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Vector Calculus - Cylindrical Co-ordinates (HELP PLEASE)

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!
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)
(edited 9 years ago)
Reply 2
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.
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
Reply 4
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?
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
Reply 6
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?
Reply 7
Original post by RigB
I got 8pi/3root3, does that sound right to you?


I( could help you but I am teaching all day

Try this link

http://madasmaths.com/archive_maths_booklets_advanced_topics.html

download file

multivariable_integration_in_polars
( long download time)
there are several cylindrical polar questions with solutions towards the end

See if it helps
Reply 8
Original post by TeeEm
I( could help you but I am teaching all day

Try this link

http://madasmaths.com/archive_maths_booklets_advanced_topics.html

download file

multivariable_integration_in_polars
( long download time)
there are several cylindrical polar questions with solutions towards the end

See if it helps


Afraid not sorry, I did read them through.
Reply 9
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.
Reply 10
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!
Reply 11
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

volume of ellipsoid is 4/3(pi)abc, so your answer is correct

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