6.5 [[[ᑖᑖᑖ

6.5 三 三 三重 重 重積 積 積 分 分 分

222 第 6 章 多變數函數的積分

(4) 0 ≤ z ≤ 2, 0 ≤ x ≤ 6 − 3z, 0 ≤ y ≤^6−x−3z₂ ˆ 2

0

ˆ 6−3z 0

ˆ ^6−x−3z₂

0

dy dx dz = ˆ 2

0

ˆ 6−3z 0

6− x − 3z 2 dx dz

= 1 2

ˆ 2 0

(6x− x²

2 − 3xz)^6−3z

0 dz

= 1 2

ˆ 2 0

6(6− 3z) −(6− 3z)²

2 − 3(6 − 3z)z dz

= 1 2

(− (6 − 3z)²+ (6− 3z)³

18 − 9z²+ 3z³)²

0

= 1

2(36− 12 − 36 + 24) = 6





 習題解答 6.5.2.









0≤ y ≤ 1

−√y≤ x ≤√y 0≤ z ≤ 1 − y









0≤ y ≤ 1 0≤ z ≤ 1 − y

−√y≤ x ≤√y









−1 ≤ x ≤ 1 0≤ z ≤ 1 − x² x²≤ y ≤ 1 − z









0≤ z ≤ 1

−√

1− z ≤ x ≤√ 1− z x²≤ y ≤ 1 − z





 習題解答 6.5.3. (1)

ˆ ˆ ˆ

Ω

x + y²+ z³dV

= ˆ 1

−1

ˆ 1

−1

ˆ 1

−1

x + y²+ z³dz dy dx

= ˆ 1

−1

ˆ 1

−1

(xz + y²z +z⁴ 4 )¹

−1dy dx

= ˆ 1

−1

ˆ 1

−1

2x + 2y²dy dx

= 2 ˆ ₁

−1

(xy +y³ 3)¹

−1dx

= 2 ˆ 1

−1

2x +2

3 dx = 4(x² 2 + 1

3x)¹

−1= 8 3

6.5. 三重積分 223

(2)

ˆ ˆ ˆ

Ω

xyz dV = ˆ 1

0

ˆ 2 0

ˆ 3 0

xyz dz dy dx

= ˆ 1

0

x dx· ˆ 2

0

y dy· ˆ 3

0

z dz = 1 2 ·4

2·9 2 = 9

2 (3)

ˆ ˆ ˆ

Ω

x dV = ˆ 1

0

ˆ 1−x 0

ˆ 1−x−y 0

x dz dy dx

= ˆ 1

0

ˆ _1−x

0

x(1− x − y) dy dx = ˆ 1

0

(

(x− x²)y− xy² 2

)^1−x

0 dx

= ˆ 1

0

x(1− x)² 2 dx =1

2 ˆ 1

0

x− 2x²+ x³dx

= 1 2

(x² 2 −2x³

3 +x⁴ 4

)¹

0= 1 2

(1 2 −2

3+ 1 4 )

= 1 24 (4)

ˆ ˆ ˆ

Ω

ye^z dV = ˆ 1

0

ˆ 1

−1

ˆ

√1−x²

−√ 1−x²

ye^z dy dx dz

= ˆ 1

0

ˆ 1

−1

e^z(y² 2 )

√1−x²

−√

1−x²dx dz = 0





 習題解答 6.5.4.

(1) 調換 x 和 y 的順序：

ˆ 1 0

ˆ 1 0

ˆ 1 x²

xze^zy2dy dx dz = ˆ 1

0

ˆ 1 0

ˆ ^√y 0

xze^zy2dx dy dz

= ˆ 1

0

ˆ 1 0

ze^zy2(x² 2

)

√y

0 dy dz = 1 2

ˆ 1 0

ˆ 1 0

zye^zy2 dy dz

= 1 2

ˆ 1 0

(e^zy2 2

)¹

0dz = 1 4

ˆ 1 0

e^z− 1 dz

= 1

4(e^z− z)¹

0= 1

4(e− 1 − 1) =e− 2 4 (2) 調換 x 和 y 的順序：

ˆ 4 0

ˆ 1 0

ˆ 2 2y

cos x²

√z dx dy dz = ˆ 4

0

ˆ 2 0

ˆ ^x₂

0

cos x²

√z dy dx dz

= ˆ 4

0

ˆ 2 0

x cos x² 2√

z dx dz =1 4

ˆ 4 0

√1 z dz·

ˆ 2 0

x cos x²dx

= 1 2 (

2z¹²)⁴

0·(sin x² 2

)²

0= 1

2· 4 ·sin 4 2 = sin 4

1

6.5. 三重積分 223

(2)

