,,,00y2
(2)相应二重积分的积分区域D:
如图10-7所示.1e,0ln.,,,,xyx
图10-7
yD亦可表示为:
01,ee,,,,,yx
eln1exd(,)dd(,)dxfxyyyfxyx,所以y,,,,100e
(3)相应二重积分的积分区域D为:
如图10-8所示.01,32,,,,,,yyxy
图10-8D亦可看成D与D的和,其中12
2D:
01,0,,,,,xyx1
113,0(3).,,,,,xyxD:
22
12,,yxx13213(3)2d(,)dd(,)dd(,)dyfxyxxfxyyxfxyy,,所以.,,,,,,y00010
x(4)相应二重积分的积分区域D为:
如图10-9所示.0,,,,,xyxπ,sinsin.2
图10-9D亦可看成由D与D两部分之和,其中12
D:
,,,,,10,2arcsinyyxπ;1
D:
01,arcsin,,,,,yyxyπarcsin.2
πsin0xyπ1π,arcsin所以d(,)dd(,)dd(,)dxfxyyyfxyxyfxyx,,x,,,,,,0sin12arcsin0arcsin,,,yy2
(5)相应二重积分的积分区域D由D与D两部分组成,其中12
D:
D:
01,02,,,,,yxy13,03.,,,,,yxy12
如图10-10所示.
图10-10
x02,3;,,,,,xyxD亦可表示为:
2
123323yyx,,所以d,dd(,)dd(,)dyfxyxyfxyxxfxyy,,,,x,,,,,,001002
7.求下列立体体积:
2222
(1)旋转抛物面z=x+y,平面z=0与柱面x+y=ax所围;
222
(2)旋转抛物面z=x+y,柱面y=x及平面y=1和z=0所围.解:
(1)由二重积分的几何意义知,所围立体的体积
2222()ddxyxy,V=其中D:
{(,)|}xyxyax,,,,D
22由被积函数及积分区域的对称性知,V=2,()ddxyxy,,,D1
其中D为D在第一象限的部分.利用极坐标计算上述二重积分得1
acos,πππacos,11334444222Vrrraa,,,,,,,,.2dd2dcosdπ,,,,000042320
(2)由二重积分的几何意义知,所围立体的体积
22Vxyxy,,()dd,,,D
2其中积分区域D为xOy面上由曲线y=x及直线y=1所围成的区域,如图10-11所示.
图10-11
2D可表示为:
,,,,11,1.xxy
112222所以Vxyxyxxyy,,,,()ddd()d2,,,,Dx,1
11111188,,23246,,,,,,,xyyxxxxxd()d.,,,,,,112333105,,x
8.计算下列二重积分:
2x1
(1)dd,:
12,;xyDxyx,,,,,,2Dyx
xy2edd,xy
(2)D由抛物线y=x,直线x=0与y=1所围;,,D
22xyxy,dd,(3)D是以O(0,0),A(1,-1),B(1,1)为顶点的三角形;,,D
cos()dd,{(,)|0xyxyDxyxxy,,,,,,π,π}(4).,,D
x222222xxxx3dddddd解:
(1)xyxyxxxx,,,,,,,1,,,,,,22111Dyyy1xx
2119,,42,,,xx.,,424,,1
(2)积分区域D如图10-12所示.
12图10-
2D可表示为:
01,0.,,,,yxy
xxx2211yyxyyy所示edddedded()xyyxyy,,,,,,,,0000Dy
2yx1111yyy,,,,,yyyyyyyyed(e1)dedd,,,,00000
1111111yyy2,,,,,,yyyyyydedeed.,,,0000220
(3)积分区域D如图10-13所示.
