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第3回 演習解答：積分定理
「一般相対性理論の学び直し」シリーズ第3回の演習解答です。記号・用語・向き付けの規約は本編に従います。線積分は ∫ C F ⋅ d l = ∫ a b F ( r ( t ) ) ⋅ r ′ ( t ) d t \displaystyle\int_C\mathbf F\cdot d\mathbf l=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt ∫ C F ⋅ d l = ∫
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r ′ ( t ) d t
∫ S F ⋅ n d S \displaystyle\int_S\mathbf F\cdot\mathbf n\,dS ∫ S F ⋅ n d S
演習 1 問題 F = ( 2 x y , x 2 ) \mathbf F=(2xy,\ x^2) F = ( 2 x y , x 2 ) F = ∇ φ \mathbf F=\nabla\varphi F = ∇ φ φ \varphi φ ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) ∫ C F ⋅ d l \int_C\mathbf F\cdot d\mathbf l ∫ C F ⋅ d l ( 0 , 0 ) → ( 1 , 0 ) → ( 1 , 1 ) (0,0)\to(1,0)\to(1,1) ( 0 , 0 ) → ( 1 , 0 ) → ( 1 , 1 ) φ ( 1 , 1 ) − φ ( 0 , 0 ) \varphi(1,1)-\varphi(0,0) φ ( 1 , 1 ) − φ ( 0 , 0 )
（a）ポテンシャルの求め方 F = ∇ φ \mathbf F=\nabla\varphi F = ∇ φ
∂ φ ∂ x = 2 x y , ∂ φ ∂ y = x 2 \frac{\partial\varphi}{\partial x}=2xy,\qquad \frac{\partial\varphi}{\partial y}=x^2 ∂ x ∂ φ = 2 x y , ∂ y ∂ φ = x 2 が成り立ちます。まず x x x y y y
φ ( x , y ) = ∫ 2 x y d x = x 2 y + g ( y ) . \varphi(x,y)=\int 2xy\,dx=x^2y+g(y). φ ( x , y ) = ∫ 2 x y d x = x 2 y + g ( y ) . ここで g ( y ) g(y) g ( y ) x x x y y y
∂ φ ∂ y = x 2 + g ′ ( y ) . \frac{\partial\varphi}{\partial y}=x^2+g'(y). ∂ y ∂ φ = x 2 + g ′ ( y ) . これが x 2 x^2 x 2 g ′ ( y ) = 0 g'(y)=0 g ′ ( y ) = 0 g g g
φ ( x , y ) = x 2 y \varphi(x,y)=x^2y φ ( x , y ) = x 2 y と取れます。確認：∂ x ( x 2 y ) = 2 x y \partial_x(x^2y)=2xy ∂ x ( x 2 y ) = 2 x y ∂ y ( x 2 y ) = x 2 \partial_y(x^2y)=x^2 ∂ y ( x 2 y ) = x 2 ∇ φ = F \nabla\varphi=\mathbf F ∇ φ = F
補足（保存場の条件）. ∇ × F \nabla\times\mathbf F ∇ × F z z z
∂ F y ∂ x − ∂ F x ∂ y = ∂ ∂ x ( x 2 ) − ∂ ∂ y ( 2 x y ) = 2 x − 2 x = 0. \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=\frac{\partial}{\partial x}(x^2)-\frac{\partial}{\partial y}(2xy)=2x-2x=0. ∂ x ∂ F y − ∂ y ∂ F x = ∂ x ∂ ( x 2 ) − ∂ y ∂ ( 2 x y ) = 2 x − 2 x = 0. 回転が消えることは勾配場の必要条件です。定義域が R 2 \mathbb R^2 R 2 F \mathbf F F