ˆ ˆ ˆ

Ω

xyz dV = ˆ 1

0

ˆ 2 0

ˆ 3 0

xyz dz dy dx

= ˆ 1

0

x dx· ˆ 2

0

y dy· ˆ 3

0

z dz = 1 2 ·4

2·9 2 = 9

2 (3)

ˆ ˆ ˆ

Ω

x dV = ˆ 1

0

ˆ 1−x 0

ˆ 1−x−y 0

x dz dy dx

= ˆ 1

0

ˆ 1−x 0

x(1− x − y) dy dx = ˆ 1

0

((x− x²)y− xy²

2 )¹₀^−xdx

= ˆ 1

0

x(1− x)² 2 dx =1

2 ˆ 1

0

x− 2x²+ x³dx

= 1 2

(x² 2 −2x³

3 +x⁴ 4

)¹

0= 1 2

(1 2 −2

3+ 1 4 )

= 1 24 (4)

ˆ ˆ ˆ

Ωye^z dV = ˆ 1

0

ˆ 1

−1

ˆ ^√1−x²

−√1−x² ye^z dy dx dz

= ˆ 1

0

ˆ 1

−1

e^z(y² 2

)

√1−x²

−√

1−x²dx dz = 0





 習題解答 6.5.4.

(1) 調換 x 和 y 的順序：

ˆ 1 0

ˆ 1 0

ˆ 1 x²

xze^zy2dy dx dz = ˆ 1

0

ˆ 1 0

ˆ ^√y 0

xze^zy2dx dy dz

= ˆ 1

0

ˆ 1 0

ze^zy2(x² 2

)

√y

0 dy dz = 1 2

ˆ 1 0

ˆ 1 0

zye^zy2 dy dz

= 1 2

ˆ 1 0

(e^zy2 2

)¹

0dz = 1 4

ˆ 1 0

e^z− 1 dz

= 1

4(e^z− z)¹

0= 1

4(e− 1 − 1) =e− 2 4 (2) 調換 x 和 y 的順序：

ˆ 4 0

ˆ 1 0

ˆ 2 2y

cos x²

√z dx dy dz = ˆ 4

0

ˆ 2 0

ˆ ^x

2 0

cos x²

√z dy dx dz

= ˆ 4

0

ˆ 2 0

x cos x² 2√

z dx dz =1 4

ˆ 4 0

√1 z dz·

ˆ 2 0

x cos x²dx

= 1 2 (

2z¹²)⁴₀·(sin x² 2

)²₀= 1

2· 4 ·sin 4 2 = sin 4

224 第 6 章 多變數函數的積分





 習題解答 6.5.5.

體積 = ˆ a

0

ˆ b(1−^xa) 0

ˆ c(1−^xa−^yb) 0

1 dz dy dx

= ˆ a

0

ˆ b(1−^xa) 0

c(1−x a− y

b) dy dx

= c ˆ a

0

( y(1−x

a)− y² 2b

)^b(1−

x a)

0 dx

= c ˆ _a

0

b(1−x

a)− b²(1−^x_a)²

2b dx =bc 2

ˆ _a

0

(1−x a)²dx

= bc 2 ·(

(−a)(1−^x_a)³ 3 )^a

0= bc 2 ·a

3 = abc 6





 習題解答 6.5.6.

體積 = ˆ a

0

ˆ _c(1−^x_a) 0

ˆ _b(1−^z_c) 0

1 dy dz dx

= ˆ a

0

ˆ c(1−^x_a) 0

b(1−z c) dz dx

= b ˆ a

0

(

(−c)(1−^z_c)² 2

)^c(1−

x a)

0 dx = bc

2 ˆ a

0

1−x² a² dx

= bc 2

( x− x³

3a² )^a

0= bc 2 ·2a

3 = abc 3

6.5.2 三重積分的變數變換





 習題解答 6.5.7.

依題意取坐標變換如下









u = x + y + z v = y + z w = x

⇔







 z = w

y = v− z = v − w

x = u− y − z = u − (v − w) − w = u − v 區域邊界的曲線對應如下：









x + y + z = d11, x + y + z = d12

y + z = d21, y + z = d22

z = d31, z = d32

⇔









u = d11, u = d12

v = d21, v = d22

w = d31, w = d32

226 第 6 章 多變數函數的積分





 習題解答 6.5.10.