图10-13
D可表示为:
01,.,,,,,xxyx
x211x,,xyy222222ddddarcsindxyxyxxyyxyx,,,,,,所以,,,,,,,,00Dx22x,,,x
11ππ1π23,,,,xxxd.,022360
ππππ(4)cos()dddcos()d[sin()]dxyxyxxyyxyx,,,,,x,,,,,Dx00
ππ,,,,,,[sin(πxxxxxx)sin2]d(sinsin2)d,,00
π11,,,,.coscos2xx,,,2,,20
9.计算下列二次积分:
1ysinx
(1)dd;yx,,0yxyy1yy1xx2
(2)dedded.yxyx,111,,,,y224
sinx解:
(1)因为求不出来,故应改变积分次序。
dx,x
积分区域D:
0?
y?
1,y?
x?
,如图10-14所示。
y
图10-14
2D也可表示为:
0?
x?
1,x?
y?
x.所以
111yxsinsinsinxxx2dddd()dyxxyxxx,,,2,,,,,000yxxxx
111,,,,(sinsin)dsindsindxxxxxxxxx,,,000
111,,,,,,,sindcosd1sin1.xxxxxxcos0,,00
y
x
(2)因为求不出来,故应改变积分次序。
积分区域D分为两部分,其中edx,
1111DyxyDyyxy:
,:
1,.,,,,,,,,124222
如图10-15所示:
图10-15积分区域D亦可表示为:
12,,,,xxyx1,.2
于是:
xyyy1y11yyx1xxx2xdeddeddeddyxyxxyx,,,xe11111,,,,,,2,yx222224x113eee1x2xx,,,,,,(ee)dxxx,,x1xee,1,1822222
10.在极坐标系下计算二重积分:
222222
(1)sindd,;xyxyD,,,,(,)|xyxyπ,,,4π,,D
22,,()xy22
(2)D为圆=1所围成的区域;edd,xyxy,,,D
x2222(3)D是由=4,=1,及直线y=0,y=x所围成的在第一象限内arctandd,xyxy,xy,,,Dy
的闭区域;
22()dd,xyxy,(4)D是由曲线=x+y所包围的闭区域。
xy,,,D
解:
(1)积分区域D如图10-16所示:
图10-16D亦可采用极坐标表示为:
π?
r?
2π,0?
θ?
2π所以
2π2π22sindddsindxyxyrrr,,,,,,,D0π
2π2,,,,,,2π6π.rrrcossin,π
(2)积分区域D可用极坐标表示为:
0?
r?
1,0?
θ?
2π.所以:
2π1122221,,,,,,()2xyrrxyrrr,,eddded2ed(),,,,,,,,,,,,D000,,211,,2,r,,,,π.1,,,,e0e,,
(3)积分区域D如图10-17所示.
图10-17D可用极坐标表示为:
π0?
θ?
1?
r?
2.4
所以:
π2x4arctanddarctan(cot)ddxyrr,,,,,,,D01yπ239ππ,,4,,d.,,,,,,0264,,2(4)积分区域D如图10-18所示,
图10-18D可用极坐标表示为:
π3π,,,,,,,,,,0cossinr44
所以:
3πcossin,,,24()ddd(cossin)dxyxyrr,,,,,,π,,,,0D,4
cossin,3π,,3r4,d(cossin),,π,,,,3043π144,,(cossin)dπ,,,,,34
3π4ππ,,44,,sind.,,,π,,,,32,,44
11.将下列积分化为极坐标形式,并计算积分值:
222aaxxax,2222
(1)d()d;
(2)dd;xxyyxxyy,,,,,,0000
122,xaay,122222(3)d()d;(4)dd.xxyyyx,,,xy,2,,,,x000
解:
(1)积分区域D如图10-19所示.
图10-19D亦可用极坐标表示为:
π,,,,,,0,02cosra2
所以:
2cosa,ππ24,2a22cosaxxa,r22322d()ddddxxyyrr,,,,,,,,,,0000040π31π344442,,,,,,4cosd4aaaπ.,,,04224
(2)积分区域D如图10-20所示.
图10-20D可用极坐标表示为:
π,,,,,,0,0secra4
于是:
asec,πππ33asecxa,ar2223444dddddsecd,,,,,,,,xxyyrr,,,,,,00000033033πaa4,,,,,,sectanln(sectan).,,2ln(21),,,,,,,,066
(3)积分区域D如图10-21所示.