（b）線積分の計算 線積分は ∫ C F ⋅ d l = ∫ a b F ( r ( t ) ) ⋅ r ′ ( t ) d t \displaystyle\int_C\mathbf F\cdot d\mathbf l=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt ∫ C F ⋅ d l = ∫ a b F ( r ( t )) ⋅ r ′ ( t ) d t
（b1）直線経路 ( 0 , 0 ) → ( 1 , 1 ) (0,0)\to(1,1) ( 0 , 0 ) → ( 1 , 1 )
端点を結ぶ直線を r ( t ) = ( t , t ) \mathbf r(t)=(t,t) r ( t ) = ( t , t ) t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ]
r ′ ( t ) = ( 1 , 1 ) , F ( r ( t ) ) = ( 2 t ⋅ t , t 2 ) = ( 2 t 2 , t 2 ) . \mathbf r'(t)=(1,1),\qquad \mathbf F(\mathbf r(t))=\bigl(2t\cdot t,\ t^2\bigr)=(2t^2,\ t^2). r ′ ( t ) = ( 1 , 1 ) , F ( r ( t )) = ( 2 t ⋅ t , t 2 ) = ( 2 t 2 , t 2 ) . F ⋅ r ′ = ( 2 t 2 ) ( 1 ) + ( t 2 ) ( 1 ) = 3 t 2 . \mathbf F\cdot\mathbf r'=(2t^2)(1)+(t^2)(1)=3t^2. F ⋅ r ′ = ( 2 t 2 ) ( 1 ) + ( t 2 ) ( 1 ) = 3 t 2 . ∫ C F ⋅ d l = ∫ 0 1 3 t 2 d t = [ t 3 ] 0 1 = 1. \int_C\mathbf F\cdot d\mathbf l=\int_0^1 3t^2\,dt=\bigl[t^3\bigr]_0^1=1. ∫ C F ⋅ d l = ∫ 0 1 3 t 2 d t = [ t 3 ] 0 1 = 1. （b2）折れ線経路 ( 0 , 0 ) → ( 1 , 0 ) → ( 1 , 1 ) (0,0)\to(1,0)\to(1,1) ( 0 , 0 ) → ( 1 , 0 ) → ( 1 , 1 )
第1区間 ( 0 , 0 ) → ( 1 , 0 ) (0,0)\to(1,0) ( 0 , 0 ) → ( 1 , 0 ) r ( t ) = ( t , 0 ) \mathbf r(t)=(t,0) r ( t ) = ( t , 0 ) t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] r ′ = ( 1 , 0 ) \mathbf r'=(1,0) r ′ = ( 1 , 0 ) y = 0 y=0 y = 0
F ( r ( t ) ) = ( 2 t ⋅ 0 , t 2 ) = ( 0 , t 2 ) , F ⋅ r ′ = 0. \mathbf F(\mathbf r(t))=\bigl(2t\cdot 0,\ t^2\bigr)=(0,\ t^2),\qquad \mathbf F\cdot\mathbf r'=0. F ( r ( t )) = ( 2 t ⋅ 0 , t 2 ) = ( 0 , t 2 ) , F ⋅ r ′ = 0. 寄与は 0 0 0 x x x F \mathbf F F x x x 0 0 0
第2区間 ( 1 , 0 ) → ( 1 , 1 ) (1,0)\to(1,1) ( 1 , 0 ) → ( 1 , 1 ) r ( t ) = ( 1 , t ) \mathbf r(t)=(1,t) r ( t ) = ( 1 , t ) t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] r ′ = ( 0 , 1 ) \mathbf r'=(0,1) r ′ = ( 0 , 1 ) x = 1 x=1 x = 1
F ( r ( t ) ) = ( 2 ⋅ 1 ⋅ t , 1 2 ) = ( 2 t , 1 ) , F ⋅ r ′ = ( 2 t ) ( 0 ) + ( 1 ) ( 1 ) = 1. \mathbf F(\mathbf r(t))=(2\cdot 1\cdot t,\ 1^2)=(2t,\ 1),\qquad \mathbf F\cdot\mathbf r'=(2t)(0)+(1)(1)=1. F ( r ( t )) = ( 2 ⋅ 1 ⋅ t , 1 2 ) = ( 2 t , 1 ) , F ⋅ r ′ = ( 2 t ) ( 0 ) + ( 1 ) ( 1 ) = 1. 寄与は ∫ 0 1 1 d t = 1 \displaystyle\int_0^1 1\,dt=1 ∫ 0 1 1 d t = 1
