微积分全英微积分全英 (74).pdf

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1、问问题题 1：222 1 2LLxyzxyzyzxzdxxdydz+=+=设是柱面与平面的交线，从 轴正向往 轴负向看去为逆时针方向，则曲线积分 222Q1:is the intersection of cylinder 1 and plane.Looking from the positive z-axis to the negative z-axis,it is counter clockwise.calculate.2+=+=LLxyzxyyxzdxxdydz 22222222Ans:1,The projection curve of on plane is 1,counter-cloc

2、kwise direction.The closed region bounded by is denoted as:1.()()22 ：+=+=+=+=+=+LLLxyxyzLxOyxyLD xyyyxzdxxdydzx xy dxxdydxdy222 =()()22 =(1)+=LDyyxxydxxdyxy dxdy 问问题题 2：2(,)|1,0,0,0 x y zxyzxyzy dS=+=设则 2Q2:(,)|1,0,0,0calculate =+=x y zxyzxyzy dS 1122200Ans:(,)|1,0,033312=+=yDDx yxyxyy dSydxdydyy dx

3、 问问题题 3：222223 (0,0)2 (2,0)4 (0,2)3(2)LLxyxxyIx ydxxxy dy+=+=+已知 是第一象限中从点沿圆周 到点再沿圆周 到点的曲线段，计算曲线积分 222223Q3:is the curve in the first quadrant from point(0,0)to point(2,0)along the circumference 2 and then point(2,0)to point(0,2)along the circumference 4calculate 3(2+=+=+LxyxxyIx ydxxxy)Ldy 111232323

4、Ans:Take the equation of the directed line as x=0,starting point(0,2)and ending point(0,0)The plane area enclosed by and is D.3(2)3(2)3(+=+=+LL LLLLIx ydxxxy dyx ydxxxy dyx ydxx111233202322)3(2)(2)(3)=23(2)(2)442+=+=+=LL LDDLxy dyx ydxxxy dyxxyx y dxdyxxdxdyx ydxxxy dyy dyI 问问题题 4：(1,0,0)(0,1,1)z 02

5、.(1)(2)LABLzz=设直线过，两点，将绕轴旋转一周得到曲面，与平面，所围成的立体为求曲面的方程求 的形心坐标 Q4:The straight line passes through(1,0,0)and(0,1,1),and is rotated around the z-axis for 360 degree to obtain the surface The solid formed by and 0,2 is.(1)Find the equation of su=LABLzzrface(2)Find the centroid coordinates of 2222222000002

6、22Ans:1(1).The straight line is 1111,:(1):221(2)Let the centroid coordinates of is(,)From Symmetry,we have 0Let(,)|221=+=+=+xxyzLxz yzxyzzxyzzxyzxyDx yxyzzdx222002200(221)10 =314(221)3757the centroid coordinates of is(0,0,)5=+=+=xDdydzdzdxdyzzdzzdxdydzzzzdzzdxdydzzdxdydz 问问题题 5：2233 (1)(1)(1)(1)zxyz

7、Ixdydzydzdxzdxdy=+=+设为曲面的上侧，计算曲面积分 2233Q5:is the upper side of the surface (1)calculate(1)(1)(1)，=+=+zxyzIxdydzydzdxzdxdy 12211332233Ans:Let is the lower side of surface1,1Let the area enclosed by and is.(1)(1)(1)3(1)3(1)11,0(1)(1)(1)+=+=+=+xyzxdydzydzdxzdxdyxydzdxdyzdzxdydzydzdxzdxdy122112220000(33

8、7)(37)4=+=+=rxdxdydzydxdydzIxydxdydzddrrrdz 问问题题 6：222222 2,(0,2,0),(0,2,0)()()LLzxyzxABIyz dxzxy dyx y dz=+已知曲线的方程:起点为终点为计算 222222Q6:2,starting point (0,2,0),ending point:(0,2,0)calculate()():：=+LL zxyzxABIyz dxzxy dyx y dz 112211222Ans:Let straight line from A to BThe projection curves of and on t

