題目
Problem
(a) C C C ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 0 , 2 ) (0,\sqrt{2}) ( 0 , 2 )
F ( x , y ) = x − 3 y x 2 + y 2 i + 3 x + y x 2 + y 2 j \displaystyle\mathbf{F}(x,y)=\frac{x-3y}{x^2+y^2}\mathbf{i}+\frac{3x+y}{x^2+y^2}\mathbf{j} F ( x , y ) = x 2 + y 2 x − 3 y i + x 2 + y 2 3 x + y j ∫ C F ⋅ d r = ( 17 ) ‾ \displaystyle\int_C \mathbf{F}\cdot d\mathbf{r} = \underline{\quad(17)\quad} ∫ C F ⋅ d r = ( 17 )
(b) S S S x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9 z = x 2 + y 2 z=\sqrt{x^2+y^2} z = x 2 + y 2 F ( x , y , z ) = x i + y j + z k ( x 2 + y 2 + z 2 ) 2 \displaystyle\mathbf{F}(x,y,z)=\frac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{(x^2+y^2+z^2)^2} F ( x , y , z ) = ( x 2 + y 2 + z 2 ) 2 x i + y j + z k ∬ S F ⋅ d S = ( 18 ) ‾ \displaystyle\iint_S \mathbf{F}\cdot d\mathbf{S} = \underline{\quad(18)\quad} ∬ S F ⋅ d S = ( 18 )
解答
解法一
思路
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(a) 曲線完全位於上半平面，避開原點；可辨認向量場為保守場，再以勢函數取兩端點差。
(b) 在半徑為 3 3 3
答題過程
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(a) 在上半平面可取連續的極角 θ ∈ ( 0 , π ) \theta\in(0,\pi) θ ∈ ( 0 , π )
ϕ ( x , y ) = ln x 2 + y 2 + 3 θ . \phi(x,y)=\ln\sqrt{x^2+y^2}+3\theta. ϕ ( x , y ) = ln x 2 + y 2 + 3 θ . 直接微分可驗證 ∇ ϕ = F \nabla\phi=\mathbf F ∇ ϕ = F
∫ C F ⋅ d r = ϕ ( 0 , 2 ) − ϕ ( 1 , 1 ) = ( 1 2 ln 2 + 3 π 2 ) − ( 1 2 ln 2 + 3 π 4 ) = 3 π 4 . \begin{align*}
\int_C\mathbf F\cdot\mathrm{d}\mathbf r
=&\,\phi(0,\sqrt2)-\phi(1,1)\\[2mm]
=&\,\left(\frac12\ln2+\frac{3\pi}{2}\right)
-\left(\frac12\ln2+\frac{3\pi}{4}\right)\\[2mm]
=&\,\frac{3\pi}{4}.
\end{align*} ∫ C F ⋅ d r = = = ϕ ( 0 , 2 ) − ϕ ( 1 , 1 ) ( 2 1 ln 2 + 2 3 π ) − ( 2 1 ln 2 + 4 3 π ) 4 3 π . 故
( 17 ) = 3 π 4 . \boxed{(17)=\frac{3\pi}{4}}. ( 17 ) = 4 3 π . (b) 在球面 x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9
F = r 3 4 , n = r 3 , \mathbf F=\frac{\mathbf r}{3^4},\qquad
\mathbf n=\frac{\mathbf r}{3}, F = 3 4 r , n = 3 r , 所以 F ⋅ n = 1 27 \mathbf F\cdot\mathbf n=\frac1{27} F ⋅ n = 27 1 0 ≤ θ ≤ π 4 0\le\theta\le\frac\pi4 0 ≤ θ ≤ 4 π
2 π ( 3 ) 2 ( 1 − cos π 4 ) = 18 π ( 1 − 2 2 ) . 2\pi(3)^2\left(1-\cos\frac\pi4\right)
=18\pi\left(1-\frac{\sqrt2}{2}\right). 2 π ( 3 ) 2 ( 1 − cos 4 π ) = 18 π ( 1 − 2 2 ) . 因此
( 18 ) = 2 π 3 ( 1 − 2 2 ) . \boxed{(18)=\frac{2\pi}{3}\left(1-\frac{\sqrt2}{2}\right)}. ( 18 ) = 3 2 π ( 1 − 2 2 ) . 