勾配定理（本編§2）より、保存場の線積分は端点のポテンシャル差に等しいはずです。
φ ( 1 , 1 ) − φ ( 0 , 0 ) = 1 2 ⋅ 1 − 0 = 1. \varphi(1,1)-\varphi(0,0)=1^2\cdot 1-0=1. φ ( 1 , 1 ) − φ ( 0 , 0 ) = 1 2 ⋅ 1 − 0 = 1. 計算方法 結果 直線経路 1 1 1 折れ線経路 1 1 1 ポテンシャル差 φ ( 1 , 1 ) − φ ( 0 , 0 ) \varphi(1,1)-\varphi(0,0) φ ( 1 , 1 ) − φ ( 0 , 0 ) 1 1 1
3通りとも一致し、勾配定理（経路非依存）が確かめられました。■ \blacksquare ■
演習 2 問題 F = r = ( x , y , z ) \mathbf F=\mathbf r=(x,y,z) F = r = ( x , y , z ) a a a V V V
∮ ∂ V F ⋅ n d S = ∫ V ( ∇ ⋅ F ) d V \oint_{\partial V}\mathbf F\cdot\mathbf n\,dS=\int_V(\nabla\cdot\mathbf F)\,dV ∮ ∂ V F ⋅ n d S = ∫ V ( ∇ ⋅ F ) d V です。左辺は境界 ∂ V \partial V ∂ V a a a
左辺（面積分） 半径 a a a x 2 + y 2 + z 2 = a 2 x^2+y^2+z^2=a^2 x 2 + y 2 + z 2 = a 2 r = ( x , y , z ) \mathbf r=(x,y,z) r = ( x , y , z ) 単位 法線は
n = r ∣ r ∣ = r a = ( x a , y a , z a ) . \mathbf n=\frac{\mathbf r}{|\mathbf r|}=\frac{\mathbf r}{a}=\left(\frac{x}{a},\ \frac{y}{a},\ \frac{z}{a}\right). n = ∣ r ∣ r = a r = ( a x , a y , a z 球面上で ∣ r ∣ = a |\mathbf r|=a ∣ r ∣ = a
F ⋅ n = r ⋅ r a = x 2 + y 2 + z 2 a = a 2 a = a \mathbf F\cdot\mathbf n=\mathbf r\cdot\frac{\mathbf r}{a}=\frac{x^2+y^2+z^2}{a}=\frac{a^2}{a}=a F ⋅ n = r ⋅ a r = a x 2 + y 2 + z 2 = a a 2 = a と、球面上のどの点でも定数 a a a 4 π a 2 4\pi a^2 4 π a 2
∮ ∂ V F ⋅ n d S = ∫ ∂ V a d S = a ⋅ 4 π a 2 = 4 π a 3 . \oint_{\partial V}\mathbf F\cdot\mathbf n\,dS=\int_{\partial V} a\,dS=a\cdot 4\pi a^2=4\pi a^3. ∮ ∂ V F ⋅ n d S = ∫ ∂ V a d S = a ⋅ 4 π a 2 = 4 π a 3 .
右辺（体積分） ∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 1 + 1 + 1 = 3. \nabla\cdot\mathbf r=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3. ∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 1 + 1 + 1 = 3. 被積分関数は定数 3 3 3 4 3 π a 3 \frac{4}{3}\pi a^3 3 4 π a 3
∫ V ( ∇ ⋅ F ) d V = ∫ V 3 d V = 3 ⋅ 4 3 π a 3 = 4 π a 3 . \int_V(\nabla\cdot\mathbf F)\,dV=\int_V 3\,dV=3\cdot\frac{4}{3}\pi a^3=4\pi a^3. ∫ V ( ∇ ⋅ F ) d V = ∫ V 3 d V = 3 ⋅ 3 4 π a 3 = 4 π a 3 .