9、he plane are:22(0),clockwise;:0,:22The area enclosed by and is,()()：，+=+LLLxOylxyxlxyllD zxIyz dxzxy dyx11112222222222 ()()()2 (12)12+=+=+=+=LL LLllllDDy dzyx dxxxy dyx y dxydyyxx ydxydyx y dxdydxdy 问问题题 7：2222=(1)23xyzIxdydzydzdxxdxdy+=+设有界区域 由平面与三个坐标平面围成，为 整个表面的外侧，计算曲面积分 2Q7:The bounded region is

10、surrounded by plane 222 and,three coordinate planes.is the outside of,calculate=(1)23+=+xyzxoy yoz zoxIxdydzydzdxxdxdy 12(1)1200012(1)00120Ans:=(21)1112 1 1323 (1)21 (1)1211121232+=+=yxxxIxdxdydzdxdydzxdxdydzdxdyxdzydxxxdyxxdxI 问问题题 8：222 1 0 LLxyzxyzxyds+=+=设为球面与平面的交线,222Q8:is the intersection of s

11、phere 1 and plane 0+=+=LLxyzxyzxyds 2222222Ans:10,have rotation symmetry1()311 ()()32 ：，+=+=+=+LLLLLLxyzxyzx y zxydsxzdsyzdsyzxyxz dsyxzxyzds21 0161 263=Lds 问问题题 9：22 1 0 zLLxyyzzdxydy+=+=+=设为柱面与平面的交线,从z轴正向往 轴负向看去为逆时针方向,则曲线积分 22Q9:is the intersection of cylinder 1 and plane 0.From the positive z-axi

12、s to the negative z-axis,its counter clockwise.Calculate+=+=+LLxyyzzdxydy222222Ans:1,The projection on xoy plane is:1,counter-clockwise direction.0The closed region enclosed by is:1.0=：+=+=+=+=+=LlDxyLl xyyzlD xyyzzyIzdxydzydxydydxdy 问问题题 10：x+y+z=1(23)xyz dxdydz+=设由平面与三个坐标平面所围成的空间区域,则 Q10:is the sp

13、ace area enclosed by plane 1 and three coordinate planes.Calculate(23)+=+xyzxyz dxdydz Ans:It can be seen from the symmetry of rotatio:=(23)6:01,(,)(),()is the plane region obtained by making a plane section perpendicular to the=+=xdvydvzdvIxyz dvzdvzx yD z D z2111223000()-axis through any point in

14、0,1.1Its area is(1).2111(1)(2)22241=4=+=D zzzzzdvzdzdxdyzzdzzzz dzI 问问题题 11：2222(,)|1,1LxdxaydyDx yxyxya=+=若曲线积分在区域 内与路径无关则 2222Q11:The curve integral is independent of the path in the region(,)|1.1Find=+LxdxaydyDx yxyxya22222222222222Ans:The curve integral is independent 1of the path in the region(

15、,)|1.=1122,(1)(1)it is independent of the path=，+=+=+LxdxaydyxyDx yxyxayPQxyxyPxyQaxyyxyxxyQx1=Pay 问问题题 12：2222222 (,)9.1);2)Szxyzxx y zxyzCCxOySM=+=+设薄片型物体 是圆锥面被柱面割下的有限部分,其上任一点的密度为记圆锥面与柱面的交线为求 在平面上的投影曲线的方程求 的质量 222222Q12:The thin object is a finite part of the cone cut by the cylinder 2.The density

16、 of any point above is(,)9.C is the intersection line of the cone and the cylinder.1)Find=+=+Szxyzxx y zxyz the equation for the projection curve of C on plane;2)Finding the mass of.xoyMS 2222222222Ans:1),Eliminate,the projection cylinder of C on xoy plane is 202.So the equation of projection curve

17、is 22)Because the point density of is(,)9:=+=+=+=+zxyCzzxzxyxxyxSx y zxyz22222222222222cos222202,So the mass of S is=9.The projection area of on plane is(,)|2.92()1()()181864+=+=+=+=SDDMxyz dSSxoyDx yxyxxyMxydxdyxyxyxy dxdydr dr 问问题题 13：22331 33,=(2)xyzIxdydzydzdxz dxdy=+设 是曲面的前侧 计算曲面积分 2233Q13:is t

