参考答案
一一一、、、填填填空空空题题题:(每题S分,共RQ分)
Q、曲面 z = 2x
y
上点(1, 0, 2)处的切平面方程为 .
解:z
x
(1, 0) = 0,z
y
(1, 0) = 0,故切平面方程为z − 2 = 0N
R、 lim
x→0
y→0
20x
3
+ 17y
3
x
2
+ y
2
= .
解:lim
x→0
y→0
20x
3
+ 17y
3
x
2
+ y
2
= lim
ρ→0
+
ρ · (20 cos
3
θ + 17 sin
3
θ) = 0.
S、设 Ω为x
2
+ y
2
+ z
2
6 1,则
˝
Ω
(x
2
+ y
2
+ z
2
)dxdydz = .
解:原式=
´
2π
0
dθ
´
π
0
dϕ
´
1
0
r
2
· r
2
sin ϕdr = 2π · 2 ·
1
5
=
4
5
π.
T、设L是y = x
2
−1上从(0, −1)到(2, 3)的有向曲线,则
´
L
ydx + xdy = N
解:曲线积分与路径无关,选择折线l :(0, −1) −→ (2, −1) −→(2, 3),
´
l
ydx + xdy = −
´
2
0
dx +
´
3
−1
2dy = −2 + 8 = 6N
另解:直接代入曲线方程,
´
L
ydx + xdy =
´
2
0
(x
2
− 1 + x · 2x)dx = 6.
U、设区域D是由y = x
2
与y = x围成的,则
˜
D
xydxdy=
.
解:
˜
D
xydxy =
´
1
0
dx
´
x
x
2
xydy =
´
1
0
1
2
(x
3
− x
5
)dx =
1
24
.
V、设曲线L的方程为x
2
+ y
2
= 1,则
¸
L
(x
2
+ 7y
2
)ds= .
解:由曲线L的对称性
¸
L
x
2
ds =
¸
L
y
2
ds,
∴
¸
L
(x
2
+ 7y
2
)ds = 4
¸
L
(x
2
+ y
2
)ds = 4
¸
L
ds = 8π.
W、微分方程 xy
0
+ y = x
2
满足y(3) = 4的特解为 .
解:∵ (xy)
0
= x
2
,xy =
1
3
x
3
+ C,由y(3) = 4可得:
12 = 9 + C,于是C = 3,∴ y =
1
3
x
2
+
3
x
.
Q
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