結論 左辺 = 4 π a 3 =4\pi a^3 = 4 π a 3 = 4 π a 3 =4\pi a^3 = 4 π a 3 ■ \blacksquare ■
演習 3 問題 F = ( − y , x , 0 ) \mathbf F=(-y,\ x,\ 0) F = ( − y , x , 0 ) x y xy x y a a a S S S
∮ ∂ S F ⋅ d l = ∫ S ( ∇ × F ) ⋅ n d S \oint_{\partial S}\mathbf F\cdot d\mathbf l=\int_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS ∮ ∂ S F ⋅ d l = ∫ S ( ∇ × F ) ⋅ n d S です。左辺は境界の循環、右辺は内部の回転の面積分です。
向き付けの確認 曲面 S S S x y xy x y n = ± z ^ \mathbf n=\pm\hat{\mathbf z} n = ± z ^ n = z ^ \mathbf n=\hat{\mathbf z} n = z ^ 反時計回り （x y xy x y
左辺（線積分） 境界 ∂ S \partial S ∂ S
r ( t ) = ( a cos t , a sin t , 0 ) , t ∈ [ 0 , 2 π ] . \mathbf r(t)=\bigl(a\cos t,\ a\sin t,\ 0\bigr),\qquad t\in[0,2\pi]. r ( t ) = ( a cos t , a sin t , 0 ) , t ∈ [ 0 , 2 π ] . 微分すると r ′ ( t ) = ( − a sin t , a cos t , 0 ) \mathbf r'(t)=(-a\sin t,\ a\cos t,\ 0) r ′ ( t ) = ( − a sin t , a cos t , 0 ) x = a cos t x=a\cos t x = a cos t y = a sin t y=a\sin t y = a sin t
F ( r ( t ) ) = ( − a sin t , a cos t , 0 ) . \mathbf F(\mathbf r(t))=(-a\sin t,\ a\cos t,\ 0). F ( r ( t )) = ( − a sin t , a cos t , 0 ) . F ⋅ r ′ = ( − a sin t ) ( − a sin t ) + ( a cos t ) ( a cos t ) = a 2 ( sin 2 t + cos 2 t ) = a 2 . \mathbf F\cdot\mathbf r'=(-a\sin t)(-a\sin t)+(a\cos t)(a\cos t)=a^2(\sin^2 t+\cos^2 t)=a^2. F ⋅ r ′ = ( − a sin t ) ( − a sin t ) + ( a cos t ) ( a cos t ) = a 2 ( sin 2 t + cos 2 t ) = a 2 . ∮ ∂ S F ⋅ d l = ∫ 0 2 π a 2 d t = a 2 ⋅ 2 π = 2 π a 2 . \oint_{\partial S}\mathbf F\cdot d\mathbf l=\int_0^{2\pi}a^2\,dt=a^2\cdot 2\pi=2\pi a^2. ∮ ∂ S F ⋅ d l = ∫ 0 2 π a 2 d t = a 2 ⋅ 2 π = 2 π a 2 .
右辺（面積分） 回転を計算します（第2回の公式 ∇ × F = ( ∂ y F z − ∂ z F y , ∂ z F x − ∂ x F z , ∂ x F y − ∂ y F x ) \nabla\times\mathbf F=\bigl(\partial_y F_z-\partial_z F_y,\ \partial_z F_x-\partial_x F_z,\ \partial_x F_y-\partial_y F_x\bigr) ∇ × F = ( ∂ y F z − ∂ z F y , ∂ z F x − ∂ x F z , ∂ x F y − ∂ y F x )
∇ × F = ( 0 − 0 , 0 − 0 , ∂ ∂ x ( x ) − ∂ ∂ y ( − y ) ) = ( 0 , 0 , 2 ) . \nabla\times\mathbf F=\left(0-0,\ 0-0,\ \frac{\partial}{\partial x}(x)-\frac{\partial}{\partial y}(-y)\right)=(0,\ 0,\ 2). ∇ × F = ( 0 − 0 , 0 − 0 , ∂ x ∂ ( x ) − ∂ y ∂ ( − y ) ) = ( 0 , 0 , 2 ) . 法線 n = z ^ = ( 0 , 0 , 1 ) \mathbf n=\hat{\mathbf z}=(0,0,1) n = z ^ = ( 0 , 0 , 1 )
( ∇ × F ) ⋅ n = 2. (\nabla\times\mathbf F)\cdot\mathbf n=2. ( ∇ × F ) ⋅ n = 2. 被積分関数が定数 2 2 2 π a 2 \pi a^2 π a 2
∫ S ( ∇ × F ) ⋅ n d S = ∫ S 2 d S = 2 ⋅ π a 2 = 2 π a 2 . \int_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=\int_S 2\,dS=2\cdot\pi a^2=2\pi a^2. ∫ S ( ∇ × F ) ⋅ n d S = ∫ S 2 d S = 2 ⋅ π a 2 = 2 π a 2 .