18、he front side of the surface 1 33 and calculates the surface integral=(2).=+xyzIxdydzydzdxz dxdy 122113322Ans:331 is the back of the part bounded by equation.0 is the solid surrounded by and.From Gauss formula,(2)(1 33).Let cos,sin,(1+=+=+=yzxxdydzydzdxz dxdyyzdxdydzyrzr21321 32223000322303122222300

19、3333)(1 3)2(1 3)1 3214Let 1 32(1 3)1 3=(2)34514(2)=0+=+=+=+=+=ryzdxdydzddrrrdxrrr drrtrrr drtt dtxdydzydzdxz dxdyI45 问问题题 14：2222244(0),44xyzzxz dxdy+=设 是曲面的上侧则 22222Q14:is the upper side of surface 44(0).calculate 44+=xyzzxz dxdy 2222222200Ans:The projection area of on plane is(,)|4.444(4)32|2sin3=

20、+=xyxyxyDDxoyDx yxyxz dxdyxxydxdyy dxdydr rdr 问问题题 15：2222(14),(),=()2()2()xyxyf xIxf xyxy dydzyf xyyx dzdxzf xyz dxdy+设 为曲面z=的下侧是连续函数 计算 2222Q15:is the lower side of(14),and()is a continuous function.Calculate=()2()2()=+zxyxyf xIxf xyxy dydzyf xyyx dzdxzf xyz dxdy 2222222222222222222222220Ans:Becau

21、se the normal vector of is(,)12=()()22()22()Let(,)|14,()()1()()12()()()+=+=+=+=+=+=Dx yzIxyzf xyxyz dSxydSxyDx yxyzzxyxyxyxyIxydxdydr21143=dr 问问题题 16：2222224,2,44LxyxyIdxdy Lxyxyxy+=+=+计算曲线积分是方向为逆时针方向.222222Q16:4Calculate,:2,and the direction is counter clockwise44+=+=+.LxyxyIdxdy L xyxyxy1122112222

22、22222222Ans:Let 4=1,the direction is clockwise;The plane area enclosed by to is marked as.44444 =4444According to:+=+LL LLLxyLLDxyxyIdxdyxyxyxyxyxyxydxdydxdyxyxyxyxy1112222222222222241Greens formula44()()0,4444444(4)()()(4),+=+=+=+=L LDLLxyxyxyxyxydxdydxdyxyxyxxyyxyxyxyIdxdyxyxyxy dxxy dyxyxydxdyxy

23、问问题题 17：222()(1)(01)0,xyzzzz+=设 是由锥面与平面围成的椎体求 的形心坐标 222Q17:is the vertebral body surrounded by cone()(1)(01)and plane 0.Finding the centroid coordinates of+=xyzzzz00002221120010Ans:Let the centroid coordinates of is(,).Because is symmetric about the plane=0when 01,Let(,)|()(1)(1)3=+=zzDxyzyozxzDx yx

24、yzzVdxdydzdzdxdyzdzydxdydzdz1210001201120000(sin)(cos,sin)(1)12(1)12141 4the centroid coordin令=+=zzzDDydxdydzdzrrdrxryzrzzdzzdxdydzdzzdxdyzzdzydxdydzyVzdxdydzzV1 1ates of is(0,).4 4 问问题题 18：2222(),1xz dvzxyzxy+=+=计算三重积分其中 是由曲面与围成的区域2222Q18:Calculate(),is the area bounded by surfaces and 1.+=+=xz dvz

25、xyzxy2122440000Ans:is symmetric about the plane011cossin2sin248()8|=+=yozxdvzdvddrrdrxz dv 问问题题 19：2222(),04yzxyz dvzxz=+=求其中 是由曲面绕 轴旋转一周而成的曲面与所围成的立体 222Q19:Calculate(),is a three-dimensional region formed 2by a surface rotating around the-axis and 4.0+=xyz dvyzzzx 422422220000Ans:256()()43+=+=zxyz

