高等数学复旦大学出版社习题答案十

导读:11??22Lxds??Lxds?)2dx1?Lxds?2?0x1???(x)???dx??0x1?(x???1x1?4x20dx??102xdx11?18?2?32?3??(1?4x)2????2x2?0??2??0?112?55?62?1?.(4)如图10-67所示,L=L1+L2+L3图10-67其中L1:y=0(0≤x≤a),从而?x2?y2xLed

高等数学复旦大学出版社习题答案十

11??22Lxds??Lxds?)2dx1?Lxds?2?0x1???(x)???dx??0x1?(x???1x1?4x20dx??102xdx11

?18?2?32?3??(1?4x)2????2x2?0??2??0?112?55?62?1?.(4)如图10-67所示,L=L1+L2+L3

图10-67

其中L1:y=0(0≤x≤a),从而

?x2?y2xLeds?dx?ea?1

1?a0eL2:x=acost, y=asint,0≤t≤

π4

ππ故

?222Lex?yds?t)2dt?2?40ea(?asint)?(acos?4a0aedt?πa4ae.

L3:y=x(0≤x≤

22a).

x2?y222故

?Leds??2a?ea?1.30e2x?2dx?e2x2a

0所以

??x2?y2Leds??2?y2Lexds?1?Lex2?y2ds?2?2?y2Lexds3ππ?

?(ea?1)?4aea?ea?1?ea???2?4a???2.(5)ds?x222t??yt??zt?dt

?(etcost?etsint)2?(etsint?etcost)2?e2tdt

?3etdt. 246

?121?x2?y2?z2ds??0e2tcos2t?e2tsin2t?e2t3etdt

??232?t?3?t?302edt??2????2e??02(1?e).?x?0(6) AB:??y?0 (0?t?2),

??z?0?x?tBC:??y?0 (0?t?1),

??z?2?x?1CD:??y?t (0?t?3).

??z?2故

?2?xyzds??ABx2yzds??2BCxyzds??CDx2yzds??200dt??12200dt??302t1?0?0dt

??302tdt?9.(7)?y2π2222Lds??20a(1?cost)?a(1?cost)??(asint)dt

5??2π32t?8a3?2π5t02a(1?cost)d0sin2dta3?2πt2π2?80sin4t2sin2dt??16a3??0??1?cos2t?2??d??t? ?cos2??

??16a3?2π???1?2cos2t4t??02?cos2??d??cost?2??2π??16a3??cost2?23cos3t1t?2563?2?5cos52???015a.(8)ds??a(cost?tsint)??2???a(sint?tcost2)??2?dt

?(atcost)2?(atsint)2dt?atdt.

?22L(x?y)ds??2π0??a2(cost?tsint)2?a2(sint?tcost)2??atdt??2π0a3(1?t2)tdt?2π2a3(1?2π2).22(9)?z2?x2?y2ds??πat22220a2cos2t?a2sin2tasint?acost?a2dt

247

??π02attdt?223πat30?23aπ.

338. 计算曲面积分??f(x,y,z)ds,其中?为抛物面z = 2-(x2+y2)在xOy面上方的部分,

?f(x, y, z)分别如下:

(1) f (x, y, z)=1; (2) f(x, y, z)=x+y; (3) f(x, y, z)=3z.

2222

解:抛物面z=2-(x+y)与xOy面的交线是xOy面上的圆x+y=2,因而曲面?在xOy面上的22

投影区域Dxy: x2+y2≤2,且ds=1?z22x?zydxdy?1?4x2?4y2dxdy

故(1)

???f(x,y,z)ds???D1?4x2?4y2dxdy?xy?2π20d??01?4r2rdr

?2π?3213?1?12(1?4r2)2???π.?03(2)??f(x,y,z222?)ds???D(x?y)1?4x?4y2dxdy

xy??2π20d??0r21?4r2?rdr?π22?1)?1]1?4r2d(4r216?0[(4r?1)31?π2160r2?1)2?(1?4r2

?[(4)2]d(4r2?1)?π?253216??5(4r2?1)2?23(1?4r2)2???149π.?030(3)??22?f(x,y,z)ds????3zds???3??2?(x2?y2D)?

xy?1?4x?4ydxdy21?3?2π0d??0(2?r2)(1?4r2)2rdr?6π?12132?0??9?(1?4r2)??(1?4r2)2d(1?4r2)

352?3π?16??9?23(1?4r2)2?25(1?4r2)2???111π.?01039. 计算??f(x2?y2?)ds,其中?是:

(1)锥面z=x2?y2及平面z=1所围成的区域的整个边界曲面;

(2)锥面z2=3(x2+y2)被平面z=0和z=3所截得的部分。 解:(1)???1??2,其中:

?1:z?1,(Dxy:x2?y2?1).ds?dxdy,

248

?2:z?ds?1?x?y,(Dxy:x?y?1)x2222222x?y?y222

dxdy?2dxdy.x?y故

???(x?y)ds?122??Dxy(x?y)dxdy?22?2π0d??r?rdr?2π?012r41?0π2,

4???因此

(x?y)ds?222??Dxy(x?y)2dxdy?222?2π0d??rdr?01322π.

???(x?y)ds?22???2(x?y)ds?122???(x?y)ds?222π2?22π?1?22π.

(2)所截得锥面为z?3x?y(Dxy:x?y?3)

222ds??1??3??????322?x?y??x2?dxdy?2dxdy. 22?x?y?2?030y2故

???(x??2?y)ds?2??2Dxy(x?y)dxdy?2?22d??rdr?9π.

340. 计算下列对面积的曲面积分: (1)???z?2x??(2)??(3)??(4)??(5)??xyz?y?ds,其中?为平面???1在第I卦限中的部分;

2343?42?2xy?2x?x?z?ds,其中??为平面2x+2y+z=6在第I卦限中的部分;

2

2

?x??y?z?ds,其中?为球面x+y+z=a上z≥h(0

22

?xy??其中?为锥面z?yz?zx?ds,

x?y被柱面x+y=2ax所截得的有限部分;

R?x?y. 2222222

222?R?x?y?ds,其中??为上半球面z?解:(1)?:z?4?2x?43y(如图10-69所示)

249

图10-69

ds??4?1?(?2)????dxdy??3?22613dxdy

?????z?2x?4??3y??ds???D4?61xy3dxdy?4613??Ddxdyxy

?46113?2?2?3?461.(2)?:z=6-2x-2y(如图10-70所示)。

图10-70

ds?1?(?2)2?(?2)2dxdy?3dxdy

????2xy?2x2?x?z?ds???D3?2xy?2x2?3x?2y?6?dxdyxy?3?33?x0dx?0?2xy?2x2?3x?2y?6?dy3

?3?0???6?3x?2x2?(3?x)?(x?1)(3?x)2??dx?3?3(3x3?10x2?9)dx270??4.(3)?:z?a2?x2?y2且其在xOy面上的投影为D22

2

h2

xy:x+y≤a-且

2ds?1???2x???2y???da?2a2?x2?y2????2a2?x2?y2?xdy?dxdy.?a2?x2?y2故

???(x?y?z)ds???D?x?y?a2?x2?y2xy?aa2dxdy

?x2?y2对称性??22Dadxdy?πa(a?h).

xy(4)z?x2?y2,Dxy:x2?y2?2ax

250

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