医用高等数学习题解答（第1,2,3,6章） - 46 -
?Fx??14?8y?4x???0?F(x,y)?15?14x?32y?8xy?2x2?10y3??(x?y?1.5)，?Fy??32?8x?30y2???0
?x?y?1.5??15y2?2x?4y?9?01?11?5y2?2y?2?0?y??0.8633 , x?0.6367 ??5?x?1.5?y第五章多元函数积分学习补充习题
1.画出积分区域，计算下列二重积分：
(1) ??xexydxdy, D为矩形：0?x?1, –1?y?0。
D(2) ??xy2dxdy, D为x2+y2?4与y轴所围成的右半区域。
D(3) ??(x2?y2)dxdy, D为：0?y?sin x , 0?x??。
D2.交换积分次序
(1) ?dy?f(x,y)dx
0 01y
(2) ?dx? 022x xf(x,y)dy
22
(3) ?dx?x2?1f(x,y)dy
?54 11?x(4) ?dx?12 01?x xf(x,y)dy
(5) ?dx?02x220f(x,y)dy?? 2dx?8?x2 0f(x,y)dy
3.用二重积分求下列曲线所围图形面积： (1) y?x , y?5x , x?1； (2) y2?x , y2?4x , x?4；
yoy(0,2)(1,0)xox2?y2?4(2,0)x(3) xy ? 4 , xy ? 8 , y ? x , y ? 2x (x > 0 , y > 0)。 (0,?1)(1,?1)(0,?2)题解
1.画出积分区域，计算下列二重积分：
(1) ??xexydxdy??dx?xexydy??(1?e?x)dx
D0?101 01yy?sinx(2) ??xydxdy??dy?D?22 24?y2 0xy2dx??y24?y2dy?0sinx222??64 15sinxo?x(3) ??(x?y)dxdy??dx?D 022? 0??11???(x?y)dy???x2y?y3?dx???x2sinx?sin3x?dx
0 033? 0????1?3??xsinxdx??sinxdx 03 0440?(?2?4)???2?
99?2y1y?x(1,1)y4y?2xy?x2(2,2)o1x医用高等数学习题解答（第1,2,3,6章） - 47 -
2.交换积分次序
(1) ?dy?f(x,y)dx??dx?f(x,y)dy
0 01y11 0x(2) ?dx? 022x xf(x,y)dy
42 2??dy?yf(x,y)dx??dy?yf(x,y)dx
0222yx2?1y?4(3) ?dx?x2?1f(x,y)dy
?54 11?x(?5,6)61?yy6y1y?x?112oy?x??1dy?-412 0 04y?1?4y?11?xf(x,y)dx??dy? 0?4y?1f(x,y)dx 11(,)2212(4) ?dx?12 0y0 xf(x,y)dy
11?y??dy?f(x,y)dx??1dy?22x22 00f(x,y)dx
o1x?51(0,?)41xy?1?xy(5) I??dx? 0f(x,y)dy??22 2dx?8?x2 0f(x,y)dy x2?y2?82??0?y?1?0?y?8?x22x积分域由两部分组成：D1:? ,D2:?0?x?2???2?x?2222y?12x??2y?x?8?y2将D?D1?D2视为Y–型区域, 则D:? ?0?y?2?D1y?5xD2I??dy? 028?y2 2yf(x,y)dx
y5210o2y22x3.用二重积分求面积
(1) y?x , y?5x , x?1；
4y2?4xS??dx?dy??4xdx?2x 0 x 015x1?2
4y?xoy2?x(2) y2?x , y2?4x , x?4；
4xS?2?dx? 042x xdy?2?4 04332xdx?x2?
3038x xo1x(3) xy ? 4 , xy ? 8 , y ? x , y ? 2x (x > 0 , y > 0)
?4y422S??dx?4dy??2 x 2 22x22 2dx?dy
y?2xy?x224???8????2x? ?dx????x?dx
2 2x???x??x?4lnx(2,2)xy?8xy?4x?2? 221????8lnx?x2?
2? 2?22o2224医用高等数学习题解答（第1,2,3,6章） - 48 -
??2?2ln2??(4ln2?2)?2ln2
第五章多元函数积分学习题题解(P162)
一、判断题题解 1.正确 2.错。缺r
3.?x?y?3?ln(x?y)?1?1?ln(x?y)?ln2(x?y) ?错 4.正确
5.如图所示,在D内有0?x+y?1?(x+y)2?(x+y)3。错 6.错。应为rdrd?
