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16.5withScannedExamples

# 16.5withScannedExamples - 16.5 CURL AND DIVERGENCE KIAM...

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Unformatted text preview: 16.5: CURL AND DIVERGENCE KIAM HEONG KWA 1. Curl Let F = P i + Q j + R k be a vector field on a solid region D ⊂ R 3 . If all first-order partial derivatives of P , Q , and R exist, then the curl of F on D is the vector field (1.1) curl F = ∂R ∂y- ∂Q ∂z i + ∂P ∂z- ∂R ∂x j + ∂Q ∂x- ∂P ∂y k . By introducing the del operator (1.2) ∇ = i ∂ ∂x + j ∂ ∂y + k ∂ ∂z and considering ∇ as a vector whose components are ∂/∂x , ∂/∂y , and ∂/∂z , we can consider curl F as the formal cross product of the operator ∇ with the vector field F : ∇ × F = i j k ∂ ∂x ∂ ∂y ∂ ∂z P Q R (1.3) = i ∂ ∂y ∂ ∂z Q R- j ∂ ∂x ∂ ∂z P R + k ∂ ∂x ∂ ∂y P Q = ∂R ∂y- ∂Q ∂z i + ∂P ∂z- ∂R ∂x j + ∂Q ∂x- ∂P ∂y k = curl F . Remark 1. When the del operator ∇ acts on a scalar function f , we obtain the gradient of f : ∇ f = i ∂f ∂x + j ∂f ∂y + k ∂f ∂z = ∂f ∂x i + ∂f ∂y j + ∂f ∂z k . Date : November 14, 2010. 1 2 KIAM HEONG KWA If the curl of a vector field is identically, then the vector field is said to be irrotational . It can be shown that if F is a conservative vector field whose components have continuous partial derivatives, then F is irrotational. In fact, if f is a potential function of F , then curl F = ∇ × ∇ f = i j k ∂ ∂x ∂ ∂y ∂ ∂z f x f y f z = i ( f zy- f yz ) + j ( f xz- f zx ) + k ( f yx- f xy ) = 0 i + 0 j + 0 k by Clairaut’s theorem = everywhere in D . More is true: as a consequence of the Stoke’s theorem in section 16.8, the converse also holds if the domain D of F is simply- connected. Hence we have: Theorem 1. Let F be a vector field on a solid region D ⊂ R 3 whose components have continuous first-order partial derivatives. If D is simply-connected, then F is irrotational if and only if F is conser- vative. Example 1 (Problem 14 in the text) . The vector field F ( x,y,z ) = xyz 2 i + x 2 yz 2 j + x 2 y 2 z k is not conservative because ∇ × F 6 = : ∇ × F = ∂ ∂y ( x 2 y 2 z )- ∂ ∂z ( x 2 yz 2 ) i + ∂ ∂z ( xyz 2 )- ∂ ∂x ( x 2 y 2 z ) j + ∂ ∂x ( x 2 yz 2 )- ∂ ∂y ( xyz 2 ) k = ( 2 x 2 yz- 2 x 2 yz ) i + ( 2 xyz- 2 xy 2 z ) j + ( 2 xyz 2- xz 2 ) k = ( 2 xyz- 2 xy 2 z ) j + ( 2 xyz 2- xz 2 ) k ....
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