综合练习六01A设z?y?f(x?1),若当y?1时,z?x,则z?().(A)x?y?1;(B)y?x?1;(C)x?y?1;(D)x?y?1.01B求函数z?arcsin(x?y2)?ln[ln(10?x2?4y2)]的定义域.01C求下列极限:(1)xlim?(x2?y2)x2?y2;(2)xlim(x21?1x?yxyy0?0y???a)(a?0);3(3)2|y|2(x,ylimx)?(0,0)x4?y2;(4)x?limxy2y2??y???(x2?y2)x.01D证明下列极限不存在:(1)limxy2x?0x2?y4;(2)x3y?xy4?x2yy?0xlimy?0?0x?y.01E证明x2y2xlimy?0?0x2?y2?0.01F讨论函数u?x?yx3?y3的连续性.02A设z?f(x,y)满足?2f?y2?2x,f(x,1)?0,?f?y?sinx,y?0求f(x,y).2202B设f(x,y)?x2arctanyx?f?f?f?fx?y2arctany,求?x,?y,?x2,?x?y.02C求函数z?ln(x?y2)的一阶和二阶偏导数.02D22设z?yxln(xy),求?z?x2,?z?x?y.02E求函数z?xyx2?y2当x?2,y?1,?x?0.01,?y?0.03时的全增量和全微分.?2xy02F考察函数f(x,y)???x2?y2,x2?y2?0在点(0,0)处可导性,??0,(x,y)?(0,0).32.连续性与可微性.?x3y?02G设f(x,y)??xy3?x2?y2,(x,y)?(0,0)??0,(x,y)?(0,0)(1)求fx(0,0);(2)求fxy(0,0).?02H设f(x,y)??x2y2?(x2?y2)3/2,x2?y2?0,证明:f(x,y)在点(0,0)?处?0,x2?y2?0,连续且偏导数存在,但不可微分.?xy(x2?y2)02I设f(x,y)???x2?y2,x2?y2?0,?求?f?0,x2?y2?0,?x,?f?y,并证明:fxy(0,0)?fyx(0,0).?xysin1x2?y2?002J设f(x,y)???x2?y2,,证明f(x,y)在原点??0,x2?y2?0(0,0)可微.02K某函数的全微分为:(x?ay)dx?ydy(x?y)2,求a值.03A通过变换??x?2y,??x?2y(y?0)一定可以把方程?2z?x2?y?2z?y?1?z22?y(y?0)化为().22(A)?z?2z0;(B)?2zz??2???2???2????2?0;(C)?2z??2z2z??2z?????0;(D)??????0.03B设u?u(x,y)为可微分的函数,且当y?x2时,有u(x,y)?1及?u?x?x;则当y?x2(x?0)时,?u?y?().(A)12;(B)?12;(C)0;(D)1..33.03C设z?x3f(xy,y)?z?2z?2zx,(f具有二阶连续偏导数),求?y,?y2,?x?y.03D设函数??f(u,v),u?u(x,y),v?v(x,y),x?x(r,?),y?y(r,?)均满足复合函数求偏导数之条件,求????r,???.03E设f(x,y)可微,且f(x,2x)?x,fx?(x,2x)?x2,求fy?(x,2x).03F求下列复合函数的二阶混合偏导数?2z?x?y:(已知f具有二阶连续偏导数).(1)z?f(x?y,xy);(2)z?f(exsiny,x2?y2).(3)u?f(xy,yz)(4)z?f[x2?y,?(xy)](?(u)二阶可导).03G设u?xy,而x??(t),y??(t)都是可微函数,求dudt.03H设z?x2yf(x2?y2,xy),其中f有连续偏导数,求?z?z?x,?y.03Im?n设u?(x?x0)m(y?y0)n,求?u?xm?yn(m,n为正整数).03J设z?f(u,v,w)具有连续偏导数,而u????,v????,w????,求?z???,z?z??,??.03K试用变换x?cost,将方程(1?x2)d2ydx2?xdydx?0中的自变量x换成t,求变换后所得的方程.03L要求通过线性变换????x??y???x??y,将方程A?2u?x2?2B?2u?2u?x?y2?C?y2?0(,B,C)2其中A为常数,且AC?B2?0化简成?u?????0.求?,?的值.03M设u?f(x,xy),v?g(x?xy),求?u?v?x??x.03N设F(x,y(x),z(x))??(x,y(x))?z(x)?(x,y(x)),其中出现的函数是连.34.