26、dvdzdrz rdrz dz 问问题题 20：2222,(,)|1z dvx y zxyz=+求其中 2222Q20:Calculate,(,)|1=+z dvx y zxyz 22222122000Ans:From symmetry of functional displacement,1()31 sin311 2 2354 15=+=z dvxyzdvddrrdr 问问题题 21：2221,()LLyxxyds=+=设平面曲线 为下半圆则 222Q21:The plane curve is the lower semicircle of 1,calculate()=+LLyxxyds 2

27、2222222Ans:cosThe parameter equation of lower semicircle 1 is,2.sin()(cossin)(sin)cos=+=+=Lxtyxtytxydsttttdt 问问题题 22：22221,(243)43lxylaxyyx ds+=+=设 为椭圆其周长为 则 2222Q22:1,its circumference is a,calculate(243)43+=+lxylxyyx ds 2222Ans:Substituting the equation of:3412 into the integral.(234)(212)212 0 12

28、+=+=+=+=+=lllllxyxyxydsxydsxydsdsa12a 问问题题 23：2:(02),LL yxxxds=已知曲线则 2Q23:Curve:(02),calculate=LL yxxxds 220222032220Ans:1(2)1 1 4(1 4)8113 (1 4)126|=+=+=+=Lxdsxx dxx dxx 问问题题 24：2229,(22)(4)LLxyxyy dxxx dy+=+设 为取正向的圆则曲线积分的值 222Q24:is the positive circle 9,calculate(22)(4)+=+LLxyxyy dxxx dy 22Ans:Fr

29、om Greens formula24(22)22 918(=(,)|9),=+DDIxxdxdydxdyDx yxy 问问题题 25：3(0,0)(,0)sin(0),(1)(2)LOAyax aLOAy dxxy dy=+在过点和的曲线族中 求一条曲线，使沿该曲线从 到 的积分的值最小.3Q25:In the curve family sin(0)passing through point(0,0)and point(,0),find a curve minimize the value of integral(1)(2)from to along the curve.=+Lyax aOA

30、Ly dxxy dyOA33302Ans:4()1sin(2sin)cos 43()4(1)01 and(1)80()takes the minimum at a=1,which is also the minimum:sin=+=+=I aaxxax ax dxaaI aaaII aL yx 问问题题 26：2(),(0)0,(0)1,()()()0,()f xffxy xyf x dxfxx y dyf x=+=设具有二阶连续导数且为一全微分方程 求及此全微分方程的通解.2Q26:()has the second-order continuous derivatives,(0)0,(0)1

31、,and()()()0 is a total differential equation.Find()and general solutions of total differential equations=+=f xffxy xyf x dxfxx y dyf x.2221212222Ans:()22()()()()cossin2(0)0,(0)12,1()2cossin2General solutions2 sincos2,2：=+=+=+=+=PQyxfxxyxxyf xfxf xxf xCxCxxffCCf xxxxx yyxyxxyCCR 问问题题 27：2(),=()xay dx

32、ydyaxy+已知为某函数的全微分 则 2Q27:()is the total differential of a function,find()+xay dxydyaxy 22243Ans:,()()()2()()2()()(2)2(2)(2)2+=+=+=xayyPQPQxyxyyxa xyxay xyyxyxyaxayyaxaya 问问题题 28：(,1)(1,)(0,0)(0,0)(,)2(,),2(,)2(,),(,)LttQ x yxOyxydxQ x y dytxydxQ x y dyxydxQ x y dyQ x y+=+设函数在平面上具有一阶连续偏导数,曲线积分与路径无关 并

33、且对任意 恒有求(,1)(1,(0,0)(0,0)Q28:The function(,)has the first order continuous partial derivative in the planecurve integral 2(,)is independent of path.For t,2(,)2(,)+=+LtQ x yxoyxydxQ x y dyxydxQ x y dyxydxQ x y dy),find(,)tQ x y 2(,1)1122(0,0)00(1,(0,0)Ans:Curve integral is independent of path2(,)(),(

