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哈爾濱工程大學(xué)本科生考試一試卷（2013年春天學(xué)期）2013-7-15課程編號(hào)：0911002課程名稱：微積分A(二)(A卷)題號(hào)一二三四五六總分分?jǐn)?shù)評(píng)卷人

得分評(píng)卷人二、單項(xiàng)選擇題（每題3分，共15分）x2y2sin2xyy2,(x,y)(0,0).1.二元函數(shù)f(x,y)x在點(diǎn)(0,0)處0,(x,y)(0,0)(A)不連續(xù)(B)偏導(dǎo)數(shù)不存在(C)偏導(dǎo)數(shù)連續(xù)(D)不能夠微2224y名姓得分評(píng)卷人一、填空題（每題3分，共15分）裝x2y2z21u1.設(shè)函數(shù)u(x,y,z)161218,單位向量n3{1,1,1},則n|(1,2,3).訂2.方程x2y2z23xyz所確立的函數(shù)zz(x,y)在點(diǎn)(1,1,1)處的全微分：線dz.號(hào)學(xué)3.設(shè)曲面是錐面zx2y2(z1),則曲面積分z2dS.4.方程xyyx2ex知足條件y|x1e的特解是.5.已知函數(shù)yC1exC2e2x為某二階線性常系數(shù)齊次微分方程的通解，則此方程：級(jí)為.班

2.設(shè)f(x,y)連續(xù),則dxf(x,y)dydyf(x,y)dx.1x1y(A)24x24xdx1f(x,y)dy(B)dxf(x,y)dy11x(C)24y22f(x,y)dxdy1f(x,y)dx(D)dy11y3.設(shè)L:x2y21,則曲線積分(xyx2y2)ds.L(A)2(B)(C)0(D)4.設(shè)an0(n1,2,),且an收斂,則級(jí)數(shù)(1)nnsin1a2n.n1n1n(A)絕對(duì)收斂(B)條件收斂(C)發(fā)散(D)斂散性沒法確立5.函數(shù)yln(x23x2)展成x的冪級(jí)數(shù)的收斂域是.(A)(1,1)(B)(1,1](C)(2,2)(D)(2,2]第1頁共8頁第2頁共8頁得分評(píng)卷人三、計(jì)算題I（每題8分，共32分）1.設(shè)zf[xy,1(x2y2)],此中f擁有二階連續(xù)偏導(dǎo)數(shù),求2z.2xy2.計(jì)算二重積分(xy)2dxdy,此中D:xx2y22x.D

3.計(jì)算曲線積分(ey2sinx)dx(xy22xyey2)dy,此中L為沿拋物線Ly2x1從點(diǎn)A(0,1)到點(diǎn)B(1,0)的一段.裝4.求微分方程yysinx的通解.訂線第3頁共12頁第4頁共12頁得分評(píng)卷人3.求冪級(jí)數(shù)xn1的收斂域及和函數(shù).四、計(jì)算題II（每題8分，共24分）n1n(n1)1.計(jì)算x3dydzy3dxdz(z31)dxdy,此中為曲面z1x2y2上側(cè).：名姓裝訂2.設(shè)函數(shù)f(x)1xx(1,1].（1）將f(x)張開為周期為2的傅里葉級(jí)數(shù)；線（2）設(shè)（1）中的傅里葉級(jí)數(shù)的和函數(shù)為S(x),計(jì)算S(2013)的值.：號(hào)學(xué)：級(jí)班第5頁共8頁第6頁共8頁得分評(píng)卷人五、應(yīng)用題（8分）求曲線C:x2y22z20上點(diǎn)到xOy面的最短距離.xy3z5