用球面坐標, 此區域相當於 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ α, 0 ≤ ρ ≤ R.

ˆ _2π

0

ˆ _α

0

ˆ _R

0

1· ρ²sin ϕ dρ dϕ dθ

= ˆ 2π

0

dθ· ˆ α

0 sin ϕ dϕ · ˆ R

0

ρ²dρ

= 2π· (− cos ϕ)^α₀·R³ 3 = 2π

3 (1− cos α)R³ 用柱面坐標, 此區域相當於 0 ≤ θ ≤ 2π, 0 ≤ r ≤ R sin α, r cot α ≤ z ≤√

R²− r². ˆ 2π

0

ˆ R sin α 0

ˆ ^√R²−r² r cot α

1· r dz dr dθ

= ˆ 2π

0

dθ·

ˆ R sin α 0

(√

R²− r²− r cot α) · r dr

= 2π·(

−(R²− r²)³²

3 − cot αr³ 3)^{R sin α}

0

= 2π·(

−cos³α 3 + R³

3 − cot αR³sin³α 3

)

= 2π·R³

3 · (1 − cos³α− cos α sin²α)

= 2π

3 (1− cos α)R³





 習題解答 6.5.11.

本題變數變換取法和上題一樣, J(u, v, w) = abc. 故積分為 ˆ ˆ ˆ

x2

a2+^y2_b2+^z2_c2≤1

x²y²dx dy dz = ˆ ˆ ˆ

u²+v²+w²≤1

a²b²u²v²· abc du dv dw

現再用球面坐標來計算, 得原式為 原式 = a³b³c

ˆ _2π

0

ˆ _π

0

ˆ ₁

0

(ρ sin ϕ cos θ)²(ρ sin ϕ sin θ)²· ρ²sin ϕ dρ dϕ dθ

= a³b³c ˆ 2π

0

(cos θ sin θ)²dθ· ˆ π

0

sin⁵ϕ dϕ· ˆ 1

0

ρ⁶dρ

= a³b³c ˆ _2π

0

1− cos 4θ 8 dθ·

ˆ _π

0 −(1 − cos²ϕ)²d(cos ϕ)·(ρ⁷ 7)¹

0

= a³b³c 7

(θ

8− sin 4θ 32 )^2π

0 ·(

− cos θ +2 cos³θ

3 −cos⁵θ 5 )^π

0

= a³b³c 7 ·π

4 · 2 · (1 −2 3+1

5) = 4

105a³b³c π

2

6.5. 三重積分 227





 習題解答 6.5.12.

依題意取坐標變換如下









u = x + y v = x− y w = ln z

⇔









x = ^u+v₂ y = ^u−v₂ z = e^w 區域邊界的曲線對應如下：









y = x− 1, y = x + 2 y =−x, y = −x + 1 z = 1, z = 2

⇔









v = 1, v =−2 u = 0, u = 1 w = 0, w = ln 2 再計算 J(u, v, w) 如下：

J(u, v, w) =      

∂x

∂u

∂y

∂u

∂z

∂u

∂x

∂v

∂y

∂v

∂z

∂v

∂x

∂w

∂y

∂w

∂z

∂w      

=     

1 2

1

2 0

1 2 −1

2 0 0 0 e^w

    

=−e^w 2

原積分經變數變換後得 ˆ ₁

0

ˆ ₁

−2

ˆ ln 2 0

uvw e^w ·e^w

2 dw dv du = 1 2

ˆ ₁

0

u du· ˆ ₁

−2

v dv· ˆ ln 2

0

w dw

= 1 2·1

2·1− 4

2 ·(ln 2)²

2 =−3(ln 2)² 16





 習題解答 6.5.13.

取坐標變換如下







 u =^x_a v = ^y_b w = ^z_c

⇔









x = au y = bv z = cw 則原區域 x²

a²+ y² b² +z²

c² ≤ 1 變成 Σ ≡ u²+ v²+ w²≤ 1. 且如前 J(u, v, w) = abc. 則原 積分成為

abc ˆ ˆ ˆ

Σ

a²u²+ b²v²dV , abc ˆ ˆ ˆ

Σ

b²v²+ c²w²dV , abc ˆ ˆ ˆ

Σ

c²w²+ a²u²dV

3