34、)is differentiable.(0,0)(,0)(,1)2(,)()()(0,0)(1,0)(1,)2(,)=+=+=+tQxQ x yxC y C yxttxydxQ x y dytC y dytC y dytxydxQ x y dy)200120021()()()()21(),()21(,)21=+=+=+=+=+ttttC y dytC y dytC y dytC y dytC t C ttQ x yxy 问问题题 29：22,(1,0),(1),.4LxdyydxILRRxy=+计算曲线积分其中 是以点为中心为半径的圆周取逆时针方向22Q29:is the circle wit

35、h(1,0)as the center and R as the radius(1);take the counter clockwise direction.Calculate.4=+LLRxdyydxIxy222222222222222Ans:=,444,(,)(0,0)(4)cosMake an ellipse small enough:(0,2,is counterclockwise)2sin0444+=+=+=+=+L CLCyxPQxyxyPyxQx yyxyxxCCyxdyydxxyxdyydxxdyydxxyxy222012=d 问问题题 30：2222LLxyxdyydx+=

36、设 为正向圆周在第一象限中的部分,则曲线积分 22Q30:is the part of the positive circle 2 in the first quadrant,and calculate curve integral 2+=LLxyxdyydx 2220220Ans:2cosParametric equation of:,:022sin2(2cos4sin)2(1 sin)3 2=+=+=LxtLtytxdyydxtt dtt dt 问问题题 31：21|(1,1)(1,0),(1,0),LLyxxxydxx dy=+=已知曲线 的方程为,起点是终点为则曲线积分 2Q31:Th

37、e equation of curve is 1|(1,1).The starting point is(1,0)and the ending point is(1,0).Calculate=+LLyxxxydxx dy 1212222Ans:Let:1(1,0),the starting point is(1,0)and the ending point is(0,1);:1(0,1),the starting point is(0,1)and the ending point is(1,0).=+=+=+LLLLyx xLyx xxydxx dyxydxx dyxydxx dy012210

38、 =(1)(1)1212 =()()02323+=xxx dxxxx dx 问问题题 32：2sin22(1),sin (0,0)(,0).LxdxxdyLyx+=计算曲线积分其中是曲线上从点到点的一段 2Q32:Calculate sin22(1),is the arc from point(0,0)to point(,0)on the curve sin.+=Lxdxxdy Lyx 22020200Ans:sin22(1)sin22(1)sin cos =sin211 =cos22 cos222|+=+Lxdxxdyxxxx dxxxdxxxxxdx2002111 =sin2sin2222

39、1 =2|+xxxdx 问问题题 33：42242,0(,)2()()(,),(,).xA x yxy xyixxyju x yu x y=+确定常数使在右半平面上的向量为某二元函数的梯度并求 42242Q33:Determining constant,let the vector(,)2()()on the right half plane 0 be the gradient of the binary function(,).Find(,).=+A x yxy xyixxyjxu x yu x y 4224242Ans:(,)2(),(,)()4()(1)0.when 1,the vect

40、or field given is a gradient field.Take any point in the half plane where 0,such as(1,0)as the starting=+=+=+=P x yxy xyQ x yxxyQPx xyxyx2(,)42(1,0)24242102point of the integration path.Take integral path(1,0)(,0)(,),we get2(,)20 =0 =arctan,.=+x yxyxx yxydxx dyu x yCxyxxdxdyCxxyyCCRx 问问题题 34：22()()(

41、),1,2Czy dxxz dyxy dzxyCzzCxyz+=+=计算曲线积分其中 是曲线从轴正向往轴负向看的方向是顺时针的.22Q34:Calculate()()(),1 is the curve.From the positive-axis to 2the negative-axis,the direction of is clockwise.+=+=Czy dxxz dyxy dzxyCzxyzzC 222222Ans:1,The projection curve on plane is:1,clockwise.2The closed region bounded by is:1,fr

42、om 22.()()()(2)(22)：+=+=+=+=+=+=+CxyCxOyl xyxyzlD xyxyzzxyIzy dxxz dyxy dzx dxxy()()(22)(322)(3 1)2+=+=llDdyxydxdyxy dxxydydxdy 问问题题 35：222222()(2)(3),2|1,LIyzdxzx dyxydzLxyzxyzL=+=+=计算其中 是平面与柱面的交线 从 轴正向看去为逆时针方向 222222Q35:Calculate()(2)(3),is the intersection line of plane 2 and cylinder|1.Viewed fr

