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Integrating Polar Equations

  1. Jan 29, 2005 #1
    Can anyone point me to some online resources on how to integrate polar equations? Thanks.
     
  2. jcsd
  3. Jan 29, 2005 #2

    dextercioby

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    What do you mean by "polar equations"??

    Daniel.
     
  4. Jan 29, 2005 #3

    arildno

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    Aren't we online here?
    What's your problem; could you post an example?
     
  5. Jan 30, 2005 #4
    In rectangular coordinates, to integrate the function f from a to b, you find the antiderivative at a and subtract it from the antiderivative at b. Is there a similar definition to find the integral of a function that is defined in polar coordinates without converting to rectangular coordinates? For example, how would I integrate r = t^2 from t = 0 to t = 2pi?
     
  6. Jan 30, 2005 #5

    dextercioby

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    I don't know,applying the fundamental theorem of calculus,maybe?? :uhh:

    Daniel.
     
  7. Jan 30, 2005 #6
    [tex]A=\int_{\alpha}^{\beta} \frac{1}{2}r^{2}\,d\theta[/tex]

    for [itex]r=f(\theta)[/itex]. I'm assuming the problem was with finding the area and not the actual process of integration, which is the same as cartesian.
     
  8. Jan 30, 2005 #7
    The fundamental theorem of calculus?? How would that work to find the area of a polar equation??? In any case, I made some stuff up and I think I got it. Thanks for the help.
     
  9. Jan 30, 2005 #8
    Oops, me and Sirus posted at the same time. Thats the same formula I got. Thanks again.
     
  10. Jan 30, 2005 #9

    HallsofIvy

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    IF you are concerned with finding area in polar coordinates, then you have to remember that dArea= r dr dθ rather than the simple dxdy of Cartesian coordinates.
    HOWEVER, once you have the integral set up, the actual integration is exactly the same: find an anti-derivative and evaluate at the limits of integration.
     
  11. Jan 31, 2005 #10
    I probably missed this, but don't you also have to express the integrand in polar coordinates as well as find the jacobian?
     
  12. Jan 31, 2005 #11
    He did. The Jacobian of the transformation from polar coordinates to Cartesian coordinates is r.
     
  13. Feb 1, 2005 #12

    dextercioby

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    It's actually the other way around.What do you make of this
    [tex] J = |\frac{\partial (x,y)}{\partial (\rho,\phi)}|=...=r [/tex]

    Daniel.
     
  14. Feb 1, 2005 #13
    That is the Jacobian I was referring to, as it implies x and y are functions of r and theta. ;) The derivation in modern terms would be as follows:
    x = r cos [tex]\theta[/tex]
    y = r sin [tex]\theta[/tex]
    gives the transformation from polar coordinates into Cartesian coordinates. We use this one because we want to find [tex]dx\wedge dy[/tex] in terms of the "old" coordinates r and theta:
    [tex]dx \wedge dy &= (\cos\theta dr - r\sin\theta d\theta)\wedge(\sin\theta dr + r\cos\theta d\theta)\\
    &= r(\cos^2\theta + \sin^2\theta) dr\wedge d\theta\\
    &= r dr\wedge d\theta\\
    [/tex]
    Say you have a linear change of coordinates, u=2x, v=2y. The determinant of this forwards transformation tells us that a 2-volume of 1 in xy-space is transformed into a 2-volume of 4 in uv-space with respect to uv-coordinates (Lay the coordinate grids on top of each other).
    Then a 2-volume of 1 in uv-space corresponds to a 2-volume of 1/4 with respect to the xy-space (The square formed by (1,0) and (0,1) becomes the square formed by (1/2,0) and (0, 1/2)). If you were integrating in uv-space over uv-coordinates, you would multiply by a factor of 1/4 in order to get the 2-volume with respect to xy-space. This is the determinant of the backwards transformation from uv-space into xy-space. The generalization to nonlinear coordinates is clear, as the determinant is then integrated if it is not constant.
     
    Last edited: Feb 1, 2005
  15. Feb 1, 2005 #14

    dextercioby

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    I stressed out the fact that the Jacobian i gave (and with which u agreed) (namely "r") is not the jacobian of the transformation from polar ----> cartesian,but from cartesian -------->polar...That's what i meant by "It's actually the other way around"...

    Daniel.
     
  16. Feb 1, 2005 #15
    In my post I showed that it is the Jacobian of the transformation from polar coordinates into Cartesian that you gave and gave the motivation for that machinery. :)
    T(r, t) = (r*cos(t), r*sin(t)) = (x,y) is a map from polar coordinates (r,t) into Cartesian (x,y). The Jacobian of this transformation is your Jacobian using the partials of x and y with respect to r and t.
    On the other hand, the transformation from Cartesian to polar coordinates is
    [tex]
    T(x,y) = (\sqrt{x^2 + y^2}, \arctan(\frac{y}{x})) = (r, \theta )
    [/tex]
    whose Jacobian is
    [tex]
    \frac{x^2}{(x^2 + y^2)^{\frac{3}{2}}} + \frac{y^2}{(x^2 + y^2)^\frac{3}{2}}\\
    = \frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{r}
    [/tex]
    as predicted by the previous post.
     
  17. Feb 1, 2005 #16
    Actually, I was suggesting that [tex]r drd\theta[/tex] is the jacobian. And that you need to transform the integrand itself into polar coordinates.
     
  18. Feb 1, 2005 #17
    Hi Chrono,
    This may just be the terminology I've been exposed to, but in the studies I've done, the Jacobian is defined as the determinant of the Jacobian matrix, which is just r. The term [tex]r dr\wedge d\theta[/tex] is the pullback of the volume form of the transformation, which can be simplified to being just the new volume form scaled by the Jacobian. I can see taking the whole thing to be the Jacobian, though, by de-emphasizing the matrix and using only exterior algebra to define the Jacobian.
     
  19. Feb 1, 2005 #18
    You may be right. I was told by my professor in class that what I had said before was the jacobian.
     
  20. Feb 1, 2005 #19

    dextercioby

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    Well,your professor was obviously wrong.Knowing that this was a pretty simple thing and yet he screwed it up,i advise you to keep an eye out,because a lot more mistakes are about to come up...
    Just for our entertainment,please come and share them with us... :tongue2:

    Daniel.
     
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