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Homework Help: To understand polar coordinates

  1. Sep 28, 2005 #1
    Please I need help about polar coordinates. I have this expression:

    [tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu.[/tex]. Now they say:
    "Let's convert to polar coordinates. Define [tex] t = r cos \theta, u = r sin \theta[/tex]. Then [tex] t^2 + u^2 = r^2[/tex] (this is OK) and [tex] dtdu = rd \theta dr[/tex] (first why?) and the limits of integration become [tex] 0 < r < \infty, 0 < \theta < \frac{\pi}{2}[/tex] (second why??). We now have:

    [tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu = \int_0^{\infty}\int_0^{\infty}re^\frac{-r^2}{2}d \theta dr [/tex] (third why???)".

    I know this may be a boring question for you. Please don't just tell me "sustitute these differentials in the integrand by those other and you are done". I would like to understand and not only to apply a mecanism.

    As Borges said (in better words, obviously): "Oh happiness of understanding, greater than that of imagining or sensing".

  2. jcsd
  3. Sep 28, 2005 #2
    The limits of integration will change according to the coordinate system you change to (substitute your old values into the new equation to solve for theta and r). Have you heard of the Jacobian? That is where the "r" comes from in your differentials.

    [tex]\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc} x_u & x_v \\ y_u & y_v\end{array}\right|[/tex]

    Edit: Take a look at the bounds of your integrals. It's basically the whole first quadrant. You can figure this out by inspection and say that r will go from 0 to infinity and theta will go from 0 to pi/2.
  4. Sep 29, 2005 #3
    No, I dont have studied jacobians yet.

    Maybe you can suggest me some book or web page to know polar coordinates from cero? Unhappily I can't follow what you explain in your post.

  5. Sep 29, 2005 #4
    If you understand why
    and so on, there's not much else to be said. The Jacobian is something you'll study later, for right now just know that you need to throw in an extra r when you change to polar coordinates(when you learn other coordinate systems like spherical you'll have a different one)

    as for the limits, you redefine them in polar coordinates. You can't change from x and y to polar coordinates and still have your limits in terms of x and y, you need them in terms of r and theta. Which probably means, especially at first, you should draw a graph of the region of integration, and ask yourself "hmm, how would I describe this graph in polar coordinates?"
  6. Sep 29, 2005 #5
    Changing to polar coordinates

    But there is not a "theorem of change of variables from cartesian coordinates to polar coordinates" ??

    I know the classical theorem of change of variables:

    [tex] \int_{g(c)}^{g(d)}f(x)dx = \int_c^df(g(t))g'(t)dt[/tex].

    When we want to change from cartesian to polar coordinates, we dont have something that can be put in terms like this?

    (Excuse bad english).
  7. Sep 29, 2005 #6
    You could always try to look at it geometrically. If you don't know about the Jacobian, this is what I would suggest you do:

    Draw out a sector of a circle (the easiest for this case). Somewhere in there take a small piece that is almost rectangular (just a portion in the middle of the sector. Now, knowing that one side of the very small rectangle is dr, you can find the other side using this formula: s=rθ. But since that part of the angle is very small, you can call the length of that side rdθ. To find the area of that rectangle, multiply your two sides and you have dA=rdrdθ.

    It's hard to explain without a diagram, but I hope you can see what I'm talking about. I'll try to make something in paintshop if you can't see it...
  8. Sep 29, 2005 #7
    Apmcavoy, I understand the following facts, maybe you could tell me if with them I can walk towards the equation stated in my first post.

    1. If x, y are the cartesian coordinates, then [tex] x = f(r, \theta) = r cos(\theta), y = g(r, \theta) = r sin(\theta), [/tex].

    2. I know some basic partial derivatives, so
    [tex] f_r = cos (\theta), f_\theta = -r sin(\theta), g_r = sin(\theta), g_\theta = -r cos(\theta) [/tex]

    Then (without using the word "jacobian", which means nothing to my ignorance) I can see that

    [tex] (f_r)(g_\theta) - (f_\theta)(g_r) = r [/tex]

    Now, what I must do?
  9. Sep 29, 2005 #8
    Multiply that by drdθ. If you look at the Jacobian I posted above, you'll see that it's really saying your determinant is equal to dxdy/drdθ. So just solve for dxdy and substitute.
  10. Sep 29, 2005 #9


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    The thing to remember is that [tex]dt\;du[/tex] is a measurement of an area. It's a little square with one side of size dt, and the other side of size du.

    When you rewrite this into polar coordinates, you're going to have to replace dt du with the polar version of area. Instead of having a little square which is parallel to the t and u axes, you're going to have a little square which will look like a tiny square in a spider web.

    The side of the spider web that is on a radial line will have length [tex]dr[/tex]. The side of the spider web that is on a circular line will have length [tex]r \; d\theta[/tex]. The reason for the extra factor of r is that the squares in the spider web are bigger in area the farther away you get from the center of the web.

    The original limits of integration defined the "first quadrant" of the t-u coordinate system. The equivalent region for polar coordinates is theta between 0 and 90 degrees, and r between 0 and infinity.

    This is a misprint. It should read something like:

    [tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}\;dt\;du = \int_{r=0}^{\infty}\int_{\theta=0}^{\pi/2}r \;e^\frac{-r^2}{2}d \theta\; dr [/tex]

    This is just the substitution of the change in [A] coordinates from (u,t) to (r,theta), the change from du dt to r dr dtheta, and [C] the change in definite integration limits from over u,t to over r,theta.

    Last edited: Sep 29, 2005
  11. Sep 29, 2005 #10
    Thanks CarlB, Apmcavoy and Schattenjaeger por all your inputs.
  12. Sep 30, 2005 #11
    I copy from a web page (x and y are the cartesian coordinates):

    "[tex] (x,y) = T(r, \theta) = (r cos (\theta), r sin (\theta) ). [/tex]

    The function [tex] T(r, \theta) [/tex] gives rectangular coordinates in terms of polar coordinates. It maps a rectangle into a circle. {Here the author puts two graphics... a rectangle divide in cell-rectangles and a circle divided in "curvy rectangles" by a spider web (using CarlB's metaphor)} (...) we need to estimate the area of each "curvy rectangle" in the circle.

    It turns out that we can approximate each "curvy rectangle" as a parallelogram with sides [tex] {\frac{\partial T}{\partial r}}
    {\Delta r} [/tex] and [tex] \frac{\partial T}{\partial \theta}
    \Delta \theta [/tex]" .

    My question is: would some of you be so kind to justify rigorously this statement about the sides of those "parallelograms"?????

    Thanks in advance.
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