(a)
원기둥좌표를 이용하여 S S S
r ( r , θ ) = r cos θ ı ^ + r sin θ ȷ ^ + r k ^ 단, 0 ≤ r ≤ 1 , 0 ≤ θ ≤ π \mathbf{r}(r, \theta) = r\cos \theta\, \hat{\boldsymbol{\imath}} +r\sin \theta\, \hat{\boldsymbol{\jmath}} + r\,\hat{\mathbf{k}} \qquad \text{단, $0\le r\le 1$, $0\le \theta \le \pi$} r ( r , θ ) = r cos θ ^ + r sin θ ^ + r k ^ 단 , 0 ≤ r ≤ 1, 0 ≤ θ ≤ π
(b)
다음을 계산한다.
∂ r ∂ r = cos θ ı ^ + sin θ ȷ ^ + k ^ ∂ r ∂ θ = − r sin θ ı ^ + r cos θ ȷ ^ n ^ d S = ± ∂ r ∂ r × ∂ r ∂ θ d r d θ = ± ( − r cos θ ı ^ − r sin θ ȷ ^ + r k ^ ) d r d θ \begin{aligned}
\frac{\partial \mathbf{r}}{\partial r}
&= \cos \theta \,\hat{\boldsymbol{\imath}} +\sin \theta \,\hat{\boldsymbol{\jmath}} +\,\hat{\mathbf{k}} \\
\frac{\partial \mathbf{r}}{\partial \theta}
&= -r \sin \theta \,\hat{\boldsymbol{\imath}} + r\cos \theta \,\hat{\boldsymbol{\jmath}} \\
\hat{\mathbf{n}}\, \mathrm{d}S &= \pm \frac{\partial \mathbf{r}}{\partial r}\times
\frac{\partial \mathbf{r}}{\partial \theta}
\ \mathrm{d}r \mathrm{d}\theta
= \pm \big(-r\cos \theta \,\hat{\boldsymbol{\imath}} - r \sin \theta \,\hat{\boldsymbol{\jmath}} + r \,\hat{\mathbf{k}}\big)\,
\mathrm{d}r \mathrm{d}\theta
\end{aligned} ∂ r ∂ r ∂ θ ∂ r n ^ d S = cos θ ^ + sin θ ^ + k ^ = − r sin θ ^ + r cos θ ^ = ± ∂ r ∂ r × ∂ θ ∂ r d r d θ = ± ( − r cos θ ^ − r sin θ ^ + r k ^ ) d r d θ
아래쪽 선속을 구하려면 음의 부호를 택한다. 따라서
∬ S v ⋅ n ^ d S = ∫ 0 π d θ ∫ 0 1 d r ( r cos θ , r sin θ , − 2 r ) ⋅ ( r cos θ , r sin θ , − r ) = ∫ 0 π d θ ∫ 0 1 d r 3 r 2 = π r 3 ∣ r = 0 1 = π \begin{aligned}
\iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S
& = \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ (r \cos \theta, r\sin \theta, -2r)\cdot
(r\cos \theta, r\sin \theta, -r) \\
&= \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ 3 r^2
= \pi r^3\Big|_{r=0}^1 = \pi
\end{aligned} ∬ S v ⋅ n ^ d S = ∫ 0 π d θ ∫ 0 1 d r ( r cos θ , r sin θ , − 2 r ) ⋅ ( r cos θ , r sin θ , − r ) = ∫ 0 π d θ ∫ 0 1 d r 3 r 2 = π r 3 r = 0 1 = π
이다.
(c) 풀이 1 :
P \mathcal{P} P ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) ( − 1 , 0 , 1 ) (-1,0,1) ( − 1 , 0 , 1 ) P \mathcal{P} P
그러면
∫ C F ⋅ d r + ∫ P F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S \int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{P} \mathbf{F} \cdot \mathrm{d}\mathbf{r}
= \iint _S { \boldsymbol{\nabla} } \times \mathbf{F} \cdot \hat{\mathbf{n}}\, \mathrm{d}S ∫ C F ⋅ d r + ∫ P F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S
이다. 이는 스토크스 정리에 의한 것이다. P \mathcal{P} P F \mathbf{F} F ∫ P F ⋅ d r = 0 \int_\mathcal{P} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0 ∫ P F ⋅ d r = 0 ∇ × F \nabla\times \mathbf{F} ∇ × F v \mathbf{v} v
∫ C F ⋅ d r = ∬ S v ⋅ n ^ d S = π \int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = \iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S = \pi ∫ C F ⋅ d r = ∬ S v ⋅ n ^ d S = π
이다.