图10-21
D也可用极坐标表示为:
π.,,,,,,,0,0sectanr4
于是:
1ππ,x,,1sectan2,22144d()dddsectandxxyyrrr,,,,,,,,2,,,,,x0000π4,,,sec21,0
(4)积分区域D如图10-22所示.
图10-22D可用极坐标表示为:
π,,,,,0,0ra2
于是:
aπ224aaya,ππr22342d()ddd.yxyxrra,,,,,,,,,,00002840*12.作适当坐标变换,计算下列二重积分:
22xyxydd
(1),其中D是由xy=2,xy=4,x=y,y=3x在第一象限所围平面区域;,,D
222dd,{1};xyD,,,
(2)(,)xyy,,xy,x,,D
12,x22d()d,xxyy,(3)令x=v,x+y=u;,,01,x
2222xy,,xy(4)dd,:
1;xyD,,,,,,,2222Dabab,,
2222dd,;xyD,xy,,9(5)(,)xyxy,,4,,,,D
2222dd,.xyD,xy,,4(6)(,)xyxyy,,2,,,,D
解:
(1)积分区域D如图10-23所示:
图10-23
y令xy=u,,则,vx
uxyuvuv,,,,,,,,(24,13)v
111vvu,,,xx,,2,(,)1xy22uvuv,,uvJ,,,,.,,yy,(,)2uvvvu
,uv22uvuv于是:
4333411281u2222xyxyuuvvuuddddddln3.,,,,,,lnv,,,,,,D12223vv231224,,u13,,v
24所示。
(2)积分区域D如图10-
图10-24令x+y=u,x-y=v,则
uvuv,,xy,,,22
且-1?
u?
1,-1?
v?
1.
11
(,)1xy22J,,,,11,(,)2uv,22于是:
4224211uuvv,,21224224()ddddd
(2)dxyxyuvuuuvvv,,,,,,,,,,,D,,1188,,,11u,,,11v
111112121,,,,423542,,dduuuvuvvuu,,,,,,,,,,,,11843535,,,,,1
1114121,,53,,.uuu,,,,445595,,,1
(3)积分区域D:
0?
x?
1,1-x?
y?
2-xxy
令x=v,x+y=u,则y=u-v
积分区域D变为D:
xyuv
0?
v?
1,1?
u?
2.
01,(,)xy且J,,,,111,,(,)uv
于是
212121,x1,,2222223d()dd(22)ddxxyyvvuvuuv,,,,,vuvuu2,,,,,,,,,01010,x3,,11137237,,,,232v,,,d.vvvvv23,,,,,,,,,023323,,,,0(4)令x=arcosθ,y=brsinθ则积分区域D变为D:
0?
θ?
2π,0?
r?
1,rθ
aarcossin,,,,(,)xyJabr,,,bbrsincos,(,)r,,,
于是:
1222π111,,xy,,234,,,xyrabrrabrrab,,,,,dddddd2πabr,,,,,,,,,,,22DD00r,2,,4ab,,0(5)令x=rcosθ,y=rsinθ.即作极坐标变换,则D变为:
0?
r?
3,0?
θ?
2π.于是:
2π32222ddddddxyrrrr,,,,xy,,4rr,,44,,,,,,DD00
2333,,,,2π(4)d(4)drrrrrr,,,,,02,,
23,,4111,,,,2442,,2ππ.,22rrrr,,,,,,,,2,,,,4402,,
(6)积分区域D如图10-25所示:
D可分为D,D?
D,D四个部分.它们可分为用极坐标1234表示为。
图10-25D:
0?
θ?
π,0?
r?
2sinθ,1
D?
D:
0?
θ?
π,2sinθ?
r?
2,23
D:
π?
θ?
2π,0?
r?
