?t2?
F(t)?3???1??2(x?2?y?2)?2x?y??dx?dy?
3????
再作旋转变换：x??
x???y??2x??2;y??
x???y??2?
,则(1)变为标准方程?1
y??22?
?1?t?3?
?1t2?1??33??
???
2????
2t2?(3x??2?y??2),记相应的区域为???，而函数为f?1?3?t2?
于是F(t)?3???1??3x??2?y??2?dx??dy??
3?????
??最后，作广义极坐标变换，即x???
1
t2t21?rcos?,y???1?rsin?333?3?r?rdr??
?t2则有：F(t)???1?3?
?t2???1?3?
其中t?
1??2?t2???0d??0??1?3??
??????
2?
2?0d??r?r3dr?
01??2??3?t2?183,而当t?3时，则有F(t)?0.
考虑到函数u?F(t)
????t????，则t?3t?3??22?(3?t),F(t)??18?0,?
§2
1.计算下列第二型曲线积分:(1)向;解：第二型曲线积分与曲面积分
?(2a?y)dx?dy，其中L为摆线x?a(t?sint),y?a(1?cost)(0?t?2?)沿t增加的方L?(2a?y)dx?dy????2a?a(1?cost)??asint?a(1?cost)?dt
L02??2?a(a?1)
(2)?xdx?ydy222,其中为圆周依逆时针方向;x?y?aL?Lx2?y2解：L的参数方程为：x?acos?,y?asin?,0???2?2??acos??(?asin?)?asin??acos??xdx?ydy
??d?所以?0La2x2?y2??sin?cos?d??0
02?(3)?xdx?ydy?zdz,其中L为从(1,1,1)到(2,3,4)的直线段;L解：L的参数方程为：x?1?t,y?1?2t,z?1?3t,0?t?1所以(4)?
Lxdx?ydy?zdz???(1?t)?2(1?2t)?3(1?3t)?dt?13
10??
L(x2?2xy)dx?(y2?2xy)dy,其中L为y?x2从(1,1)到(?1,1)的一段;解：(5)??
L(x2?2xy)dx?(y2?2xy)dy=?(x2?2x?x2)?(x4?2x3)?2xdx?
1?1??2
15Lydx?xdy?(x2?y2)dz,其中L为曲线x?et,y?e?t,z?at从(1,1,0)到(e,e?1,a);L解：ydx?xdy?(x?y)dz=?e?t?et?et?(?e?t)?(e2t?e?2t)?adt
0221????2?a(e2t?e?2t)dt
01???2?
a2(e?e?2)2?2?asinh2
(6)?
L(x2?y2)dx?(x2?y2)dy,其中L为以A(1,0),B(2,0),C(2,1),D(1,1)为顶点的正方形沿逆时针方向.解：L?AB?BC?CD?DA
其中AB:y?0,1?x?2,起点对应x?1;BC:x?2,0?y?1,起点对应y?0;CD:y?1,1?x?2,起点对应x?2;DA:x?1,0?y?1,起点对应y?1.
所以?(x
L2?y2)dx?(x2?y2)dy
110021=?
21x2dx??(4?y2)dy??(x2?1)dx??(1?y2)dy=22.计算曲线积分?
L(y2?z2)dx?(z2?x2)dy?(x2?y2)dz.222(1)L为球面三角形x?y?z?1,x?0,y?0,z?0的边界线，从球的外侧看去，L的方向为逆时针方向;(2)L是球面x?y?z?a和柱面x?y?ax(a?0)的交线位于oxy平面上方的部分，从x轴上(b,0,0)(b?a)点看去，L是顺时针方向.解：(1)L?L1?L2?L3222222L1:x?0,z?1?y2,0?y?1,起点对应y?1;L2:y?0,z?1?x2,0?x?1,起点对应x?0;L3:z?0,y?1?x2,0?x?1,起点对应x?1.
所以?(y
L2?z2)dx?(z2?x2)dy?(x2?y2)dz
?1??2x?dy????(1?x2)?x2?
0?21?x2????dx?
??dx?
??
01??2y?(1?y2)?(?y2)??21?y2?
0??2x???(1?x2)?(?x2)?121?x2?
?0
(2)L?L1?L2,其中L1:y?ax?x2,z?a2?ax,0?x?a,起点x?a;
L2:y??ax?x2,z?a2?ax,0?x?a,起点x?0.
所以I?
?
L(y2?z2)dx?(z2?x2)dy?(x2?y2)dz
=?
L1(y2?z2)dx?(z2?x2)dy?(x2?y2)dz??(y2?z2)dx?(z2?x2)dy?(x2?y2)dz
L2??
0a??a?2x?a22222?(2x?ax)??2ax?x?a?(a?ax?x)?dx222ax?x2a?ax??
?a?2x?a?a
???2ax?x2?a2?(a2?ax?x2)?(2x2?ax)??dx0222ax?x2a?ax????(a2?ax?x2)?
0a2x?aax?x
2dx
令a?x?t,则x?0时,t???;x?a时,t?0.xa,21?t0x?dx??
2at
dt,代入上式得22(1?t)?
1
?
2
??dt??dt?2?
??0?13
I??2a??
???1?t21?t2?
3???1?t??1?t?22324???dt??1???2a??3?01?t2?01?t2?
3??2dt??
??0?1?t?1
23?1?t?1
24?dt???
??2a3(I1?3I2?I3?2I4)
其中In??
??0?1?t?dt
2n,n?1,2,3,4.
??2n?31t2n?3
由递推公式In???I?In?1n?12(n?1)(1?t2)n?102(n?1)2(n?1)最后得3
I???a323.求闭曲线L上的第二型曲线积分222?Lydx?xdy
.x2?y2(1)L为圆x?y?a,逆时针方向;x2y2(2)L为椭圆2?2?1,顺时针方向;ab(3)L为以(0,0)为中心,边长为a,对边平行于坐标轴的正方形,顺时针方向;(4)L是以(?1,?1),(1,?1),(0,1)为顶点的三角形,顺时针方向.解：(1)L的参数方程为:x?acos?,y?asin?,0???2?,起点对应??0
所以?Lydx?xdy1
=222ax?y2?0??asin??(?asin?)?acos??acos??d?2?0??d???2?(2)L为椭圆，其参数方程为：x?acos?,y?bsin?,0???2?,起点对应??2?.所以?L0bsin??(?asin?)?acos??bcos?ydx?xdy?d?222222?2?acos??bsin?x?y
??
2?0ab
d(tan?)
a2?b2tan2?ab
d(tan?)
a2?b2tan2??4?
?20?b??2
?4arctan?tan???2?0a??
(3)L?L1?L2?L3?L4,其中aaaa
L1:x??,??y?,起点y??;
2222aaaa
L2:y?,??x?,起点x??;
2222aaaa
L3:x?,??y?,起点y?;
2222aaaa
L4:y??,??x?,起点x?.
2222所以?Lydx?xdy?ydx?xdy????????L2?L3?L4?x2?y2??L1x2?y2a2a?2??
a2?y241
ady2+?
a2a?2aaadx?dy?dxaa??22+?a222+?a222aaa22222x??yx?
444?4?
a2a?22ya2?2y?
d?4arctan?2???2a?a2?2y??a?
1????a?
数学分析简明教程答案21
