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(1)

微分積分

II

補助演習問題

No. 12

解答例

1 6

つの平面

x = 0, y = 0, z = 0, x = 1, y = 1, z = 1

で囲まれる立方体の領域を

G,

その表面を

S

とする

.

ベクトル場

a(x, y, z) = xy i + y j + z 2 k

に対して

div a = y + 1 + 2z

なので

,

ガウスの発散定理より

∫∫

S

a · n dS =

∫∫

G

div a dxdydz

=

∫ 1 0

(∫ 1 0

(∫ 1 0

(y + 1 + 2z) dz )

dy )

dx =

∫ 1 0

(∫ 1 0

(y + 2) dy )

dx = 5 2

を得る

.

2

曲面

S

で囲まれる円柱領域を

G

とする

.

ベクトル場

a(x, y, z) = (2x + y + 2z) i + (4y + x − 3z) j + ( − 5z + 2x − 3y) k

に対して

,

div a = 2 + 4 − 5 = 1

となるので

,

ガウスの発散定理より

∫∫

S

a · n dS =

∫∫

G

div a dxdydz =

∫∫

G

dxdydz = πa 2 h

となる

.

3

曲面

S : z = x 2 + y 2 , z ≤ 4

の他に

,

曲面

S 0 : z = 4, x 2 + y 2 ≤ 4

を考えて

, S

と

S 0

とで囲まれる領域を

G

とする

. S 0

上での単位法線ベクトルを

n = k

とする

.

ベクトル場

a(x, y, z) = xz i − 3yz j + z 2 k

に対して

,

div a = z − 3z + 2z = 0

であることに注意して

,

ガウスの発散定理を用いることで

,

∫∫

S

a · n dS +

∫∫

S

0

a · n dS =

∫∫∫

G

div a dxdydz = 0

(2)

を得る

.

また

, D : x 2 + y 2 ≤ 4

として

,

∫∫

S

0

a · n dS =

∫∫

S

0

z 2 dS =

∫∫

D

16 dxdy = 16 × 4π = 64π

をえる

.

従って

,

求める面積分は

∫∫

S

a · n dS = − 64π

となる

.

参照

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