Let it flow
I'm thinking of a real number n such that the flux of the vector field F ( x , y , z ) = ( x 2 + y 2 + z 2 ) n ( x , y , z ) through the surface ( x − 2 ) 2 + ( y − 3 ) 2 + ( z − 4 ) 2 = 1 is 0. Find the sum of all possible values of n .
(From a calculus exam I recently gave.)
The answer is -1.5.
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1 solution
Very nicely explained; thank you!
We might add that d i v ( F ) = ( 3 + 2 n ) ( x 2 + y 2 + z 2 ) n is either always positive or always negative when n = − 2 3 , so that the flux will be positive or negative, but never zero.
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Yeah, that's what I tried to communicate in my last sentence, but not very eloquently. I will edit it to make it more readable.
Recall the divergence theorem:
∫ ∫ S F ⋅ d S = ∫ ∫ ∫ V ∇ ⋅ F d V
The divergence is (using the product rule and chain rule): ∇ ⋅ F = ( x 2 + y 2 + z 2 ) n + 2 n x 2 ( x 2 + y 2 + z 2 ) n − 1 + ( x 2 + y 2 + z 2 ) n + 2 n y 2 ( x 2 + y 2 + z 2 ) n − 1 + ( x 2 + y 2 + z 2 ) n + 2 n z 2 ( x 2 + y 2 + z 2 ) n − 1 = 3 ( x 2 + y 2 + z 2 ) n + 2 n ( x 2 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) n − 1 = ( 3 + 2 n ) ( x 2 + y 2 + z 2 ) n
The divergence is globally zero when n = − 2 3 . For this value of n , the flux through the test surface is correspondingly zero. For other values of n , the divergence is either always positive or always negative, and thus it will not integrate to zero in those cases.