題目
Problem
10. Let F ( x , y , z ) = x x 2 + y 2 + z 2 i + y x 2 + y 2 + z 2 j + z x 2 + y 2 + z 2 k \mathbf{F}(x, y, z) = \frac{x}{x^2 + y^2 + z^2}\mathbf{i} + \frac{y}{x^2 + y^2 + z^2}\mathbf{j} + \frac{z}{x^2 + y^2 + z^2}\mathbf{k} F ( x , y , z ) = x 2 + y 2 + z 2 x i + x 2 + y 2 + z 2 y j + x 2 + y 2 + z 2 z k ∬ S F ⋅ d S \iint_S \mathbf{F} \cdot \mathrm{d}\mathbf{S} ∬ S F ⋅ d S S S S x 2 + y 2 + z 2 = 225 x^2 + y^2 + z^2 = 225 x 2 + y 2 + z 2 = 225
解答
思路
展開
令位置向量 r = ⟨ x , y , z ⟩ \mathbf{r} = \langle x, y, z \rangle r = ⟨ x , y , z ⟩ r = ∣ r ∣ = x 2 + y 2 + z 2 r = |\mathbf{r}| = \sqrt{x^2+y^2+z^2} r = ∣ r ∣ = x 2 + y 2 + z 2
向量場可寫成：
F = r r 2 \mathbf{F} = \frac{\mathbf{r}}{r^2} F = r 2 r
積分曲面 S S S R = 225 = 15 R = \sqrt{225} = 15 R = 225 = 15
球面上的單位外法向量為：
n = r r \mathbf{n} = \frac{\mathbf{r}}{r} n = r r
計算 F ⋅ n \mathbf{F} \cdot \mathbf{n} F ⋅ n F ⋅ n = r r 2 ⋅ r r = r 2 r 3 = 1 r \mathbf{F} \cdot \mathbf{n} = \frac{\mathbf{r}}{r^2} \cdot \frac{\mathbf{r}}{r} = \frac{r^2}{r^3} = \frac{1}{r} F ⋅ n = r 2 r ⋅ r r = r 3 r 2 = r 1
由於在球面 S S S r = 15 r = 15 r = 15 F ⋅ n = 1 15 \mathbf{F} \cdot \mathbf{n} = \frac{1}{15} F ⋅ n = 15 1
答題過程
展開
令位置向量 r = x i + y j + z k \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} r = x i + y j + z k r = x 2 + y 2 + z 2 r = \sqrt{x^2+y^2+z^2} r = x 2 + y 2 + z 2 F \mathbf{F} F
F = r r 2 \mathbf{F} = \frac{\mathbf{r}}{r^2} F = r 2 r 曲面 S S S R = 15 R = 15 R = 15 S S S
n = r r \mathbf{n} = \frac{\mathbf{r}}{r} n = r r 計算向量場在法向上的投影：
F ⋅ n = ( r r 2 ) ⋅ ( r r ) = r ⋅ r r 3 = r 2 r 3 = 1 r \mathbf{F} \cdot \mathbf{n} = \left( \frac{\mathbf{r}}{r^2} \right) \cdot \left( \frac{\mathbf{r}}{r} \right) = \frac{\mathbf{r} \cdot \mathbf{r}}{r^3} = \frac{r^2}{r^3} = \frac{1}{r} F ⋅ n = ( r 2 r ) ⋅ ( r r ) = r 3 r ⋅ r = r 3 r 2 = r 1 因為在球面 S S S r = R = 15 r = R = 15 r = R = 15
F ⋅ n = 1 15 \mathbf{F} \cdot \mathbf{n} = \frac{1}{15} F ⋅ n = 15 1 計算曲面積分：
∬ S F ⋅ d S = ∬ S ( F ⋅ n ) d S = ∬ S 1 15 d S = 1 15 ∬ S 1 d S \iint_S \mathbf{F} \cdot \mathrm{d}\mathbf{S} = \iint_S (\mathbf{F} \cdot \mathbf{n}) \,\mathrm{d}S = \iint_S \frac{1}{15}\,\mathrm{d}S = \frac{1}{15} \iint_S 1\,\mathrm{d}S ∬ S F ⋅ d S = ∬ S ( F ⋅ n ) d S = ∬ S 15 1 d S = 15 1 ∬ S 1 d S = 1 15 ⋅ Area ( S ) = \frac{1}{15} \cdot \text{Area}(S) = 15 1 ⋅ Area ( S ) 球面的表面積公式為 Area ( S ) = 4 π R 2 \text{Area}(S) = 4\pi R^2 Area ( S ) = 4 π R 2
∬ S F ⋅ d S = 1 15 ⋅ 4 π ( 15 ) 2 = 4 π ⋅ 15 = 60 π \iint_S \mathbf{F} \cdot \mathrm{d}\mathbf{S} = \frac{1}{15} \cdot 4\pi (15)^2 = 4\pi \cdot 15 = 60\pi ∬ S F ⋅ d S = 15 1 ⋅ 4 π ( 15 ) 2 = 4 π ⋅ 15 = 60 π 結論：
曲面積分值為 60 π 60\pi 60 π
