06至10年浙江省高等数学(竞赛 工科类试题)(2)

解:

?a0dx?e022bmax{b2x2,a2y2}dy???eD22max{b2x2,a2y2}d?, 其中D如右图

???emax{bxD122,a2y2}d????emax{bxD222,a2y2}d? by????eayd????ebxd?

D1D2D1bxa0bxa??dy?0bayb0ea2y2dx??dx?ebxdy

0a22D2aba2y2bab2x2yedy?xedx b?0a?01ba2y21ab2x222?ed(ay)?ed(b2x2) ??2ab02ab01a2b2?(e?1). ab25、计算??(x?y)dS,其中S为圆柱面x2?y2?4,(0?z?1). ?Saz解:

2(x???y)dS?S122(x?y)dS???ydS ??2SS ?14dS???ydS 2??SS22??x???x??8????y1??????dydz

??y???z?Dyzy2?8????y1??02dydz?8? 24?yDyz被积函数关于y是奇函数,积分区域关于z对称,

二、(20分)设un?1?oyx12112112?????????, 234563n?23n?13nvn?111u????,求: (1)10;(2) limun.

n??n?1n?23nv10解: (1)un?12??1????? 3k?23k?13k?k?1?n12112112?1??????????,

234563n?23n?13n 6

3n11n1 vn??????

k?1n?kk?1kk?1k2n ??1???111111111??11????????????????1????? 23456n3n?23n?13n??2n?n12?3n1n1?1un?vn??????????

3k?13k?k?1kk?1kk?1?3k?2?21?n1 ?????????0

3k?k?1kk?1?3k?un?1; vv11??1?????

n??n?1n?23n??n(2) limun?limvn?lim?n??n????1?1111??lim?????? (图来说明积分上下) n??n12k2n??1?1?1??1??nnn??n12n1?lim? n??nkk?11?n??1dx?ln3. 01?x2三、(满分20分)有一张边长为4?的正方形纸(如图),C、D分别为AA?、BB?的中点，E为DB?的中点，现将纸卷成圆柱形，使A与A?重合，B与B?重合，并将圆柱垂直放在xOy平面上，且B与原点O重合，D若在y轴正向上，求：

（1） 通过C，E两点的直线绕z轴旋转所得的旋转曲面方程；

（2） 此旋转曲面、xOy平面和过A点垂直于z轴的平面所围成的立体体积. 解:

C(0,4,4?)2ACA?zA

MyDQN4?7 E(2,2,0)BDEB?x?2y?2z?? LCE :2?2?4?旋转曲面上任意取一点M(x,y,z)

x?z?x??20?2??z??2 ， Q(0,0,z) 则N(x0,y0,z0)的坐标为：?y0?2???z0?z????z??z?MQ?x?y?NQ???2????2?

?2???2??2222z2?8， 化简得：所求的旋转曲面方程为：x?y?2?222（2）A(0,0,4?)，故过A(0,0,4?)垂直z轴的平面方程为：z?4?

z2?8， 令x?0，解得在坐标面yoz上的曲线方程为：y?22?2图中所求的旋转体的体积为： V??4?0?z????8?dz ?2?2???22z?z?????2?8?dz 0?2??4?2z?4???4?0zdz?32?2 2?2yB32?2128?22??32??.

33xz2x?y?2?82?22x2?yz222四、(20分) 求函数f(x,y,z)?2,在D?{(x,y,z)1?x?y?z?4}的最大22x?y?z值、最小值.

2x(x2?y2?z2)?2x(x2?yz)2xy2?2xz2?2xyz解： fx?(x,y,z)? ?(x2?y2?z2)2(x2?y2?z2)2z(x2?y2?z2)?2y(x2?yz)zx2?z3?2yx2?y2z fy?(x,y,z)? ?22222222(x?y?z)(x?y?z)

8

fy(x2?y2?z2)?2z(x2?yz)yx2?y3?2zx2?z2yz?(x,y,z)?(x2?y2?z2)2?(x2?y2?z2)2 由于x,y具有轮换对称性,令x?y, x?0或y?z?0 解得驻点: (0,y,y)或(x,0,0)

对f(0,y,y)?x2?yz1x2?yzx2?y2?z2?2, f(x,0,0)?x2?y2?z2?1,

在圆周x2?y2?z2?1上,由条件极值得: 令F(x,y,z)?x2?yz??(x2?y2?z2?1)

Fx?(x,y,z)?2x?2?x?0

Fy?(x,y,z)?z?2?y?0 Fz?(x,y,z)?y?2?z?0

F??(x,y,z)?x2?y2?z2?1?0

解得: (0,22,22),(0,22,?22),(0,?22,?22),(0,?22,22),(1,0,0),(?1,0,0) f(0,22,212)?2,

f(0,2212,?2)??2,

f(0,?22,?212)?2f(0,?22,22)??12,f(1,0,0)?1,f(?1,0,0)?1; 在圆周x2?y2?z2?4上,由条件极值得: 令F(x,y,z)?x2?yz??(x2?y2?z2?4)

Fx?(x,y,z)?2x?2?x?0

Fy?(x,y,z)?z?2?y?0 Fz?(x,y,z)?y?2?z?0

F2??(x,y,z)?x2?y2?z?4?0

解得: (0,2,2),(0,2,?2),(0,?2,?2),(0,?2,2) ,(2,0,0),(?2,0,0)

,

9

111,f(0,2,?2)??,f(0,?2,?2)?, 2221f(0,?2,2)??,f(2,0,0)?1,f(?2,0,0)?1;

2f(0,2,2)?x2?yz222,在f(x,y,z)?2D?{(x,y,z)1?x?y?z?4}的最大值为1,最小值22x?y?z为?1. 2五、(15分)设幂级数数的和函数.

?axnn?0??n的系数满足a0?2,nan?an?1?n?1,n?1,2,3,?,求此幂级

证明：S(x)??axnn?0n?S?(x)??nanxn?1?n??n?1??an?1xn?1?n?1??(n?1)xn?1

n?1? ??ax??nxnn?0n?0n?S(x)??nxn

n?0?而

?nxn?0?n?x?nxn?0?n?1x??n???1????x??x??x??x??x??, ?21?x(1?x)??n?0?n?0??n即: S?(x)?S(x)?x 一阶非齐次线性微分方程---常数变易法,

(1?x)2x求S?(x)?S(x)?0的通解: S(x)?ce, 令S(x)?c(x)ex代入S?(x)?S(x)?x得: 2(1?x)c(x)ex?c?(x)ex?c(x)ex?x,

(1?x)2xxe?x1?1???x?x?dx??xedx??xedx 即: c(x)??????2x??(1?x)e1?x1?x?1?x?xe?xxe?x?x?????e?dx??e?x?c 1?x1?x?xe?x?xx1?x故S?(x)?S(x)?的通解为: S(x)??e?c?e??cex, ??2(1?x)1?x?1?x?由于S(0)?0,解得c??1, 故

?anxn的和函数S(x)?n?0?1?ex. 1?x10

百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，免费范文网，提供经典小说综合文库06至10年浙江省高等数学(竞赛 工科类试题)(2)在线全文阅读。