高数2008 2009试题第二学期期终参高等数学

A和平面S平行，并且和AB垂直的直线方程.6x2yz0.(2)ABsAB(−130),从而所求直线的方s=sABn=(3120).因此由点向式得直线方程为x−1=y2=x−2.xy x2+y2,

（12分）f(xy

x2+y2=

f(xy)(00)的连续性解(1)limf(x,y)=limxy= k.由于极限与k的取值有关,因此极限不存在

x→0x2

x→0x2(1+k2

f(xy(00)处不连续(2)f0(00)=limf(x,0)−f(0,0)=limx2x·20−0 yy

x)x)(∆x)2+

(∆x)2+(∆y)2=

= ∆x

(∆x)2+(k∆x)2 ∆x→0(1+k0∞k00.从而其极限不存在(00)处不可微y1zxyarctanxy

=y ,=y ,

=x+x2+x,dz=y1−x dx+x1+x1y 2+x 2+x

=f0+f0+yf0,z0

=f00−f00+xf00+f00−f00+xf00+f0+yf00−yf00+xyf00

x 1 2

1 2 2

2 3 3 3f00+(x+y)f00−f00+(x−y)f00+xyf00+f1 1 2 2 3 3r2dxrx2f(xy)dy的积分次序 r r r r r r解1dx2−xf(xy)dy=0dy2−yf(xy)dx+1dy√yf(xD

=(x,y)|0≤x≤1,x2≤y≤1,D

,y)|0≤x≤1,0≤y

D

x |y−x2|dxdy

(y−x2)dxdy

(−y+w w

w w

(−y+xw011

+x4−x2dx=0 5、求(xy)2dxdy,D=(xy|x2y21x0.D!Dx轴对称D

xydxdy= (x+y)2dxdy

(x2+y2)dxdy

w w2

r3dr=–2 –2 6zx2y2z=xy所围图形的体积 而所求体积是在区域D=(x,y) x−12+y−12≤1

V (x+y)−(x2+y2)Dx=1+rcos2y=1+rsin2√2此时，dσrdrdθD1=(rθ)0θ2π0r≤2.2"V

..

11

rdr= λ22λ10,1为二重根y(C1C2 8f(xy=x22y2x2y2D=(xy|x2y24y0上的最大值解(1)先讨论区域内可能的的最值点.f0(xy)=2x(1−y2)=0,f0(xy)=4y1−1x2= 2,1).且f 2,1)=2.(2)再讨论边界上可能的最值点.y=0−2≤x≤2时,f(xy)=x2,(±20)4.−2≤x≤2y2=4x2时,f(xy)=x2−25+7(02)8.综合 （12分）∞1xn的收敛域与和函数∞1的值n=1

解级数的收敛半径为R=lim(n+1)2n+1=2.当x=2时，级数为 12n=∞1为调和级数 n=1 n=1发散x=−2∞1(−2)n=∞(−1)n1为交错级数并由莱布尼兹定理知其收敛n=1 故级数的收敛域为[−22).令S(x)=∞1xn,则S(0)=0.S0(x)= xn0=1∞xn−1=,

n=11.S(x)=S(x)−S(0)=rxS0(x)dx=rx1dx=ln2.∞1=S(1)=ln 0 n=1（8分）yy(x(−ππ√π√πx0, (1)x01=y.x2y2C2. 1√π,√πC2π2.(2)x0y00yx的特征方程为λ210,1 λ=±i.y∗=x.