# Vector Calculus - Cylindrical Co-ordinates (HELP PLEASE)Watch

#1
I am really struggling on how to do these, please may someone guide me through? Its the limits I struggle on. I would really appreciate it. Thanks!
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4 years ago
#2
Cylindrical rθz coordinates: r2 = x2 + y2, θ = arctan(y/x), z = z

The r and θ formulae should be rather familiar.
(If you are good at manipulating algebra then it should all be a walk in the park)
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#3
I understand the basis, I just dont understand how to get the individual limits for each integral for each question. Example 4.11a) I got z=sqrt((1-r^2)/3) to -((1-r^2)/3) and R=1 to -1 and theta=2pi to 0 but upon solving that, in the second integral r=1 and r=-1 cause the integral to become 0. Hence why I realise I have made a mistake.
0
4 years ago
#4
(Original post by RigB)
I understand the basis, I just dont understand how to get the individual limits for each integral for each question. Example 4.11a) I got z=sqrt((1-r^2)/3) to -((1-r^2)/3) and R=1 to -1 and theta=2pi to 0 but upon solving that, in the second integral r=1 and r=-1 cause the integral to become 0. Hence why I realise I have made a mistake.
Yeah that's the thing with solving integrals. The limits are correct but doing 2 * the integral with the limits r=0 and r=1 would get an answer that's not 0
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#5
(Original post by shawn_o1)
Yeah that's the thing with solving integrals. The limits are correct but doing 2 * the integral with the limits r=0 and r=1 would get an answer that's not 0
So if I did the integral with all bottom limits=0 and taking the scale factor 2 out for both top and bottom half of the ellipsoid I would get the answer as 4pi/3root3. Does that sound okay to you?
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4 years ago
#6
(Original post by RigB)
So if I did the integral with all bottom limits=0 and taking the scale factor 2 out for both top and bottom half of the ellipsoid I would get the answer as 4pi/3root3. Does that sound okay to you?
No, just the one integral involving -1/1 requires the 0/1, scale by two treatment
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#7
(Original post by shawn_o1)
No, just the one integral involving -1/1 requires the 0/1, scale by two treatment
I got 8pi/3root3, does that sound right to you?
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4 years ago
#8
(Original post by RigB)
I got 8pi/3root3, does that sound right to you?

multivariable_integration_in_pol ars
there are several cylindrical polar questions with solutions towards the end

See if it helps
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#9
(Original post by TeeEm)

multivariable_integration_in_pol ars
there are several cylindrical polar questions with solutions towards the end

See if it helps
Afraid not sorry, I did read them through.
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4 years ago
#10
(Original post by RigB)
Afraid not sorry, I did read them through.
I am sorry I could not be more helpful.

Explaining the limits particularly for volume/surface integrals in the more "exotic" coordinate systems is very hard to do without somebody explaining "face to face".

Hopefully somebody else will be more helpful.
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#11
That's okay and yes I understand that. Luckily I drew them out and it made more sense so I believe I have got them in the bag. I have a spherical which I'm uncertain on but I should be okay now. Thanks for your help all!
0
4 years ago
#12
(Original post by RigB)
So if I did the integral with all bottom limits=0 and taking the scale factor 2 out for both top and bottom half of the ellipsoid I would get the answer as 4pi/3root3. Does that sound okay to you?
the ellipsoid has a=1 b=1 c= 1/√3

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