??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?22?22（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?2?22图中所求的旋转体的体积为：
?z2?4? V?????8?dz
0?2?2???2z????4?0?z2???2?8?dz ?2??z2dz?32?2 2?Bz?4?4?y032?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)? ?22222222(x?y?z)(x?y?z)z(x2?y2?z2)?2y(x2?yz)zx2?z3?2yx2?y2z fy?(x,y,z)? ?22222222(x?y?z)(x?y?z)y(x2?y2?z2)?2z(x2?yz)yx2?y3?2zx2?z2y fz?(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)
x2?yz1x2?yz对f(0,y,y)?2?, f(x,0,0)?2?1, 2222x?y?z2x?y?z 16
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: (0,22222222,),(0,,?),(0,?,?),(0,?,),(1,0,0),(?1,0,0) 22222222221,)?222,
f(0,f(0,221,?)??222,
f(0,?221,?)?222,
f(0,?221,)??,f(1,0,0)?1,f(?1,0,0)?1; 222在圆周x2?y2?z2?4上,由条件极值得: 令F(x,y,z)?x?yz??(x?y?z?4)
2222Fx?(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?4?0
解得: (0,2,2),(0,2,?2),(0,?2,?2),(0,?2,2) ,(2,0,0),(?2,0,0)
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D?{(x,y,z)1?x?y?z?4}的最大值为1,最小值f(x,y,z)?222x?y?z为?1. 217
五、(15分)设幂级数?axn?n的系数满足a0?2,nan?an?1?n?1,n?1,2,3,?,求此幂级
n?0数的和函数.
??证明：S(x)???ann?1?1nx?S?(x)??n?1?
n?0?nanxn?1?an?1xn?1?(n?1)xn?n?1?n?? ??annx??S(x)?n?0?nxn?0?nxn
n?0???而
?nxn?xn?1?x??xn???x??n???1??xn?0?nxn?0n?0???xn?0???x??1?x???(1?x)2, 即: S?(x)?S(x)?x(1?x)2 一阶非齐次线性微分方程---常数变易法, 求S?(x)?S(x)?0的通解: S(x)?cex, 令S(x)?c(x)ex代入S?(x)?S(x)?x(1?x)2得: c(x)ex?c?(x)ex?c(x)ex?x(1?x)2,
即: c(x)??x?1???xxe?x1?x?(1?x)2exdx????1?x???xedx?1?x??1?x?xe?dx ?xe?x1?x????e?x?dx?xe?x1?x?e?x?c 故S?(x)?S(x)?x(1?x)2的通解为: S(x)???xe?x?x?x1?1?x?e?c???e?1?x?cex, ?由于S(0)?0,解得c??1, 故
?anxn的和函数S(x)?1n?1?x?ex. 0??1???1?x?????x?? ?1?x???1?1?x?2 法二：naan?1n?an?1?n?1?a?1，同学们自行完成。 n?1?1n
18
六、(15分)已知f(x)二阶可导,且f(x)?0,f??(x)f(x)??f?(x)?2?0,x?R, (3) 证明:f(x?x1?x2?1)f(x2)?f2??2??,?x1,x2?R.
(4) 若f(0)?1,证明f(x)?ef?(0)x,x?R.
证明: (1) 要证明f(x?x1?x2?1)f(x2)?f2??2??,?x1,x2?R,
只需证明
112lnf(x1)?2lnf(x2)?lnf??1?2x?12x?12??,?x1,x2?R,
也即说明F(x)?lnf(x)是凹函数,
?lnf(x)???f?(x)f(x), ?lnf(x)??????f?(x)??f(x)f??(x)??f?(x)?2?f(x)???f2(x)?0, 故F(x)?lnf(x)是凹函数, 即证. (2) F(x)?F(0)?F?(0)x?F??(?)22x ?lnf(0)?f?(0)f(x)f??(x)??f?(x)?2f(0)x?2f2(x)x2?f?(0)x,
x??即: f(x)?ef?(0)x,x?R.
2009年浙江省高等数学（微积分）竞赛试题
一、计算题（每小题12分，满分60分）
1.求极限
解 =====
19
2．计算不定积分
解 ==
3．设，求解
=
4．设，，求此曲线的拐点
解 ，
，
令得
20