高數(shù)2008 2009試題第二學(xué)期期終參高等數(shù)學(xué)

A和平面S平行，并且和AB垂直的直線方程.6x2yz0.(2)ABsAB(?130),從而所求直線的方s=sABn=(3120).因此由點(diǎn)向式得直線方程為x?1=y2=x?2.xy x2+y2,

（12分）f(xy

x2+y2=

f(xy)(00)的連續(xù)性解(1)limf(x,y)=limxy= k.由于極限與k的取值有關(guān),因此極限不存在

x→0x2

x→0x2(1+k2

f(xy(00)處不連續(xù)(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.從而其極限不存在(00)處不可微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的積分次序 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軸對(duì)稱D

xydxdy= (x+y)2dxdy

(x2+y2)dxdy

w w2

r3dr=–2 –2 6zx2y2z=xy所圍圖形的體積 而所求體積是在區(qū)域D=(x,y) x?12+y?12≤1

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

..

11

rdr= λ22λ10,1為二重根y(C1C2 8f(xy=x22y2x2y2D=(xy|x2y24y0上的最大值解(1)先討論區(qū)域內(nèi)可能的的最值點(diǎn).f0(xy)=2x(1?y2)=0,f0(xy)=4y1?1x2= 2,1).且f 2,1)=2.(2)再討論邊界上可能的最值點(diǎn).y=0?2≤x≤2時(shí),f(xy)=x2,(±20)4.?2≤x≤2y2=4x2時(shí),f(xy)=x2?25+7(02)8.綜合 （12分）∞1xn的收斂域與和函數(shù)∞1的值n=1

解級(jí)數(shù)的收斂半徑為R=lim(n+1)2n+1=2.當(dāng)x=2時(shí)，級(jí)數(shù)為 12n=∞1為調(diào)和級(jí)數(shù) n=1 n=1發(fā)散x=?2∞1(?2)n=∞(?1)n1為交錯(cuò)級(jí)數(shù)并由萊布尼茲定理知其收斂n=1 故級(jí)數(shù)的收斂域?yàn)閇?22).令S(x)=∞1xn,則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的特征方程為λ210,1 λ=±i.y?=x.