北大版高数第七章习题解答

à ¼ ùÒ ¿ f(x,y)g(x,y)dσ
.
m ≤ D g(x,y)dσ ≤ M .
, (x0, y0) ∈ D
D
f (x,y)g(x,y)dσ
f (x0, y0) = D g(x,y)dσ ,
D
ø ¿ º 4.
f (x, y)
f (x, y)g(x, y)dσ = f (x0, y0) g(x, y)dσ.
=
−
2v u
.
I=
2 1
du
3 1
(u
+
v)
2v u
dv
=
8
+
52 3
ln 2.
23. ydxdy,
Ǒ« D
x2 + y2 ≤ x + y.
D
µ Ã 1.
x
=
1 2
+
r cos θ,
y
=
1 2
+ r sin θ.
D(x,y) D(u,v)
=
r.
µ Ã I =
√1
0 2 dr
2π 0
(
1 2
+
r
sin θ)rdθ
= 2π ·
4 3
·
b5 −a5 5
=
8π 15
(b5
−
a5).
8. (x2 + z2)dV , Ω : x2 + y2 ≤ z ≤ 1.
Ω
I=
2π 0
dθ
1 0
rdr
1 r2
dz(r2 cos2 θ
+ z2)
=
π 12
+
π 4
=
π 3
.
9. z2dV , Ω : x2 + y2 + z2 ≤ R2, x2 + y2 ≤ Rx.
=
4π 15
.
Ω
Ω
3
2.
x2y2zdV ,
ðΩ 2z = x2 + y2, z = 2
.
Ω
I=
2π 0
dθ
2 0
rdr
2
r2
dz(r cos θ)2(r sin θ)2z
=
2
2π 0
(sin2
θ
cos2
θ)dθ
·
2 0
r5(2
−
r4 8
)dr
=
π 4
·
128 15
=
32π 15
.
3.
x2 sin xdxdydz,
.
I=
arctan β arctan α
dθ
b a
(r
1 cos
θ)2
rdr
=
(β
−
α)
ln
b a
.
18. rdσ,
ð ± « D
r = a(1 + cos θ)
r = a (a > 0)
í
õD
.
I=
π
2
−
π 2
dθ
a(1+cos θ) a
rrdr
=
ûþÝ 19.
Ä (α ≤ θ ≤ β)
D
π
2
−
π 2
x
+
x
5 2
−
3 2
x4
)
=
33 140
.
8.
π 0
dx
π x
sin y
y
dy
=
π 0
dy
y 0
dx
sin y
y
=
π 0
dy
sin
y
=
2.
9.
2 0
dx
2 x
2y
sin(xy)dy
=
2 0
dy
2 0
dx2y
sin(xy)
=
2 0
dy2(1−cos
2y)
=
4−sin
4.
1
10. y2√1 − x2dxdy, D = {(x, y) | x2 + y2 ≤ 1}.
x=
u+v 3
,
y
=
u−2v 3
,
D(x,y) D(u,v)
=
−
1 3
.
I=
7 4
du
−21(uv)
1 3
dv
=
33 4
.
22.
(
y x
+
√xy)dxdy,
D xy = 1, xy = 9, y = x y = 4x .
D
µ Ã .
u=
y x
,
v
=
√xy.
x=
v u
,
y
= uv,
D(x,y) D(u,v)
=
√π 2
.
Ý. (1) [I(a)]2 =
a 0
e−x2
dx
a 0
e−y2 dy
=
e−x2−y2 dxdy.
Ra
(2) Da ⊂ Ra ⊂ D√2a,
e−x2−y2 dxdy ≤ e−x2−y2 dxdy ≤
e−x2−y2 dxdy.
Da
Ra
D√2a
(3) J(a) =
π
2
0
dθ
a 0
e−r2
rdr
Ω
4
I=
2π 0
dθ
π
2
0
dϕ
cos 0
ϕ
r2
sin
ϕdrr
=
π 10
.
13. z2dV , Ω : 3(x2 + y2) ≤ z ≤ 1 − x2 − y2.
Ω
I=
2π 0
dθ
π
6
0
dϕ
1 0
r2
sin
ϕdrr2
cos2
ϕ
=
2π 15
(1
−
√
3 8 3 ).
14.
√ zdV , Ω x2 + y2 + z2 = 2az .
D
D
D
, êÃ,
Ý f (x, y)dxdy = 0.
¨ î f (x, y) = 0, (x, y) ∈ D .
D
Ý¿.P èǑfêÃ,Ëff
Ǒ à ¿¸ Ù Õ , f
1
Ù Ùð 2
f
(P
).
P ∈D
0.
f
f (x, y)dxdy > 0, .
,
D
ËÃ7.2
ǯ .
3. ydxdy, D y = 0 y = sin x (0 ≤ x ≤ π) .
π
2
0
dθ
1 0
r2rdr
=
π 8
.
14.