得分評(píng)卷人六、證明題（6分）已知f(u)導(dǎo)函數(shù)連續(xù),證明曲面zyf(x)的全部切平面過必定點(diǎn).y裝訂線第7頁共12頁第8頁共12頁微積分A（二）（A卷）參照答案及評(píng)分標(biāo)準(zhǔn)

y22yey2......3分(12ye)dxdyBOOAD2013年7月15日一、填空題（每題3分，共15分）1.1;32.dxdy;3.2.2：4.yxex;名5.yy2y0;姓二、單項(xiàng)選擇題（每題3分，共15分）1.D2.C3.A4.A5.B三、計(jì)算題I（每題8分，共32分）裝1.zyf1xf2...................................................................................4分解答：x2zf1y(xf11yf12)x(xf21yf22)............................................4分訂xy2.解答：（方法1）原式(x2y2+2xy)dxdy(x2y2)dxdy.................2分DD：線2cosr3dr...........................................................4分2d號(hào)2cos學(xué)152cos4d45.................................................2分2016（方法2）原式2cosrsin)2rdr2d2cos2sincos)r3dr2d(rcos(1cos2cos2

0x1dy或10dx0sinx)dx1..........................4分dx0dy21(1y2dy10y101分1cos13（方法2）原式(ey2sinx)dx(y22xyey2)dyxdyLL設(shè)Pey2sinx，Qy22xyey2，QP，故(ey2sinx)dx(y22xyey2)dyxyL與路徑?jīng)]關(guān)，取折線AO和OB作為積分路徑.................................................3分原式01(1sinx)dx01)dy............................................4分y2dy(y21011分1cos13（方法3）原式10(y21y2)dy1x1ex11dx(ex1sinx)dx2x0102x1....................................4分ex1111(xex1ex1)11cos1...................................4分cosx003034.解答：特點(diǎn)方程為r210，解得r1,2i，原方程對(duì)應(yīng)齊次方程的通解為：YC1cosxC2sinx.......................................4分（方法1）設(shè)原方程的特解為y*x(AcosxBsinx)..................................2分代入原方程得A1,B0，因此原方程的特解為y*1xcosx.....1分22原方程的通解為yYy*C1cosxC2sinx1xcosx....................1分2（方法2）設(shè)原方程的特解為y*Axcosx2分152(cos442152cos4d20

....................................................4分代入原方程得A1，因此原方程的特解為y*1xcosx................1分5.....................................................2分222sincos)d1xcosx45.....................................................................原方程的通解為yYy*C1cosxC2sinx....................1分2分四、計(jì)算題II（每題8分，共24分）232：223.解答：（方法1）補(bǔ)線BO和OA，D:1x0,0yx1，則1.解答：補(bǔ)面:z0,xy1，方向向下。..........................................2分級(jí)L(ey2sinx)dx(xy22xyey2設(shè)為由曲面和圍成的閉地區(qū)，D:x2y21，則有班)dy第9頁共8頁第10頁共8頁3(x2y2z2)dv1dxdy22d2d1dr1dxdy.............................................33r4sin000D11152.1f(x)2F(x)a01x)dx21(11an1x)cosnxdx0....................................................................1(11bn1x)sinnxdx21xdcosnx(1n012xcosnx2cosnxdx2(1)n......................................211n0n0nF(x)x1f(x)12(1)n1x1)....................................2nsinnx(k12S(2013)S(1)F(10)F(10)021..............................................2223.Rlim(n1)(n2)1nxn1x1nn(n1)1n(n1)[1,1]..21S(x)xnxn11S(x)S(x)1....2n1nn1xS(x)(0,x)S(x)ln(1x)(1x1).....2S(x)(0,x)xn1(1x)ln(1x)x,1x1,S(x)1)1,x..................2n1n(n1.2S(x)xnln(1x)(1x1)..........4S(x)n1n

S(x)(0,x)S(x)xn1(1x)ln(1x)x,1x1,1)1,x................2n1n(n1.8x2y22z20zz2xy3z5...2L(x,y,z,,)z2(x2y22z2)(xy3z5)L2x0xL2y0yL2z4z30........................................4zLx2y22z20Lxy3z50(1,1,1)(5,5,5)(1,1,1)CxOy1.......................................................26F(x,y,z)yf(x)zzyf(x)yyn{Fx,Fy,Fz}{f,fx,1}..................