s10 - Solution to Set 10 1. (a) ∇ · F = 2 ∇ × F =...

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Unformatted text preview: Solution to Set 10 1. (a) ∇ · F = 2 ∇ × F = (∂x y − ∂y x)k = 0 (b) ∇ · F = x2 y2 +2 = (x2 + y 2 )3/2 (x + y 2 )3/2 (x2 −xy xy +2 2 )3/2 +y (x + y 2 )3/2 1 x2 + y2 ∇×F = (c) ∇ · F = 0 k=0 ∇ × F = −2k (d) F is not differentiable on the circle x2 + y 2 = 1. 0 for x2 + y 2 < 1 ∇·F = 0 for x2 + y 2 > 1 ⎧ for x2 + y 2 < 1 ⎨ −2k 2 2 2 2 ∇×F = x −y x −y ⎩ −2 k = 0 for x2 + y 2 > 1 (x2 + y 2 )2 (x + y 2 )2 2. ∇ · f F = 3ex+y+z ∇ × fF = 0 3. Let F1 , F2 , F3 be the components of F . The x component of ∇ × (φ∇F ) is ∂y (φF3 ) − ∂z (φF2 ) = φ(∂y F3 − ∂z F2 ) + (φy F3 − φz F2 ). It is the same as the x component of φ∇ × F + ∇φ × F . Similarly, the same can be shown for the y and z components. 4. ∇ · F = ∇ · f r = f ∇ · r + (∇f ) · r Since ∇ · r = 3 and ∇f = f ∇r = f ∇ · F = 3f + rf = 0 ln |f | = Therefore, A r ˆ r2 where A is an arbitrary constant and r = r/r is the unit vector along the direction of r . ˆ F= and r , r d 3 f = ln |f | = − . f dr r ⇒ f= ±eC . r3 3 − dr = −3 ln r + C r 1 ...
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This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.

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