0 −1
dx
0√ − 1−x2
√2 dy
1+ x2+y2
=
3π
2
π
dθ
1 0
2 1+r
rdr
=
π(1
−
ln
2).
√
15.
2 0
dx
0
1−(x−1)2 3xydy =
π
2
0
dθ
2 0
cos θ
3r
cos
θr
sin
θrdr
=
π 2
0
dθ12
cos5
θ
sin
θ
=
2.
I=
1 0
dy
y2 0
x
dxe y
=
1 0
dyyey
=
1.
6.
1 0
dy
1
1
y3
√ 1
−
x4dx
=
1 0
dx
x3 0
√ dy 1
−
x4
=
1 0
dxx3
√ 1
−
x4
=
1 6
.
7. (x2 + y)dxdy, D y = x2, x = y2 .
D
√
I=
1 0
dx
x x2
dy(x2
+
y)
=
1 0
dx(
1 2
.
µ. S =
2π 0
dθ
a(1+cos θ) 0
rdr
=
2π 0
dθ
1 2
a2
(1
+
cos
θ)2
=
3 2
πa2
.
ǯ .
21. (2x2 − xy − y2)dxdy,
D
y=x+1 .
D y = −2x + 4, y = −2x + 7, y = x − 2,
2
µ Ã .
u = 2x + y, v = x − y.
√
16.
R 0
dx
0 R2−x2 ln(1+x2+y2)dy =
π
2
0
dθ
R 0
ln(1+r2)rdr
=
π 4
[(1+R2)
ln(1+
R2) − R2].
17.
ð 1
x2
dxdy
,
D
y
=
αx,
y
=
βx
(
π 2
>
β
>
α
>
0),
x2 + y2
=
a2,
D
x2 + y2 = b2 (b > a > 0)
¿½èæ
=
2π
1 0
drr2
|r+2|−|r−2| r
=
2 3
π.
17. (x3 + sin y + z)dV , Ω x2 + y2 + z2 ≤ 2az, x2 + y2 ≤ z .
rdr
√ 8−r2
r2
dz
=
√
2π 16
2−14 3
.
2
11. (x2 + y2)dV , Ω z = R2 − x2 − y2 z = x2 + y2 .
Ω
I=
2π 0
dθ
π
4
0
dϕ
R 0
r2
sin
ϕdrr2
sin2
ϕ
=
2π(
2 15
−
√
ຫໍສະໝຸດ Baidu2 12
)R5.
12.
x2 + y2 + z2dV , Ω z = x2 + y2 + z2 = z .
D
I=
π 0
dx
sin x 0
dyy
=
π 0
dx
sin2 2
x
=
π 4
.
4. xy2dxdy, D x = 1, y2 = 4x .
D
I=
2 −2
dy
1 y2/4
dxxy2
=
2 −2
dy
1 2
(1
−
y4 16
)y
2
=
32 21
.
5.
e
x y
dxdy,
D y2 = x, x = 0, y = 1 .
D
dθa3(cos θ
+
cos2
θ
+
1 3
cos3
θ)
=
(
22 9
+
π 2
)a3.
:
θ = α, r = β
1 2
β α
[r(θ)2
]dθ.
r = r(θ)
Ý. S =
dxdy =
β α
dθ
r(θ) 0
rdr
=
1 2
β α
[r(θ)2
]dθ.
D
± 20.
r = a(1 + cos θ) (a > 0, 0 ≤ θ < 2π)
Ô ½ Ëõ
ËÃ7.1
ø ¿ º ¿ êà ¿ 3.
f (x, y)
Ý ¿ ¿è . : D
D
, g(x, y) D
, g(x, y) f (x, y)g(x, y) D
(x0, y0) f (x, y)g(x, y)dσ = f (x0, y0) g(x, y)dσ.
Ý. Ǒ ¿ m, M f D ¢,
3(1
− r)r3dr
=
3π 10
.
7. (y2 + z2)dV , Ω : 0 ≤ a2 ≤ x2 + y2 + z2 ≤ b2.
Á Ω x = r cos ϕ, y = r sin ϕ cos θ, z = r sin ϕ sin θ.
I=
2π 0
dθ
π 0
dϕ
b a
r2
sin
ϕdr(r
sin
ϕ)2
ΩǑ
z = 0, y + z = 1
y = x2
Ω
.
Ω³ÙOyz
, ø ð³Ùx ø ,
Ǒ0.
4. zdxdydz, Ω x2 + y2 = 4, z = x2 + y2 z = 0
Ω
I=
2π 0
dθ
2 0
rdr
r2 0
dzz
=
2π
·
16 3
=
32π 3
.
5. (x2 − y2 − z2)dV , Ω : x2 + y2 + z2 ≤ a2.
Ω
x2 +y 2 +z 2
I=
2π 0
dθ
π
2
0
dϕ
2a 0
cos
ϕ
r2
sin ϕdr cos ϕ
=
16π 15
.
15.
Ǒ 2xy+1
x2 +y 2 +z 2
dV
,
Ω
x2 + y2 + z2 = 2a2 az = x2 + y2
z≥0
.
