令xy=u,y?v,则
xx?u,y?uv,(2?u?4,1?v?3) v?x12?v??y?v2v1?uvvuv1v?u?12uv2?. 2vu2uv?x?(x,y)?uJ???(u,v)?y?u于是：
3141281u32xydxdy?u?dudv?dvudu???ln3. lnv??D????122v2v232312?u?42221?v?334(2)积分区域D如图10-24所示。
图10-24
令x+y=u,x-y=v,则
x?u?vu?v, y? 22且 -1≤u≤1, -1≤v≤1.
1?(x,y)2J???(u,v)12112?? 12?2于是：
16
4224??222u?2uv?vD(x?y)dxdy?dudv?1??1du?1(u4?2u2v2?v4)dv?1???1??uv?88?11?111?18?1??1??u4v?23u2v3?15v5?11??du?
?14???1?21??u4?3u2?5??du1?1?4?121?14?5u5?9u3?5u???.?145（3）积分区域Dxy: 0≤x≤1, 1-x≤y≤2-x 令x=v, x+y=u, 则y=u-v 积分区域Dxy变为Duv:
0≤v≤1, 1≤u≤2.
且J??(x,y)01?(u,v)?1?1??1
于是
2?1dx2?x220?1?(x?y)dy??1dv?2(2v2?2uv?u2)du??1?1?x0??2v2u?vu201?3u3??dv11
??1???2v2?3v?7???237?3?dv???3v3?2v203?3v???.02(4)令x=arcosθ, y=brsinθ则积分区域D变为 Drθ： 0≤θ≤2π, 0≤r≤1,
J??(x,y)?arsin??(r,?)?acos?bsin?brcos??abr
于是：
?x2y2?1??D??a2?b2??dxdy???Dr2abrdrd??2πd?1abr3dr?2???14??1r??0?0??4abr??πab02(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：
17
??Dx?y?4dxdy???Dr?4rdrd???0d??0r2?4rdr22223?2π???(4r?r3)dr??(r3?4r)dr?2?0?3??214?2?14?412??2π?2r?r??r?2r???π.??4?0?4?2?2??2π3
（6）积分区域D如图10-25所示：D可分为D1,D2∪D3,D4四个部分.它们可分为用极坐标表示为。
图10-25
D1: 0≤θ≤π, 0≤r≤2sinθ, D2∪D3: 0≤θ≤π, 2sinθ≤r≤2, D4: π≤θ≤2π, 0≤r≤2 于是：
??Dx2?y2?2ydxdy???Dx2?y2?2ydxdy???D?Dx2?y2?2ydxdy???Dx2?y2?2ydxdy1234??d??0π2sin?02sin?(2rsin??r2)rdr??d??0π22sin?2(r2?2rsin?)rdr??d??(r2?2rsin?)rdrπ02π2??d??0ππ0(2r2sin??r3)dr??d??02sin?ππ2sin?(r3?2r2sin?)dr??d??(r3?2r2sin?)drπ0222π22π?r4?r423??2??d?????rsin??d?????r3sin??d?00π?43?2sin??43?0π?2π?4π16416????sin4?d????4?sin??sin4??d????4?sin??d?0π30333????π?2π?8π1616????sin4?d????4?sin??d????4?sin??d?0π3033?????23r4??3rsin??4???081?3162π1??????sin2??sin4???8π??sin?d?34?2308?023??π?8π?0?9π.32
18
π
13. 求由下列曲线所围成的闭区域的面积：
b2(1)曲线y?x,y?bx所围（a>0,b>0）;
aa2(2)曲线xy=a2, xy=2a2, y=x, y=2x所围（x>0,y>0）.
b2解：（1）曲线y?x,y?bx(a?0,b?0)所围的图形
2D如图10-26
aa所示：
图10-26
D可以表示为：
??a2a?2y?x?y ?bb?0?y?b所求面积为：
S???dxdy??bady?bydx??b?aa?1D0ab2y0??by?b2y2??dy?6ab. (2)曲线xy=a2,xy=2a2,y=x,y=2x(x>0,y>0)所围图形D10-27所示：
图10-27
如图19
所求面积为
S???dxdy
D令xy = u,y?v,则
xx?u,y?uv,(a2?u?2a2,1?v?2) vJ??(x,y)1 ??(u,v)2v于是
222a212a1a2S???dxdy???dudv??dv?2du??dv?ln2 D1a12v2v2v2a2?u?2a21?v?214. 证明： (1)?ady?a(y?x)Dbynf(x)dx??1ba1f(x)(b?x)n?1dx; n?1(2)??f(x?y)dxdy???1f(u)du,D为|x|+|y|≤1; (3)??f(ax?by?c)dxdy?2??1D11?u2f?ua2?b2?c?du,其中D为x
2
+y2≤1
且 a2+b2≠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：
20
高等数学下册复旦大学出版社答案黄立宏著