二、选择题题解
1.函数相同且关于,x,y轴对称，而D1是对称区域, D2是其中四分之一， 故I1=4I2。(C)
2.因为|x|?1,|y|?1所以x+1?0。(D) 3.如图交换积分次序(D)。
y1oD3xy1?10dx?1?x 0f(x,y)dy??dy?011?y 0f(x,y)dx
oyD11x2?y2?14.积分区域如图，将之化为极坐标：
?2 0?10dx?1?x2x 011?x2?y2dy
??d??r1?rdr?012??1???1?r2?3?322?????(C) ?06y22x?y?2?05.积分区域的面积为1，如图，选择(A)。 6.积分区域为矩形。
x?yxyedxdy?edxe????dy?eDo1x11x1000?ey10?(e?1)2 (A)
yy?1?x1oDo11?y00三、填空题题解
1x1xI??dy??1 01?y0f(x,y)dx??dy?f(x,y)dx??dx?011?xx?1f(x,y)dy 1x?1?y四、解答题题解
1. 证明：因为(1,0)在圆周(x?2)2+(y?1)2=2上,圆周上的导数为：2(x?2)+2(y?1)y?=0, y(x?2)2?(y?1)2?2y?x?1??1，故圆周上(1,0)处的切线方程为：x+y=1。而切线上方有：x+y?1，
y?0那么在区域D内也有：x+y?1。? (x+y)2?(x+y)3
?
2. 列出两个变量先后次序不同积分 (1)区域D如图：
y1?x23(x?y)dxdy?(x?y)dxdy ????DDDo13xx?y?1??Df(x,y)d???dx?011?y1 0f(x,y)dy
1yy?xy2?4x4Df(x,y)dx
o??dy?0 01xo4x医用高等数学习题解答（第1,2,3,6章） - 49 -
(2)区域D如图：
??Df(x,y)d???dx?042x xf(x,y)dy??dy?104yy2f(x,y)dx 4y(0,a)x2?y2?a2(3)区域D如图：
??f(x,y)d???dx?D?aaa2?x2 0f(x,y)dy??dy?02xaa2?y2?a2?y2f(x,y)dx (?a,0)o(a,0)x(4)区域D如图：
??f(x,y)d???dx?D1221y1xf(x,y)dy yx?221oy?xy2y?2x??1dy?1f(x,y)dx??dy?f(x,y)dx
2y12xy?1y?x3.改变积分次序
12x1(1,1)(1) ?dx?f(x,y)dy??dy?yf(x,y)dx??dy?yf(x,y)dx 0 x12x1y22 0212o1y?x(2) ?dy? 0 12y 0f(x,y)dx??dy? 133?y 0f(x,y)dx
y3y?x?3??dx?1 023?xxf(x,y)dy
24.计算二重积分 12y?x(2,1)(1) ??sinxsinydxdy D:0?x?D?2, 0?y?? ?o23xo?x?20??sinxdx?sinydy?cosx02?cosy0?(0?1)?(?1?1)?2
0??y1y?x21??(2) ??(x?y)dxdy??dx?2(x?y)dy???x2y?y2?dx
0 x02?x2?D21x21xy?x7?5??13213?3342?dx??x2?x2?x5?????x?x?x??7?140 0?22410??x2??0x1 1oxx1x(3) ??xcos(x?y)dxdy??xdx?cos(x?y)dy??xsin(x?y)0dx??(xsin2x?xsinx)dx
D?x??000 0?1?1??(xcos2x)0??cos2xdx?? ? 0xsin2xdx??2? 0xd(cos2x)??2? 0??111?????sin2x0???
242??????xsinxdx??xd(cosx)??(xcosx)?cosxdx?? ? 0? 0? 00???y(?,?)o(?,0)x???sinx0??
?yx?221oy?xyy?xy2?x1xy?112xo1x医用高等数学习题解答（第1,2,3,6章） - 50 -
3? ??xcos(x?y)dxdy???
2D2x1x2(4) ??2dxdy??x2dx?12dy
1yxyD22?1?9?1214?3? x2??dx?(?x?x)dx??x?x?????y?114?14?2??xx2??1(5) ??D1sinyy1siny11siny1dxdy??dy?2dx??(y?y2)dy??(siny?ysiny)dy??cosy0??yd(cosy)
0y000yyy111??cos1?1?(ycosy)0??cosydy?1?siny0?1?sin1
0yyy?2x2y2?11?(6) ??(x2?y2?x)dxdy??dy?y(x2?y2?x)dx???x3?y2x?x2?dy
0 032?y?2Dy?x221?13?19332??19???y?y?dy??y4?y3?? 0248?8?06??9622o2x5.用极坐标计算二重积分
(1) ??eD2?x?y22dxdy D:x?y?4
r222y222x?y?4y2x2?y2?41r22?2??e??d??erdr???(e4?1)
0002(2) ??x3y2dxdy D:x2?y2?4,x?0,y?0
2Do2xo2x72??2d??(rcos?)(rsin?)rdr??2cos?sin?d??rdr?000072323226????20cos3?sin2?d?
2?77???D20128?sin?sin??2256??(1?sin?)sin?d(sin?)??? ??7?35?01052235(3) ??ln(1?x2?y2)dxdy D:x2?y2?1,x?0,y?0
?0??2d??rln(1?r2)dr?01?1?22?01ln(1?r2)d(1?r2)?1??22122?(1?r)ln(1?r)0??(1?r)dln(1?r)??
4?0?0?44?(4) ??(4?2x?3y)dxdy D:x2?y2?9
D?1?????2ln2?1? 2ln2?2rdr????y?2?x2?y2?4?2??d??(4?2rcos??3rsin?)rdr
002?3o?2?x