結論 左辺 = 2 π a 2 =2\pi a^2 = 2 π a 2 = 2 π a 2 =2\pi a^2 = 2 π a 2 ■ \blacksquare ■
演習 4 問題 F = ( − y x 2 + y 2 , x x 2 + y 2 , 0 ) \mathbf F=\left(\dfrac{-y}{x^2+y^2},\ \dfrac{x}{x^2+y^2},\ 0\right) F = ( x 2 + y 2 − y , x 2 + y 2 x , 0 ) ( x , y ) ≠ ( 0 , 0 ) (x,y)\neq(0,0) ( x , y ) = ( 0 , 0 ) ∇ × F = 0 \nabla\times\mathbf F=\mathbf 0 ∇ × F = 0 ∮ C F ⋅ d l \oint_C\mathbf F\cdot d\mathbf l ∮ C F ⋅ d l
この場は極座標の角度 θ = atan2 ( y , x ) \theta=\operatorname{atan2}(y,x) θ = atan2 ( y , x ) ∇ θ \nabla\theta ∇ θ 0 0 0
（a）回転が消えることの確認 回転の z z z x y xy x y F \mathbf F F z z z 0 0 0 x , y x,y x , y 0 0 0
( ∇ × F ) z = ∂ F y ∂ x − ∂ F x ∂ y . (\nabla\times\mathbf F)_z=\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}. ( ∇ × F ) z = ∂ x ∂ F y − ∂ y ∂ F x . F y F_y F y x x x F y = x x 2 + y 2 F_y=\dfrac{x}{x^2+y^2} F y = x 2 + y 2 x x x x x 2 + y 2 x^2+y^2 x 2 + y 2
∂ F y ∂ x = 1 ⋅ ( x 2 + y 2 ) − x ⋅ 2 x ( x 2 + y 2 ) 2 = x 2 + y 2 − 2 x 2 ( x 2 + y 2 ) 2 = y 2 − x 2 ( x 2 + y 2 ) 2 . \frac{\partial F_y}{\partial x}=\frac{1\cdot(x^2+y^2)-x\cdot 2x}{(x^2+y^2)^2}=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}. ∂ x ∂ F y = ( x 2 + y 2 ) 2 1 ⋅ ( x 2 + y ( x 2 + y 2 ) 2 x 2 + y ( x 2 + y 2 ) 2 y 2 − x F x F_x F x y y y F x = − y x 2 + y 2 F_x=\dfrac{-y}{x^2+y^2} F x = x 2 + y 2 − y
∂ F x ∂ y = ( − 1 ) ⋅ ( x 2 + y 2 ) − ( − y ) ⋅ 2 y ( x 2 + y 2 ) 2 = − ( x 2 + y 2 ) + 2 y 2 ( x 2 + y 2 ) 2 = y 2 − x 2 ( x 2 + y 2 ) 2 . \frac{\partial F_x}{\partial y}=\frac{(-1)\cdot(x^2+y^2)-(-y)\cdot 2y}{(x^2+y^2)^2}=\frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}. ∂ y ∂ F x = ( x 2 + y 2 ) 2 ( − 1 ) ⋅ ( x 2 + ( x 2 + y 2 ) 2 − ( x 2 + y ( x 2 + y 2 ) 2 y 2 − x 両者が等しいので ( ∇ × F ) z = 0 (\nabla\times\mathbf F)_z=0 ( ∇ × F ) z = 0 ( x , y ) ≠ ( 0 , 0 ) (x,y)\neq(0,0) ( x , y ) = ( 0 , 0 ) ∇ × F = 0 \nabla\times\mathbf F=\mathbf 0 ∇ × F = 0
（b）単位円まわりの循環 r ( t ) = ( cos t , sin t ) , t ∈ [ 0 , 2 π ] , r ′ = ( − sin t , cos t ) . \mathbf r(t)=(\cos t,\ \sin t),\qquad t\in[0,2\pi],\qquad \mathbf r'=(-\sin t,\ \cos t). r ( t ) = ( cos t , sin t ) , t ∈ [ 0 , 2 π ] , r ′ = ( − sin t , cos t ) . 単位円上では x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1