Ω
√
I=
x2
1 +y2
+z2
dV
=
2π 0
dθ
π
4
0
dϕ
0
2a
r2
Ω
z2dV =
2π 0
dθ
π 0
dϕ
a 0
r2
sin
ϕdr(r
cos
ϕ)2
=
4π 15
a5.
Ω
y2dV
=
4π 15
a5
.
I
=
−
4π 15
a5
.
Ω
.
Ò x2dV = Ω
6. (x2 + y2)dV , Ω : 3 x2 + y2 ≤ z ≤ 3.
Ω
I=
2π 0
dθ
1 0
rdr
3 3r
dzr2
=
2π
1 0
D
I =4
1 0
dx
√ 1−x2
0
dyy2√1
−
x2
=
4
1 0
dx
1 3
(1
−
x2)2
=
32 45
.
11. (|x| + y)dxdy, D = {(x, y) | |x| + |y| ≤ 1}.
D
I=
|x|dxdy +
ydxdy = 4
1 0
dx
1−x 0
dyx
+
0
=
4
1 0
dxx(1
−
x)
=
2 3
=
π 4
.
2.
x
=
u
+
1 2
,
y
=
v
+
1 2
.
D(x,y) D(u,v)
=
1.
Ǒ« D′
u2
+ v2
≤
1 2
,
I=
D′ (v
+
1 2
)dudv
=
0+
1 2
·
π 2
=
π 4
.
24. (x2 + y2)dxdy,
Ǒ « D
x2 a2
+
y2 b2
≤ 1.
D
µ Ã .
x = ar cos θ, y = br sin θ.
.
D
D
12. (x + y)dxdy,
ǑD x2 + y2 = 1, x2 + y2 = 2y
Ñè .
D
I=
xdxdy +
ydxdy = 0 + 2
√
3
2
0
dx
√
1√−x2 1− 1−x2
ydy
=
2
√ 3
√
2
0
dx(
1 − x2 −
D
√
D
1 2
)
=
π 3
−
3 4
.
ÁÇ ¯ ù .
13.
√
1 0
dx
0 1−x2 (x2 + y2)dy =
ÃD
D
. mg(x, y) ≤ f (x, y)g(x, y) ≤ M g(x, y).
mg(x, y)dσ ≤ f (x, y)g(x, y)dσ ≤ M g(x, y)dσ.
à D
D
D
g(x, y)dσ = 0,
f (x, y)g(x, y)dσ = 0,
è óÃ (x0, y0) ∈ D
D
D
Ω
√
I
=
4
π
2
0
dθ
R 0
cos
θ
2rdr
0 R2−r2 dzz2
=
4
π 2
0
dθ
1 15
R5
(1
−
sin5
θ)
=
2 15
(π
−
16 15
)R5
.
10. (1 + xy + yz + zx)dV ,
ΩǑ
x2 + y2 = 2z x2 + y2 + z2 = 8
Ω
z≥0
.
I=
Ω
1dV =
2π 0
dθ
2 0
sin
ϕdr
1 r2
+
2π 0
dθ
π 2 π 4
Ω
dϕ
a cos ϕ
sin2 ϕ
0
r2
sin
ϕdr
1 r2
√
√
= 2πa( 2 − 1 + ln 2).
16.
√ dV
, Ω : x2 + y2 + z2 ≤ 1.
Ω
x2+y2+(z−2)2
I=
2π 0
dθ
1 0
dr
π 0
dϕr2 sin ϕ √ 1
r2−4r cos ϕ+4
D(x,y) D(u,v)
=
abr.
π 4
I=
1 0
dr
(a2 + b2)ab.
2π 0
(a2r2
cos2
θ
+
b2r2 sin2 θ)abrdθ
=
1 0
drπ(a2
r2
+
b2r2)abr
=
û¹ 26. a > 0,
I(a) =
a 0
e−x2 dx,
J (a)
=
e−x2−y2 dxdy,
Da = {(x, y) |
Ý x2 + y2 ≤ a2, x ≥ 0, y ≥ 0}.
Da
(1) [I(a)]2 = e−x2−y2 dxdy,
Ra = {(x, y) | 0 ≤ x ≤ a, 0 ≤ y ≤ a};
Ra
√
(2) J(a) ≤ [I(a)]2 ≤ J( 2a);
(3)
10 ´Â
lim
a→+∞
a 0
e−x2
dx
=
π 4
(1
−
e−a2
).
π 4
.
±ùÒ,
lim I(a)
a→+∞
=
√π 2
.
√
lim J(a) = lim J( 2a) =
a→+∞
a→+∞
ËÃ7.3
Ç ¯Ò .
1. (z + z2)dV ,
ΩǑ
x2 + y2 + z2 ≤ 1.
Ω
I=
zdV +
z2dV = 0 +
2π 0
dθ
π 0
dϕ
1 0
drr2 sin ϕ(r cos ϕ)2