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Homework Help: Polar coordinates and integral

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    I have a problem I hope you guys can help me with.
    It's quite simple I think, but there is one thing that I can't figure out.


    2. Relevant equations
    I have to use polar coordinates to evaluate this integral:
    See image


    3. The attempt at a solution
    I really don't have an attempt, because I can't remember, and it doesn't say so in my book, how I turn the dx limits into polar coordinates (theta).
    The dy limits are:
    x = r*cos(theta)
    y=r*sin(theta)

    , where you just insert them at the x and y. But it's the dx limits that annoy me. Do I just leave them as they are (Which sound unlogical to me), or do I change them into angles ? If yes, how ?


    Hope you can help.


    - Ylle
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2008 #2
    Hi,

    If you let [tex]x=r*cos(\theta) \ and \ y=r*sin(\theta)[/tex]

    then [tex]\frac{dx}{d\theta}=-r*sin(\theta) and \frac{dy}{d\theta}=r*cos(\theta)[/tex]

    You can then rearrange these to get [tex]dx=-r*sin(\theta)*d\theta \ and \ dy=r*cos(\theta)*d\theta[/tex]

    Your limits on the integals can be found by subbing the x and y values into the new expressions for x and y. Remember that the inside integral is with respect to y and the outside integral is with respect to x. Also r=sqrt(x^2+y^2).
     
  4. Nov 20, 2008 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Have you draw a graph of this?

    You have two circles, both with center (0,0), radius 1 and 2, and the straight line y= x.

    the line y= x crosses the circle x+ y2= 1 at [itex]x= 1/\sqrt{2}[/itex] and y= 0 crosses the circle at x= 1. That's the reason for the limits of integration in the first integral. That line y= x crosses the circle x2+ y2= 4 at [itex]x= \sqrt{2}[/itex]. That's where the limits on the second integral come from. The line y= 2 crosses the circle at y= 2 which is where the final limits come from. Look closely at your graph and you should be able to see why it is divided into 3 integrals.

    To change the order of integration, look at your graph. What are the least and greatest values of y? The least is, of course, y= 0. The greatest occurs where the line y= x crosses the larger circle which is [itex]\sqrt{2}[/itex]. Those will be the limits on the outer, dy, integral: y= 0 and [itex]y= \sqrt{2}[/itex]. Now, the limits on the inner, dx, integral for each y will be the left and right hand x-values of the horizontal line at y. You should be able to see that there will be two distinct integrals because the left-hand curve changes where y= x intersects the smaller circle: at [itex]y= 1/\sqrt{2}[/itex].
     
  5. Nov 21, 2008 #4
    But when I have the limits, do I then just integrate x*y with the limits I found, and then replace x and with x=r*cos(theta) and y=r*sin(theta) for dr dtheta ?
     
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