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Compensation factor for converting dy dx to cylindrical coordinates?

  1. Aug 3, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the compensation factor for converting dy dx to cylindrical coordinates?

    2. Relevant equations
    None that I know of besides the bottom ones as part of the attempt

    3. The attempt at a solution
    So I know that the conversion formulas for going from Cartesian (x,y,z) coordinates to Cylindrical (r, θ, z) coordinates are as follows.
    ##x^2 + y^2 = r^2##
    ##x = r cos(θ)##
    ##y = r sin(θ)##
    ##z = z##
    "When evaluating an integral, the integrating factors have a slightly different formula." - My professor regarding these formulas at the above question.
    Last edited: Aug 3, 2013
  2. jcsd
  3. Aug 3, 2013 #2


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    Don't forget your jacobian.
  4. Aug 3, 2013 #3
    Since my class pretty much rests on the idea of giving us stuff we haven't really covered, I don't know how to use a Jacobian. Be right back, researching that.
  5. Aug 3, 2013 #4


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    You accidentally posted another thread I see. What is it you don't understand? I've already told you what you were missing.

    This should help you : http://en.wikipedia.org/wiki/Jacobian
  6. Aug 3, 2013 #5
    Thanks. I know how to find the Jacobian Determinant now, even the 3D ones. But I still don't know how to apply them to this problem to find a compensation factor.
    Last edited: Aug 3, 2013
  7. Aug 4, 2013 #6


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    What is a compensation factor? I've never heard this in connection with integrals.

    You find the area element in arbitrary coordinates [itex](q^1,q^2)[/itex] by using the Jacobian determinant, because this gives the area of the surface elements spanned by the coordinate lines,
    [tex]d^2 \vec{x}=\mathrm{det} \frac{\partial (x^1,x^2)}{\partial (q^1,q^2)} \mathrm{d}x^1 \mathrm{d} x^2,[/tex]
    [tex]\frac{\partial (x^1,x^2)}{\partial (q^1,q^2)} = \begin{pmatrix}
    \partial x^1/\partial q^1 & \partial x^1/\partial q^2\\
    \partial x^2/\partial q^1 & \partial x^2/\partial q^2
    For polar coordinates you can even read it off from the geometry of the coordinate lines which are radial straight lines and circles around the origin. Obviously this gives immediately
    [tex]\mathrm{d^2} \vec{x}=\mathrm{d}r \mathrm{d} \varphi r,[/tex]
    which you can easily check using the Jacobian.
  8. Aug 4, 2013 #7


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    One way to approach this problem is purely "geometric": Imagine two points with coordinates [itex](rcos(\theta), r(sin(theta), z)[/itex] and [itex]((r+dr)cos(\theta+ d\theta), (r+ dr)sin(\theta+ d\theta), z+dz)[/itex]. Drawing the appropriate "coordinate lines" (lines from (0, 0, z) to [itex](rcos(\theta), r(sin(theta), z)[/itex], [itex](rcos(\theta), r(sin(theta), z+dz)[/itex], [itex]((r+ dr)cos(\theta+d\theta), (r+dr)sin(theta+d\theta), z+dz)[/itex], the cylinders at r and r+ dr, we have a curved figure that is almost a rectangular solid. The "height" is dz, the "width" is dr and the smaller "length" (the curved sides) is [itex]rd\theta[/itex] (the larger is (r+ dr)d\theta). The difference between those two lengths won't matter in the limit so we can take as the volume the product of those: [itex](dz)(dr)(r+ dr)d\theta= rdrd\theta dz+ (dr)^2d\theta dz[/itex]. As those all go to 0, the term with [itex]dr^2[/itex] will go to 0 faster than the first term so in the limit we have [itex]rdrd\theta dz[/itex].

    (Of course, since "cylindrical coordinates" is just "polar coordinates" with z added, the differential of volume is just the differential of area in polar coorrdinates, [itex]rdrd\theta[/itex] with "dz" multiplied.)
  9. Aug 4, 2013 #8


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    I'm assuming what your professor is trying to show you is this :

    ##\int \int \int_R f(x,y,z) dxdydz = \int \int \int_{R'} f(r, \theta, z) |J| drdθdz##

    This equivalence is actually based off a much simpler relationship. Try convincing yourself of this first ( and then after convincing yourself, you will be able to show a much more general case and then even extend it to higher dimensions yourself ). This method is a good way to understand this, but if you don't know about curves and such it might not be the best.

    Suppose that ##R## and ##R'## are bounded, connected regions such that Green's Theorem holds. Then :

    Area of a region R = ##\int \int_R 1dxdy = \int \int_{R'} |J|dudv## where |J| is the Jacobian.

    How do we show this?

    Suppose the boundary of ##R## is a smooth curve ##C## given by ##P(t) = (x(t), y(t))## for ##t \in [a,b]## such that :

    ##\int_C x(t) dy = \int \int_R 1dxdy##

    Now, take the boundary of ##R'## to be another smooth curve ##C'## oriented so that the transformation ##u = h(x,y), v = k(x,y)##
    maps ##C → C'##.

    Then we have ( Note I'm using the chain rule here and the subscripts for the functions are the derivatives of that function with respect to that variable ) :

    ##\int_C xdy = \int_{a}^{b} x(t)y'(t) dt = \int_{a}^{b} f(u,v) \frac{∂}{∂t} (g(u,v)) dt = \int_{C'} f(u,v) g_u(u,v) du + f(u,v) g_v(u,v) dv##

    Continuing ( I'll drop the (u,v)'s for the functions to make this easier to read ) :

    ##\int_{C'} fg_u du + fg_v dv = ± \int \int_{R'} \frac{∂}{∂u}(fg_v) - \frac{∂}{∂v} (fg_u) dudv = ± \int \int_{R'} f_ug_v - f_vg_u dudv = ± \int \int_{R'} J dudv = \int \int_{R'} |J| dudv ##

    Hence the result is shown. Now using this result, you can show how to compute the area for any general function in any dimension.
    Last edited: Aug 4, 2013
  10. Aug 5, 2013 #9

    Thanks. I just looked through your post and it looks like I got the answer right.
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