F ( r ( t ) ) = ( − sin t 1 , cos t 1 , 0 ) = ( − sin t , cos t , 0 ) . \mathbf F(\mathbf r(t))=\left(\frac{-\sin t}{1},\ \frac{\cos t}{1},\ 0\right)=(-\sin t,\ \cos t,\ 0). F ( r ( t )) = ( 1 − sin t , 1 cos t , 0 ) = ( − sin t , cos t , 0 ) . F ⋅ r ′ = ( − sin t ) ( − sin t ) + ( cos t ) ( cos t ) = sin 2 t + cos 2 t = 1. \mathbf F\cdot\mathbf r'=(-\sin t)(-\sin t)+(\cos t)(\cos t)=\sin^2 t+\cos^2 t=1. F ⋅ r ′ = ( − sin t ) ( − sin t ) + ( cos t ) ( cos t ) = sin 2 t + cos 2 t = 1. ∮ C F ⋅ d l = ∫ 0 2 π 1 d t = 2 π . \oint_C\mathbf F\cdot d\mathbf l=\int_0^{2\pi}1\,dt=2\pi. ∮ C F ⋅ d l = ∫ 0 2 π 1 d t = 2 π . 幾何学的な意味. この線積分は、原点のまわりを1周したときの角度の増分 Δ θ = 2 π \Delta\theta=2\pi Δ θ = 2 π F = ∇ θ \mathbf F=\nabla\theta F = ∇ θ ∮ ∇ θ ⋅ d l = Δ θ \oint\nabla\theta\cdot d\mathbf l=\Delta\theta ∮ ∇ θ ⋅ d l = Δ θ
（c）矛盾でない理由 一見すると、「回転が消えるのに閉曲線の線積分が 0 0 0 単連結性 という前提があります。
領域 U U U U U U ∇ × F = 0 \nabla\times\mathbf F=\mathbf 0 ∇ × F = 0 F = ∇ φ \mathbf F=\nabla\varphi F = ∇ φ ∮ = 0 \oint=0 ∮ = 0
この F \mathbf F F U = R 2 ∖ { 0 } U=\mathbb R^2\setminus\{\mathbf 0\} U = R 2 ∖ { 0 } 穴 があります。原点を囲む単位円は U U U U U U 単連結ではありません 。
ストークスの定理から見ると、単位円 C C C S S S U U U U U U ∇ × F = 0 \nabla\times\mathbf F=\mathbf 0 ∇ × F = 0 ∮ C = 0 \oint_C=0 ∮ C = 0
また F \mathbf F F ∇ θ \nabla\theta ∇ θ θ = atan2 ( y , x ) \theta=\operatorname{atan2}(y,x) θ = atan2 ( y , x ) θ \theta θ 2 π 2\pi 2 π 多価関数 です。U U U
まとめ. 回転が消えることは局所的 な性質、閉曲線で ∮ = 0 \oint=0 ∮ = 0 大域的 な性質です。両者が一致するのは定義域が単連結のときに限られます。∮ = 2 π ≠ 0 \oint=2\pi\neq 0 ∮ = 2 π = 0 ■ \blacksquare ■
発展演習：発散定理の数値確認（F = ( x 3 , y 3 , z 3 ) \mathbf F=(x^3,y^3,z^3) F = ( x 3 , y 3 , z 3 ) 問題 F = ( x 3 , y 3 , z 3 ) \mathbf F=(x^3,\ y^3,\ z^3) F = ( x 3 , y 3 , z 3 ) 12 π / 5 12\pi/5 12 π /5
解析計算（目標値の導出） ∇ ⋅ F = ∂ ∂ x ( x 3 ) + ∂ ∂ y ( y 3 ) + ∂ ∂ z ( z 3 ) = 3 x 2 + 3 y 2 + 3 z 2 = 3 r 2 , \nabla\cdot\mathbf F=\frac{\partial}{\partial x}(x^3)+\frac{\partial}{\partial y}(y^3)+\frac{\partial}{\partial z}(z^3)=3x^2+3y^2+3z^2=3r^2, ∇ ⋅ F = ∂ x ∂ ( x 3 ) + ∂ y ∂ ( y 3 ) + ∂ z ∂ ( z 3 ) = 3 x 2 + 3 y 2 + 3 z 2 = 3 r 2 , ただし r 2 = x 2 + y 2 + z 2 r^2=x^2+y^2+z^2 r 2 = x 2 + y 2 + z 2
右辺（体積分）. 球座標 ( r , θ , ϕ ) (r,\theta,\phi) ( r , θ , ϕ ) d V = r 2 sin θ d r d θ d ϕ dV=r^2\sin\theta\,dr\,d\theta\,d\phi d V = r 2 sin θ d r d θ d ϕ r ∈ [ 0 , 1 ] r\in[0,1] r ∈ [ 0 , 1 ]
