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 Calculus Calculus Math Forum

 August 5th, 2015, 02:38 PM #1 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 Why no r cancelation in divergence hello For cylindrical coords... for a small area the flux in the radial direction sums up to approximately: [F(r+Δr,θ ,z)-F(r,θ ,z)]r*Δθ*Δz/ΔV but the 'r' multiplication is always then put inside the [ ] with the function so that: [r*F(r+Δr,θ ,z)-r*F(r,θ ,z)]Δθ*Δz/ΔV when the limit is taken it ends up being (1/r)*[d(r*F)/dr] instead of simply dF/dr. Why? Last edited by Kinroh; August 5th, 2015 at 02:45 PM. August 10th, 2015, 10:27 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I suspect that you have not gotten any responses because no one can determine exactly what question you are asking! What cancellation is it that you think should happen? August 10th, 2015, 11:54 AM #3 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 I see, Well I am confused because isn't ΔV=r*Δθ*Δr*Δz so if we are dividing: [F(r+Δr,θ ,z)-F(r,θ ,z)]r*Δθ*Δz/ΔV that is: [F(r+Δr,θ ,z)-F(r,θ ,z)](r*Δθ*Δz)/(r*Δθ*Δr*Δz) =[F(r+Δr,θ ,z)-F(r,θ ,z)]/Δr everything cancels but the Δr So then the small radial flux is approximately: [F(r+Δr,θ ,z)-F(r,θ ,z)]/Δr, which assume in the limit would be =dF/dr But then then it is always given as (1/r)*d(r*F)/dr August 10th, 2015, 12:50 PM #4 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 I think part of my problem is that a polar area actually has another polar term that disappears in the limit, but is significant before the limit. But I am not sure which. Last edited by Kinroh; August 10th, 2015 at 01:03 PM. August 10th, 2015, 01:47 PM #5 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 ok so MIGHT no what I'm thinking wrong: So the function at the far radial end would be: F(r+Δr,θ ,z) But the small surface area would be: ΔS=(r+Δr)*Δθ*ΔZ NOT ΔS=r*Δθ*ΔZ as this is the case for the inner radial end: so the flux of outer end is : F(r+Δr,θ ,z)*(r+Δr)*Δθ*ΔZ and on the inner end is: -F(r,θ ,z)*r*Δθ*ΔZ ( negative for opposite direction) so the Sum of the two radial flux is: [F(r+Δr,θ ,z)*(r+Δr)*Δθ*ΔZ]-[F(r,θ ,z)*r*Δθ*ΔZ] In which case only the Δθ*ΔZ can factor out, and not a radial term so: [F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*Δθ*ΔZ the small Volume is r*Δθ*Δr*ΔZ then the radial flux per volume is: [F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*Δθ*ΔZ/r*Δθ*Δr*ΔZ =F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*(1/r)*(1/Δr) Which in the limit would perhaps be (1/r)*d(r*F)/dr r*Δθ*Δr*Δz) August 10th, 2015, 02:14 PM #6 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 since it the flux per volume, there are two flux's summed up in two different locations.. Maybe the volume should really be the average volume... Meaning ΔV=(r+Δr/2)*Δθ*ΔZ*Δr but in the limit the it simply becomes r*dr*dθ*dZ...and the answer is the same Tags cancelation, divergence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Probability and Statistics 9 March 27th, 2015 07:40 AM razzatazz Real Analysis 8 May 10th, 2013 11:30 PM walter r Real Analysis 4 May 4th, 2013 07:34 PM math221 Calculus 3 April 8th, 2013 06:39 AM mathbalarka Calculus 3 September 2nd, 2012 01:30 AM

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