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Homework Help: Convert Cartesian To Cylindrical Limits

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data


    I am converting from Cartesian co-ordinates to cylindrical co-ordinates systems and can do the conversion fine, but am unsure about how to convert the limits.


    Region V is given in Cartesian by: [tex]F(x,y,z) = xi+yj+z(x^{2}+y^{2})k[/tex]

    Where [tex]x^{2}+y^{2} \leq 1[/tex]
    and [tex]0 \leq z \leq 2[/tex]

    2. Relevant equations

    [tex] x = \rho cos(\varphi), y = \rho sin(\varphi), z = z[/tex]
    [tex] cos^{2}(x)+sin^{2}(x) = 1[/tex]

    3. The attempt at a solution

    [tex] F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{\rho} + \rho sin(\varphi)\widehat{\varphi} + z\rho^{2}\widehat{z}[/tex]

    I know that z has stayed the same as it has not changed, but im not sure how to do the [tex]\rho[/tex] and [tex]\varphi[/tex].

    If you could help me on how to do this id really appreciate it.

    Many thanks in advance
  2. jcsd
  3. Apr 29, 2010 #2


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    It looks as though you've magically replaced [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] with [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]....are you really allowed to do that?:wink:

    What are [itex]\hat{\mathbf{x}}[/itex] and [itex]\hat{\mathbf{y}}[/itex] in terms of [tex]\hat{\mathbf{\rho}}[/itex] and [tex]\hat{\mathbf{\varphi}}[/itex]?

    Well, your other boundary is [itex]x^2+y^2\leq 1[/itex]....so, what is [itex]x^2+y^2[/itex] is cylindrical coordinates?
    Last edited: Apr 29, 2010
  4. Apr 29, 2010 #3
    Sorry, Im unsure what you mean here?

    Ok, I think I see what you mean here... Would it be:

    [tex]\rho \leq 1[/tex]
    [tex]x^{2}+y^{2} \leq 1[/tex]
    [tex] = \rho cos^{2}(\varphi) + \rho sin^{2}(\varphi)[/tex]

    If this is correct, Im still unsure how to get [tex]\varphi[/tex]?

    Thanks for the speedy response!
  5. Apr 29, 2010 #4


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    When you substitute [itex]x=\rho\cos\varphi[/itex] and [itex]y=\rho\sin\varphi[/itex] into [itex]\textbf{F}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z(x^2+y^2)\hat{\mathbf{z}}[/itex], you get [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{x}}+\rho\sin\varphi\hat{\mathbf{y}}+z\rho^2\hat{\mathbf{z}}[/itex], not [itex]\textbf{F}=\rho\cos\varphi\hat{\mathbf{\rho}}+\rho\sin\varphi\hat{\mathbf{\varphi}}+z\rho^2\hat{\mathbf{z}}[/itex]

    [tex]\hat{\mathbf{x}}\neq\hat{\mathbf{\rho}} \;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; \hat{\mathbf{y}}\neq\hat{\mathbf{\varphi}}[/itex]

    In your text, you should have derived expressions of the form [itex]\hat{\mathbf{x}}=f(\rho,\theta)\hat{\mathbf{\rho}}+g(\rho\theta)\hat{\mathbf{\varphi}}[/itex] and [itex]\hat{\mathbf{y}}=F(\rho,\theta)\hat{\mathbf{\rho}}+G(\rho\theta)\hat{\mathbf{\varphi}}[/itex] ...use those

    Well, you know that [itex]0\leq\rho\leq 1[/itex], so these are your integration limits for [itex]\rho[/itex]. As for [itex]\varphi[/itex], there are no further restrictions on its value, so you just go with the conventional [itex]0\leq\varphi\leq 2\pi[/itex].
  6. Apr 30, 2010 #5

    I have searched my notes, and my text books, but do not have anything which seems to cover this. Is there a particular name for this operation that I may be able to search up and check out some examples?

    Last edited: Apr 30, 2010
  7. Apr 30, 2010 #6


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    I have a hard time believing that your course textbook doesn't cover unit vector conversions. What text are you using for this course?

    Anyways, see here
  8. Apr 30, 2010 #7
    Thats amazing, thanks. I hope I have now got it correct.

    [tex]F(\rho,\varphi,z) = \rho cos(\varphi)\widehat{i} + \rho sin(\varphi)\widehat{j} + z\rho^{2}\widehat{z}[/tex]


    [tex]\widehat{\rho} = \widehat{i} cos(\varphi) + \widehat{j} sin(\varphi)[/tex]

    Subbing that in Im just left with:
    [tex]F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}[/tex]

    Is that looking to be correct?

    Thanks for all your help, you have been very good to me.

    BTW, I am using KA Stroud, Advanced Engineering Maths, but cannot find it in there still.
  9. Apr 30, 2010 #8


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    Looks good to me!:approve:
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