∫ V 3 r 2 d V = 3 ∫ 0 2 π ∫ 0 π ∫ 0 1 r 2 ⋅ r 2 sin θ d r d θ d ϕ . \int_V 3r^2\,dV=3\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^1 r^2\cdot r^2\sin\theta\,dr\,d\theta\,d\phi. ∫ V 3 r 2 d V = 3 ∫ 0 2 π ∫ 0 π ∫ 0 r 2 sin θ d r d θ d ϕ . r r r ∫ 0 1 r 4 d r = 1 5 \displaystyle\int_0^1 r^4\,dr=\frac{1}{5} ∫ 0 1 r 4 d r = 5 1 θ \theta θ ∫ 0 π sin θ d θ = 2 \displaystyle\int_0^{\pi}\sin\theta\,d\theta=2 ∫ 0 π sin θ d θ = 2 ϕ \phi ϕ ∫ 0 2 π d ϕ = 2 π \displaystyle\int_0^{2\pi}d\phi=2\pi ∫ 0 2 π d ϕ = 2 π
∫ V 3 r 2 d V = 3 ⋅ 1 5 ⋅ 2 ⋅ 2 π = 12 π 5 . \int_V 3r^2\,dV=3\cdot\frac{1}{5}\cdot 2\cdot 2\pi=\frac{12\pi}{5}. ∫ V 3 r 2 d V = 3 ⋅ 5 1 ⋅ 2 ⋅ 2 π = 5 12 π . 左辺（面積分）. 単位球面上で r = 1 r=1 r = 1 n = ( x , y , z ) \mathbf n=(x,y,z) n = ( x , y , z ) r / r \mathbf r/r r / r
F ⋅ n = x 3 ⋅ x + y 3 ⋅ y + z 3 ⋅ z = x 4 + y 4 + z 4 . \mathbf F\cdot\mathbf n=x^3\cdot x+y^3\cdot y+z^3\cdot z=x^4+y^4+z^4. F ⋅ n = x 3 ⋅ x + y 3 ⋅ y + z 3 ⋅ z = x 4 + y 4 + z 4 . 球の対称性より ∮ x 4 d S = ∮ y 4 d S = ∮ z 4 d S \displaystyle\oint x^4\,dS=\oint y^4\,dS=\oint z^4\,dS ∮ x 4 d S = ∮ y 4 d S = ∮ z 4 d S x 4 x^4 x 4 d S = sin θ d θ d ϕ dS=\sin\theta\,d\theta\,d\phi d S = sin θ d θ d ϕ
∮ x 4 d S = ∫ 0 2 π ∫ 0 π ( sin θ cos ϕ ) 4 sin θ d θ d ϕ . \oint x^4\,dS=\int_0^{2\pi}\!\!\int_0^{\pi}(\sin\theta\cos\phi)^4\sin\theta\,d\theta\,d\phi. ∮ x 4 d S = ∫ 0 2 π ∫ 0 π ( sin θ cos ϕ ) 4 sin θ d θ d ϕ . ϕ \phi ϕ ∫ 0 2 π cos 4 ϕ d ϕ = 3 π 4 \displaystyle\int_0^{2\pi}\cos^4\phi\,d\phi=\frac{3\pi}{4} ∫ 0 2 π cos 4 ϕ d ϕ = 4 3 π θ \theta θ ∫ 0 π sin 5 θ d θ = 16 15 \displaystyle\int_0^{\pi}\sin^5\theta\,d\theta=\frac{16}{15} ∫ 0 π sin 5 θ d θ =
∮ x 4 d S = 3 π 4 ⋅ 16 15 = 4 π 5 . \oint x^4\,dS=\frac{3\pi}{4}\cdot\frac{16}{15}=\frac{4\pi}{5}. ∮ x 4 d S = 4 3 π ⋅ 15 16 = 5 4 π . ∮ ( x 4 + y 4 + z 4 ) d S = 3 ⋅ 4 π 5 = 12 π 5 . \oint(x^4+y^4+z^4)\,dS=3\cdot\frac{4\pi}{5}=\frac{12\pi}{5}. ∮ ( x 4 + y 4 + z 4 ) d S = 3 ⋅ 5 4 π = 5 12 π . 両辺とも 12 π 5 ≈ 7.5398 \dfrac{12\pi}{5}\approx 7.5398 5 12 π ≈ 7.5398
数値計算（Python） コピー import numpy as np
# F = (x^3, y^3, z^3), div F = 3(x^2+y^2+z^2) = 3 r^2
# 単位球と単位球面で発散定理を数値的に確認する
# --- 体積分:∫_V div F dV = ∫ 3 r^2 dV ---
Nr, Nt, Np = 60, 60, 120