43、om the positive direction of axis,is counterclockwise=+=+=LIyzdxzx dyxydzLxyzxyzL 222222Ans:2,the projection curve on plane is:|1,counterclockwise,|1The closed region bounded by is:|1,from 22()(2)(3)：+=+=+=+=+=+CxyzLxOylxyxylDxyxyzzxyIyzdxzxdyxydz2222222222 (2)2(2)(3)()(44442)(823884)(2212)24=+=+=+=

44、llDyxydxxyx dyxydxdyxyxyxy dxxyxyxy dyxydxdy 问问题题 36：2222,2zdSzxyxyx=+计算曲面积分其中 为锥面在柱体内的部分 2222Q36:Calculate,is the part of conical surface in cylinder 2.=+zdSzxyxyx 22222cos222202320Ans:The projection area of on plane is:2.1()()2221632 =2cos239+=+=+=xyDxOyD xyxdSzzddzdSxyddr drd 问问题题 37：:|1,(|)xyzxy

45、 dS+=+=设曲面则 Q37:|1,calculate(|)+=+xyzxy dS 2Ans:is symmetric with respect to plane 0.is rotate symmetric with respect to,|111|(|)33318(2)sin4 323|=+=yOzxdSx y zy dSz dSx dSy dSxyz dSdSSSy dS433=问问题题 38：2223331,xyzx dydzy dzdxz dxdy+=+设 为曲面的外侧 计算曲面积分 222333Q38:is the outside of surface 1,calculate+=+

46、xyzx dydzy dzdxz dxdy 2222122000Ans:From Gauss formula,and use spherical coordinates to calculate triple integral.=3()(is the area enclosed by)12 =3sin5+=Ixyzdvddrr dr 问问题题 39：2222222=0,.:(1).Sz axyzaSVx yz dydzxy z dzdxzxyz dxdyV+=设空间区域 由曲面与平面围成 其中为正常数.记 表面的外侧为的体积为证明 2222222Q39:The space region is

47、bounded by a surface=and a plane=0,where a is a positive constant.Note that the outside of surface is,and the volume of is.Proof:(+z axyzSVx yz dydzxy z dzdxz 1).+=Sxyz dxdyV 222222Ans:From Gauss formula,(1)=(1 2)2A is symmetric with respect to plane,is an odd function of y on.=0+=+Sx yz dydzxy z dz

48、dxzxyz dxdyxyz dxdydzVxyzdxdydzxOz xyzxyzdxdydzx yz dydz22(1)+=Sxy z dzdxzxyz dxdyV 问问题题 40：323232222()()(),.Ixazdydzyaxdzdxzaydxdyzaxy=+=计算曲面积分其中 为上半球面的上侧 323232222Q40:Calculate()()(),is the upper side of the upper hemisphere.=+=Ixazdydzyaxdzdxzaydxdyzaxy 222323232323232222Ans:Let is the lower side

49、 of plane 0(),is the space enclosed by and.()()()()()()=3()+=+=+SSSzxyaSIxazdydzyax dzdxzaydxdyxazdydzyax dzdxzaydxdyxyzdv222222423200000555 =3sinsin6129 =5420 +=xyaaaay dxdyddr dradr draaa 问问题题 41：222222,=2xzdydzyzdzdxz dxdyzxyzxy+=+计算其中 是由曲面与所围立体的表面外侧.22222Q41:Calculate2,is the outside of the surf

50、ace bounded by and=2.+=+xzdydzyzdzdxz dxdyzxyzxy222234000Ans:2,2,2,.From Gauss formula,2 =sincos =+=+=Pxz Qyz RzPQRPQRzzzzxyzxyzxzdydzyzdzdxz dxdyzdxdydzddr dr 2=问问题题 42：22(2),(01),Sxz dydzzdxdySzxyzz+=+计算曲面积分其中 为有向曲面其法向量与 轴正向夹角为锐角.22Q42:Calculate(2),is a directed surface(01),and the angle between i