关于区域D

证明定理1： 笔记来源于：Path independence for line integrals | Multivariable Calculus | Khan Academy

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证明定理2：

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两个问题： 1.我们如何知道一个给定的向量场F是否是保守的? 2.如果F是保守的，我们如何找到一个势函数ƒ（以便 F = ∇ f \boldsymbol{F}=\nabla f F=∇f）

回顾 F \boldsymbol{F} F 的三维旋度中： c u r l   F ( x , y , z ) = ∇ × F   ( ∂ ∂ x ∂ ∂ y ∂ ∂ z ) × ( M ( x , y , z ) N ( x , y , z ) P ( x , y , z ) )   i = [ 1 0 0 ] j = [ 0 1 0 ] k = [ 0 0 1 ]   d e t ( i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z M ( x , y , z ) N ( x , y , z ) P ( x , y , z ) )     ( ∂ ∂ y ∂ ∂ z N ( x , y , z ) P ( x , y , z ) ) i − ( ∂ ∂ x ∂ ∂ z M ( x , y , z ) P ( x , y , z ) ) j + ( ∂ ∂ x ∂ ∂ y M ( x , y , z ) N ( x , y , z ) ) k   ( ∂ P ∂ y − ∂ N ∂ z ) i − ( ∂ P ∂ x − ∂ M ∂ z ) j + ( ∂ N ∂ x − ∂ M ∂ y ) k   ( ∂ P ∂ y − ∂ N ∂ z ) i + ( ∂ M ∂ z − ∂ P ∂ x ) j + ( ∂ N ∂ x − ∂ M ∂ y ) k   c u r l   F ( x , y , z ) = ( ∂ P ∂ y − ∂ N ∂ z   ∂ M ∂ z − ∂ P ∂ x   ∂ N ∂ x − ∂ M ∂ y ) curl\,\boldsymbol{F}(x,y,z)=\nabla×\boldsymbol{F}\\ ~\\ \begin{pmatrix} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z} \end{pmatrix}× \begin{pmatrix} M(x,y,z)\\ N(x,y,z)\\ P(x,y,z) \end{pmatrix}\\ ~\\ \boldsymbol{i}=\left[ \begin{array}{l} 1\\ 0\\ 0 \end{array} \right] \quad\boldsymbol{j}=\left[ \begin{array}{l} 0\\ 1\\ 0 \end{array} \right] \quad\boldsymbol{k}=\left[ \begin{array}{l} 0\\ 0\\ 1 \end{array} \right]\\ ~\\ det\begin{pmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ M(x,y,z) & N(x,y,z) & P(x,y,z) \end{pmatrix} ~\\ ~\\ \begin{pmatrix} \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ N(x,y,z) & P(x,y,z) \end{pmatrix}\boldsymbol{i}- \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial z}\\ M(x,y,z) & P(x,y,z) \end{pmatrix}\boldsymbol{j}+ \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y}\\ M(x,y,z) & N(x,y,z) \end{pmatrix}\boldsymbol{k}\\ ~\\ (\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z})\boldsymbol{i}-(\frac{\partial P}{\partial x}-\frac{\partial M}{\partial z})\boldsymbol{j}+(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\boldsymbol{k}\\ ~\\ (\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z})\boldsymbol{i}+(\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x})\boldsymbol{j}+(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\boldsymbol{k}\\ ~\\ curl\,\boldsymbol{F}(x,y,z)= \begin{pmatrix} \frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}\\ ~\\ \frac{\partial M}{\partial z}-\frac{\partial P}{\partial x}\\ ~\\ \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \end{pmatrix} curlF(x,y,z)=∇×F ⎝ ⎛​∂x∂​∂y∂​∂z∂​​⎠ ⎞​×⎝ ⎛​M(x,y,z)N(x,y,z)P(x,y,z)​⎠ ⎞​ i=⎣ ⎡​100​⎦ ⎤​j=⎣ ⎡​010​⎦ ⎤​k=⎣ ⎡​001​⎦ ⎤​ det⎝ ⎛​i∂x∂​M(x,y,z)​j∂y∂​N(x,y,z)​k∂z∂​P(x,y,z)​⎠ ⎞​  (∂y∂​N(x,y,z)​∂z∂​P(x,y,z)​)i−(∂x∂​M(x,y,z)​∂z∂​P(x,y,z)​)j+(∂x∂​M(x,y,z)​∂y∂​N(x,y,z)​)k (∂y∂P​−∂z∂N​)i−(∂x∂P​−∂z∂M​)j+(∂x∂N​−∂y∂M​)k (∂y∂P​−∂z∂N​)i+(∂z∂M​−∂x∂P​)j+(∂x∂N​−∂y∂M​)k curlF(x,y,z)=⎝ ⎛​∂y∂P​−∂z∂N​ ∂z∂M​−∂x∂P​ ∂x∂N​−∂y∂M​​⎠ ⎞​ 当且仅当 F \boldsymbol{F} F 是保守场时，下面这个向量为0（即旋度为0） ( ∂ P ∂ y − ∂ N ∂ z ) i + ( ∂ M ∂ z − ∂ P ∂ x ) j + ( ∂ N ∂ x − ∂ M ∂ y ) k (\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z})\boldsymbol{i}+(\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x})\boldsymbol{j}+(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\boldsymbol{k}\\ (∂y∂P​−∂z∂N​)i+(∂z∂M​−∂x∂P​)j+(∂x∂N​−∂y∂M​)k 即： ∂ P ∂ y − ∂ N ∂ z = 0   ∂ M ∂ z − ∂ P ∂ x = 0   ∂ N ∂ x − ∂ M ∂ y = 0 \frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}=0\\ ~\\ \frac{\partial M}{\partial z}-\frac{\partial P}{\partial x}=0\\ ~\\ \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=0 ∂y∂P​−∂z∂N​=0 ∂z∂M​−∂x∂P​=0 ∂x∂N​−∂y∂M​=0 即： ∂ P ∂ y = ∂ N ∂ z   ∂ M ∂ z = ∂ P ∂ x   ∂ N ∂ x = ∂ M ∂ y \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}\\ ~\\ \frac{\partial M}{\partial z}=\frac{\partial P}{\partial x}\\ ~\\ \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y} ∂y∂P​=∂z∂N​ ∂z∂M​=∂x∂P​ ∂x∂N​=∂y∂M​

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