24
于是:
22222222ddddddddxyxyxyxy,,,xyyxyyxyyxyy,,,,,,,,2222,,,,,,,,DDDDD,1234
π2sinπ22π2222,,,,,,,,,,,,d(2sin)dd(2sin)dd(2sin)drrrrrrrrrrrr,,,,,,0002sinπ0,
2π2π2sinπ2,322332rrrr,,d(2sin)d,,,,,,d(2sin)dd(2sin)drrrrr,,,,,,,,,,π00002sin,,2sin22444ππ2π,,,,,,222rrr333,,,dddrrrsinsinsin,,,,,,,,,,,,,,,,,,00π344343,,,,,,02sin0,
ππ2π416416,,,,44,,,sinddd4sinsin4sin,,,,,,,,,,,,,,,,,00π3333,,,,
π2ππ81616,,,,4,,dd,,,sind4sin,,4sin,,,,,,,,,,,0π033,,,,3
π2π811631,,,,,,8πsind,,sin2sin4,,,,,,,,034328,,0
23,,,,,π8π09π.32
13.求由下列曲线所围成的闭区域的面积:
2bb2yxyx,,,
(1)曲线所围(a>0,b>0);aa
22
(2)曲线xy=a,xy=2a,y=x,y=2x所围(x>0,y>0).
2bb2yxyxab,,,,,(0,0)解:
(1)曲线所围的图形D如图10-26所示:
aa
图10-26D可以表示为:
aa,2yxy,,,2bb,
0,,yb,
所求面积为:
abyb1aa,,2bSxyyxyab,,,,ddddd.yy,,,a,,,,,200Dy62bb,,b
22
(2)曲线xy=a,xy=2a,y=x,y=2x(x>0,y>0)所围图形D如图10-27所示:
图10-27所求面积为
Sxy,dd,,D
y,v令xy=u,,则x
u22xyuvauav,,,,,,,,(2,12)v
(,)1xyJ,,,(,)2uvv于是
222a22211aa,,,,,Sxyuvvuvdddddddln22,,,,,,,Da112222vvv22,,aua2,,v12
14.证明:
byb1,1nnyyxfxxfxbxx,,,d()()d()()d;
(1),,,aaan,1
1fxyxyfuu()dd()d,,
(2),D为|x|+|y|?
1;,,,,1D
122222faxbycxyufu()dd21d,,,,(3),其中D为x+y?
1且,,uabc,,,,,,1D
22a+b?
0.
解:
(1)题中所给累次积分的积分区域D为
a?
y?
b,a?
x?
y.如图10-28所示:
图10-28D也可表示为a?
x?
b,x?
y?
b,于是:
bbybbb1nn,1nyyxfxxxyxfxyx,,,,d()()dd()()ddfxyx,()(),,,,,aaaxan,1x
b1,1n,,fxbxx()()d.,an,1
(2)令x+y=u,x-y=v,则
uvuv,,xy,,,,且-1?
u?
1,-1?
v?
122
(,)1xy,于是,,,(,)2uv
11111fxyxyfuuvufuvfuu,,,,()dd()ddd()d()d.,,,,,,,,,,,11122Du,,,11v,,,11
aubvbuav,,(3)令,则xy,,,2222abab,,
22faxbycfuabc()(),,,,,
ab,
222222,(,)xyababab,,J,,,,,12222ba,,,(,)uvabab
2222abab,,
22当x+y?
1时,
22222222aubvbuav,,,,,,()()abuabv,,,22,,,,,uv1.,,,,222222ab,abab,,,,,,
于是
22faxbycxyfuv()dddd,,,,,uabc,,,,,,22Duv,,1
2,u1122,ddufv,,uabc,,2,,,,,u112,u1122,fvud,,uabc,,2,,1,,u1
1222,,21d.ufu,,uabc,,,,122222215.求球面x+y+z=y含在圆柱面x+y=ax内部的那部分面积。
解:
如图10-29所示:
图10-29
222zaxy,,,上半球面的方程为,由
,,,zxzy,,,222222,,xyaxyaxy,,,,
得
22,za,,,z,,1,,,,,,,222,y,x,,,,axy,,
由对称性知
22,za,,,z,,Axyxy,,,,41dd4dd,,,,,,,,222DD,y,x,