(c) 풀이 2:
L \mathcal{L} L ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( − 1 , 0 , 1 ) (-1,0,1) ( − 1 , 0 , 1 )
R = { ( x , y , z ) } x 2 + y 2 ≤ 1 , y ≥ 0 , z = 1 \begin{aligned}
\mathcal{R}=\left\{(x,y,z)\right\}{x^2+y^2\le1, y\ge 0, z=1}
\end{aligned} R = { ( x , y , z ) } x 2 + y 2 ≤ 1 , y ≥ 0 , z = 1
이라 하자. 그림은 다음과 같다. L \mathcal{L} L R \mathcal{R} R
그러면
∫ C F ⋅ d r + ∫ L F ⋅ d r = ∬ R ∇ × F ⋅ ( − k ^ ) d S \int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{L} \mathbf{F} \cdot \mathrm{d}\mathbf{r}
= \iint _{\mathcal{R}} { \boldsymbol{\nabla} } \times \mathbf{F} \cdot (-\hat{\mathbf{k}})\, \mathrm{d}S ∫ C F ⋅ d r + ∫ L F ⋅ d r = ∬ R ∇ × F ⋅ ( − k ^ ) d S
이다. 이는 스토크스 정리에 의한 것이다. L \mathcal{L} L F = ȷ ^ \mathbf{F}=\hat{\boldsymbol{\jmath}} F = ^ − ı ^ -\hat{\boldsymbol{\imath}} − ^ ∫ L F ⋅ d r = 0 \int_\mathcal{L} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0 ∫ L F ⋅ d r = 0 ∇ × F \nabla\times \mathbf{F} ∇ × F v \mathbf{v} v
∫ C F ⋅ d r = − ∬ R v ⋅ k ^ d S = ∬ R 2 z d S = 2 ∬ R d S = 2 넓이 ( R ) = π \int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = -\iint _\mathcal{R} \mathbf{v} \cdot \hat{\mathbf{k}}\, \mathrm{d}S
= \iint _\mathcal{R} 2z \, \mathrm{d}S
= 2\iint _\mathcal{R} \, \mathrm{d}S
=2\,\text{넓이}(\mathcal{R})
= \pi ∫ C F ⋅ d r = − ∬ R v ⋅ k ^ d S = ∬ R 2 z d S = 2 ∬ R d S = 2 넓이 ( R ) = π
이다.
원본 교재 풀이(영어) (a)
We parametrize S S S
r ( r , θ ) = r cos θ ı ^ + r sin θ ȷ ^ + r k ^ with 0 ≤ r ≤ 1 , 0 ≤ θ ≤ π \mathbf{r}(r, \theta) = r\cos \theta\, \hat{\boldsymbol{\imath}} +r\sin \theta\, \hat{\boldsymbol{\jmath}} + r\,\hat{\mathbf{k}} \qquad \text{with $0\le r\le 1$, $0\le \theta \le \pi$} r ( r , θ ) = r cos θ ^ + r sin θ ^ + r k ^ with 0 ≤ r ≤ 1, 0 ≤ θ ≤ π
(b)
We compute
∂ r ∂ r = cos θ ı ^ + sin θ ȷ ^ + k ^ ∂ r ∂ θ = − r sin θ ı ^ + r cos θ ȷ ^ n ^ d S = ± ∂ r ∂ r × ∂ r ∂ θ d r d θ = ± ( − r cos θ ı ^ − r sin θ ȷ ^ + r k ^ ) d r d θ \begin{aligned}
\frac{\partial \mathbf{r}}{\partial r}
&= \cos \theta \,\hat{\boldsymbol{\imath}} +\sin \theta \,\hat{\boldsymbol{\jmath}} +\,\hat{\mathbf{k}} \\
\frac{\partial \mathbf{r}}{\partial \theta}
&= -r \sin \theta \,\hat{\boldsymbol{\imath}} + r\cos \theta \,\hat{\boldsymbol{\jmath}} \\
\hat{\mathbf{n}}\, \mathrm{d}S &= \pm \frac{\partial \mathbf{r}}{\partial r}\times
\frac{\partial \mathbf{r}}{\partial \theta}
\ \mathrm{d}r \mathrm{d}\theta
= \pm \big(-r\cos \theta \,\hat{\boldsymbol{\imath}} - r \sin \theta \,\hat{\boldsymbol{\jmath}} + r \,\hat{\mathbf{k}}\big)\,
\mathrm{d}r \mathrm{d}\theta
\end{aligned} ∂ r ∂ r ∂ θ ∂ r n ^ d S = cos θ ^ + sin θ ^ + k ^ = − r sin θ ^ + r cos θ ^ = ± ∂ r ∂ r × ∂ θ ∂ r d r d θ = ± ( − r cos θ ^ − r sin θ ^ + r k ^ ) d r d θ
To calculate the downward flux, we use the minus sign. We find