r = np.linspace(0, 1, Nr)
th = np.linspace(0, np.pi, Nt)
ph = np.linspace(0, 2*np.pi, Np)
R, T, P = np.meshgrid(r, th, ph, indexing='ij')
integrand = 3 * R**2 # div F
dV = R**2 * np.sin(T) # 体積要素 r^2 sinθ
vol = np.trapezoid(np.trapezoid(np.trapezoid(integrand * dV, ph, axis=2), th, axis=1), r)
# --- 面積分:∮_S F·n dS,単位球面で n=(x,y,z),F·n = x^4+y^4+z^4 ---
T2, P2 = np.meshgrid(th, ph, indexing='ij')
x = np.sin(T2)*np.cos(P2); y = np.sin(T2)*np.sin(P2); z = np.cos(T2)
Fn = x**4 + y**4 + z**4
dS = np.sin(T2) # 単位球面の面積要素 sinθ
flux = np.trapezoid(np.trapezoid(Fn * dS, ph, axis=1), th)
exact = 12*np.pi/5
print("体積分 ∫ div F dV =", vol)
print("面積分 ∮ F·n dS =", flux)
print("12π/5 (解析値) =", exact)
数値計算（Julia） コピー function trapezoid(y, x)
s = 0.0
for i in 1:(length(x) - 1)
s += (y[i] + y[i + 1]) * (x[i + 1] - x[i]) / 2
end
return s
end
function integrate3(f, r, th, ph)
Nr, Nt, Np = length(r), length(th), length(ph)
s = zeros(Nr, Nt)
for ir in 1:Nr, it in 1:Nt
y = [f(r[ir], th[it], ph[ip]) for ip in 1:Np]
s[ir, it] = trapezoid(y, collect(ph))
end
s2 = zeros(Nr)
for ir in 1:Nr
y = [s[ir, it] for it in 1:Nt]
s2[ir] = trapezoid(y, collect(th))
end
return trapezoid([s2[ir] for ir in 1:Nr], collect(r))
end
function integrate2(f, th, ph)
Nt, Np = length(th), length(ph)
s = zeros(Nt)
for it in 1:Nt
y = [f(th[it], ph[ip]) for ip in 1:Np]
s[it] = trapezoid(y, collect(ph))
end
return trapezoid([s[it] for it in 1:Nt], collect(th))
end
Nr, Nt, Np = 60, 60, 120
r = range(0, 1, length=Nr)
th = range(0, π, length=Nt)
ph = range(0, 2π, length=Np)
vol = integrate3((R, T, P) -> 3 * R^2 * R^2 * sin(T), r, th, ph)
flux = integrate2((T, P) -> begin
x = sin(T) * cos(P)
y = sin(T) * sin(P)
z = cos(T)
(x^4 + y^4 + z^4) * sin(T)
end, th, ph)
exact = 12 * π / 5
println("体積分 ∫ div F dV = ", vol)
println("面積分 ∮ F·n dS = ", flux)
println("12π/5 (解析値) = ", exact)
出力と確認 コピー 体積分 ∫ div F dV = 7.54164985843735
面積分 ∮ F·n dS = 7.536851444843106
12π/5 (解析値) = 7.5398223686155035
vol と flux はともに解析値 12 π / 5 ≈ 7.5398 12\pi/5\approx 7.5398 12 π /5 ≈ 7.5398 F = ( x 3 , y 3 , z 3 ) \mathbf F=(x^3,y^3,z^3) F = ( x 3 , y 3 , z 3 ) ■ \blacksquare ■
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赤神 青空（Sora Akagami）