续可微的,试计算?Fd??y?dx??F???z??.03O设u?f(x?y,y?z,t?z),求?u?x??u?u?u?y??z??t.03P设x?ucosvv?u?v?u?vu,y?usinu,求?x,?x,?y,?y.04A设z?z(x,y)是由方程F(x?az,y?bz)?0所定义的隐函数,其中F(u,v)是变量u,v的任意可微函数,a,b为常数,则必有().(A)b?z?x?a?z?y?1;(B)a?z?x?b?z?y?1;(C)b?z?z?x?a?z?y?1;(D)a?x?b?z?y?1.04B设x?eucosv,y?eusinv,z?uv,试求?z?x和?z?y.04C设u?f(x,y,z),?(x2,ey,z)?0,y?sinx,其中f,?都具有一阶连续偏导数,且???z?0,求dudx.04D设y?g(x,z),而z是由方程f(x?z,xy)?0所确定的x,y的函数,求dzdx.04E设函数z(x,y)由方程F(x?zzy,y?x)?0确定,证明x?z?x?y?z?y?z?xy.04F设由方程F(x?y,y?z,z?x)?0确定隐函数z?z(x,y),求?z?z?x,?y及dz.04G设函数z?z(x,y)是由方程xyz?x2?y2?z2?2所确定的,求z在点(1,0,?1)处的全微分.04H设{u?v?x?y?0xu?yv?1?0,求?u?x,?v?u?v?x,?y,?y.04I设方程组.35.u?excotvy?x?u?extanvy?y确定函数u?u(x,y),v?v(x,y),试求在点x?1,y?1,u?0,v??4处的全微分du和dv.04J从方程组{x?y?u?v?1?2u?v?2x2?y2?u2?v2?2中求出?u?x,?x2,?x,v?x2.04K设{u?f(x?ut,y?ut,z?ut),求?u?ug(x,y,z)?0,?x,?y.05A函数u?sinxsinysinz满足x?y?z??2(x?0,y?0,z?0)的条件极值是().(A)1;(B)0;(C)1/6;(D)1/8.05B求由方程2x2?y2?z2?2xy?2x?2y?4z?4?0所确定的函数z?z(x,y)的极值.05C求z?x2?y2?5在约束条件y?1?x下的极值.05D某工厂生产两种产品,总成本函数为C?Q221?2Q1Q2?Q2?5,两种产品的需求函数分别为Q1?26?P1,Q2?10?14P2,试问当两种产品的产量分别为多少时,该工厂获得最大利润,并求出最大利润.05E某公司可通过电台及报纸两种方式做销售某种商品的广告,根据统计资料,销售收入R(万元)与电台广告费用x1(万元)及报纸广告费用x2(万元)之间的关系如下经验公式：R?15?14x1?32x2?8x1x2?2x21?10x22(1)在广告费用不限的情况下,求最优广告策略;(2)若提供的广告费用为1.5(万元),求相应的最优广告策略.06A估计积分I?100?cos2x?cos2yxdy的值,则正确的是().x???1dy?10(A)12?I?1.04;(B)1.04?I?1.96;.36.(C)1.96?I?2;(D)2?I?2.14.06B设f(x,y)是有界闭区域D:x2?y2?a2上的连续函数,则当a?0时,1?a2??f(x,y)dy的极限().D(A)不存在;(B)等于f(0,0);(C)等于f(1,1);(D)等于f(1,0).06C判断下列积分值的大小:ji???e?(x2?y2)dxdy,i?1,2,3其中DiD1?{(x,y)|x2?y2?R2},D2?{(x,y)|x2?y2?2R2},D3?{(x,y)|x|?R,|y|?R}.则J1,J2,J3之间的大小顺序为().(A)J1?J2?J3(B)J2?J3?J1(C)J1?J3?J2(D)J3?J2?J106D设D是有界闭区域,若f(x,y)在D上连续,??f2(x,y)d??0,则Df(x,y)?0((x,y)?D)06E设D是有界闭区域,若f(x)在D上连续,f(x,y)?0((x,y)?D),则??f(x,y)d??0.D06F利用重积分的性质判断下列积分的符号:(1)I???ln(x2?y2?12)dxdy;|x|?|y|?12(2)I???31?x2?y2dxdy,其中D?{(x,y)x2?y2?4}.D(3)I?|x??(x?1)dxdy.|?1|y|?106G计算I???(21?x?1D?e)ey2dxdy,其中D:x?1,y?1.06H求I???y2dxdy,D是由xy?2,y?x?1,y?x?1所围成的区域.D07A将坐标系中的累次积分转换成直角坐标系中的累次积分或相反:.37.?20d?cos?0f(rcos?,rsin?)rdr?(10).