∬ S v ⋅ n ^ d S = ∫ 0 π d θ ∫ 0 1 d r ( r cos θ , r sin θ , − 2 r ) ⋅ ( r cos θ , r sin θ , − r ) = ∫ 0 π d θ ∫ 0 1 d r 3 r 2 = π r 3 ∣ r = 0 1 = π \begin{aligned}
\iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S
& = \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ (r \cos \theta, r\sin \theta, -2r)\cdot
(r\cos \theta, r\sin \theta, -r) \\
&= \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ 3 r^2
= \pi r^3\Big|_{r=0}^1 = \pi
\end{aligned} ∬ S v ⋅ n ^ d S = ∫ 0 π d θ ∫ 0 1 d r ( r cos θ , r sin θ , − 2 r ) ⋅ ( r cos θ , r sin θ , − r ) = ∫ 0 π d θ ∫ 0 1 d r 3 r 2 = π r 3 r = 0 1 = π
(c) Solution 1 :
Let P \mathcal{P} P ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) ( − 1 , 0 , 1 ) (-1,0,1) ( − 1 , 0 , 1 ) P \mathcal{P} P
Then
∫ C F ⋅ d r + ∫ P F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S \int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{P} \mathbf{F} \cdot \mathrm{d}\mathbf{r}
= \iint _S { \boldsymbol{\nabla} } \times \mathbf{F} \cdot \hat{\mathbf{n}}\, \mathrm{d}S ∫ C F ⋅ d r + ∫ P F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S
by Stokes' Theorem. Along P \mathcal{P} P F \mathbf{F} F ∫ P F ⋅ d r = 0 \int_\mathcal{P} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0 ∫ P F ⋅ d r = 0 ∇ × F \nabla\times \mathbf{F} ∇ × F v \mathbf{v} v
∫ C F ⋅ d r = ∬ S v ⋅ n ^ d S = π \int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = \iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S = \pi ∫ C F ⋅ d r = ∬ S v ⋅ n ^ d S = π
(c) Solution 2:
Let L \mathcal{L} L ( 1 , 0 , 1 ) (1,0,1) ( 1 , 0 , 1 ) ( − 1 , 0 , 1 ) (-1,0,1) ( − 1 , 0 , 1 )
R = { ( x , y , z ) } x 2 + y 2 ≤ 1 , y ≥ 0 , z = 1 \begin{aligned}
\mathcal{R}=\left\{(x,y,z)\right\}{x^2+y^2\le1, y\ge 0, z=1}
\end{aligned} R = { ( x , y , z ) } x 2 + y 2 ≤ 1 , y ≥ 0 , z = 1
Here is a sketch. L \mathcal{L} L R \mathcal{R} R
Then
∫ C F ⋅ d r + ∫ L F ⋅ d r = ∬ R ∇ × F ⋅ ( − k ^ ) d S \int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{L} \mathbf{F} \cdot \mathrm{d}\mathbf{r}
= \iint _{\mathcal{R}} { \boldsymbol{\nabla} } \times \mathbf{F} \cdot (-\hat{\mathbf{k}})\, \mathrm{d}S ∫ C F ⋅ d r + ∫ L F ⋅ d r = ∬ R ∇ × F ⋅ ( − k ^ ) d S
by Stokes' Theorem. Along L \mathcal{L} L F = ȷ ^ \mathbf{F}=\hat{\boldsymbol{\jmath}} F = ^ − ı ^ -\hat{\boldsymbol{\imath}} − ^ ∫ L F ⋅ d r = 0 \int_\mathcal{L} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0 ∫ L F ⋅ d r = 0 ∇ × F \nabla\times \mathbf{F} ∇ × F v \mathbf{v} v
∫ C F ⋅ d r = − ∬ R v ⋅ k ^ d S = ∬ R 2 z d S = 2 ∬ R d S = 2 Area ( R ) = π \int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = -\iint _\mathcal{R} \mathbf{v} \cdot \hat{\mathbf{k}}\, \mathrm{d}S
= \iint _\mathcal{R} 2z \, \mathrm{d}S
= 2\iint _\mathcal{R} \, \mathrm{d}S
=2\,\text{Area}(\mathcal{R})
= \pi ∫ C F ⋅ d r = − ∬ R v ⋅ k ^ d S = ∬ R 2 z d S = 2 ∬ R d S = 2 Area ( R ) = π