(B)(D)10(1)0R2e?y2dy2x?1y0e?x2dx?RR2e?y2dyR2?y20e?xdx;2(A)(C)dy010y?y2f(x,y)dx;dy1?y20x?00x2f(x,y)dx;f(x,y)dy.1?x2(2)I?2131dxxxeydy;?xdy?2y422x210dyf(x,y)dy;10dy(3)).dxsindxsinx?xdy.2yu07B设f(x,y)是连续函数,则二次积分dxf(x,y)dy?(x?1x?1(A)1dyy?1f(x,y)dx?22y)dx;0?11dyy?1?1f(x,(B)1y?10dy?1f(x,y)dx;(C)12y2?10dyy?1?1f(x,y)dx?1dy??1f(x,y)dx;(D)2y2?10dy??1f(x,y)dx.07C交换下列二次积分的次序:(1)41dy2(y?4)f(x,y)dx.0?4?y(2)1dy2yf(x,y)dx?3dy3?yf(x,y)dx.0010(3)1dx1?1?x2f(x,y)dy.0x?(4)2d?asin2?f(r,?)dr.00(5)I?0?2dx4?x2f(x,y)dy?2?x22?xdx402?xf(x,y)dy22(6)1dy1?y2f(x,y)dx.01?y(7)1dx4?x2f(x,y)dy?3dx4?x2f(x,y)dy.01?1?x21107D计算下列累次积分:.38.07E设f(x,y)在D上连续,证明:duf(t)dt?(x?u)f(u)du.00007Fadyyem(a?x)f(x)dx?a证明(a?x)em(a?x)f(x)dx.00008A由曲线x2?y2?2x,x2?y2?4x,y?x,y?0所围成的图形的面积S?().(A)14(2??);(B)12(2??);(C)34(2??);(D)2??.08B计算下列二重积分:(1)I?xdxdy,其中D?{(x,y)|y2?2x,0?x?2};D1?x2?y2(2)I?(1?x)1?cos2ydxdy,其中D是y?x?3,y?x?5,?D22y?2,y???2围成.(3)I???ydxdy,其中D是由曲线xxa?b?1及x轴,y轴所围成的闭区D域.08C计算下列二重积分:(1)??yd?,其中D是由曲线r?2(1?cos?)的上半部分与极轴所围成的区域;D(2)??x2?y2dxdy,其中D由y?x与y?x4围成;D(3)??R2?x2?y2dxdy,其中D:x2?y2?Rx;D.39.(4)??(x?y)dxdy.x2?y2?x?y08D设D?{(x,y)|x2?y2?x},求??xdxdy.D09A计算下列二重积分:(1)0???|cos(x?y)|dxdy;x??0?y??(2)??|sin(x?y)|d?,其中D:0?x?y?2?;D(3)??|x?y|?2dxdy,其中D:0?x?2,?2?y?2;D(4)??|xy|dxdy;|x|?|y|?11(5)??(|x|?|y|)2d?,其中D:0?x?2,|y|?1;D(6)??sinxsiny?max{x,y}dxdy,其中D:0?x??,0?y??.D09B设f(x,y)?{1,0?x?1,0?y?10,其它,D是由x?0,y?0及x?y?t所围区域.计算F(t)???f(x,y)dxdy.D09C计算I???(|x|?|y|)dxdy,其中D:|x|?|y|?1.D10A设f(t)在[1,??)上有连续的二阶导数,f(1)?0,f?(1)?1,且二元函数z?(x2?y2)f(x2?y2)满足?2z?x2??2z?y2?0,求f(t)在[1,??)上的最大值.10B证明:若函数u?u(x,y),满足拉普拉斯方程?2u?2?x2?u?y2?0,则函数v?u(xyx2?y2,x2?y2)也满足上述拉普拉斯方程..40.10C设u?f(r),r?lnx2?y2?z2满足方程?2u?2u?2?u22?x2?y2??z2?(x?y2?z)?3/2,求f(x).10D设p(x),f(x),g(x)是[a,b]上的连续函数,且在[a,b]上,p(x)?0,f(x),g(x)为单调递增,试证:bap(x)f(x)dxbap(x)g(x)dx?bap(x)dxbap(x)f(x)g(x)dx.10E试证:抛物面z?1?x2?y2上任意点处的切平面与抛物面z?x2?y2所围成立体的体积是一定值.10F设函数f(x)连续,f(0)?1,令F(t)???f(x2?y2)dxdy(t?0),求F??(0).x2?y2?t210G设f(x,y)在区域D:0?x?1,0?y?1上有定义f(0,0)?0,且在(0,0)x2dtt处f(x,y)可微,求0xf(t,u)duxlim?0?4.1?e?x410H记D(R)?{(x,y)|x2?y2?R2},求Rlim?????e?(x2?y2)dxdy.D(R)10I证明：??e?x2dx??.??.41.
