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Polar coordinates

  1. Jun 17, 2005 #1
    Hi, I'm having problems with converting to polar coordinates when evaluating integrals. Here is an example, it comes down to writing the following equality. After that the evaluation of the intergral is straight forward.

    [tex]
    \int\limits_0^2 {\int_{ - \sqrt {4 - y^2 } }^{\sqrt {4 - y^2 } } {x^2 y^2 dxdy = } } \int\limits_0^\pi {\int\limits_0^2 {\left( {r^4 \cos ^2 \theta \sin ^2 \theta } \right)} } rdrd\theta
    [/tex]

    Looking at the original integral I can see that if I equate "x" to the upper or lower limit of the inner integral then I get a x^2 + y^2 = 4, which is the equation of a circle of radius 2. But I'm not sure how to convert to polar coordinates the right way.

    Firstly I see that dy corresponds to the outer integral(of the LHS) from y = 0 to y = 2 but what can I do with that? Not much, at least now anyway so I then observe that dx corresponds to the upper and lower limits of the inner integral. So I figured that means a point on the circle would travel from x = -2 to x = 2, then the outer part of the 'polar' integral would have limits 0 to pi/2. But that's wrong.

    Actually...I can sort of see why the second integral is correct now. The main problem is that I don't completely understand how the conversions of the upper and lower limits of integration proceed. I kind of understand the r = 0 to r = 2 part but I don't really understand outer integral with zero and pi. Can someone please explain it to me?
     
  2. jcsd
  3. Jun 17, 2005 #2
    You converted the integrand correctly. you are right in identifying the domain of integration as a circle of radius 2, but you didn't reflect that accurately in your final itnegration limits. Identify the limits of 'r', how far does r go in a circle of radius 2? it starts at the origin and goes out to 2, thas correct. To sweep out the circle what angle do you ahve to cover? 0 to Pi covers the positive y region of the circle (upper half), but you have neglected the bottom half, how would you amend this?
     
  4. Jun 17, 2005 #3
    The answer has a diagram which only shows the upper part of the circle and having thought about the question more, I can't see much wrong with it(in other words I now agree with the RHS of the equation in my first post). Here is why I think so.

    In the original integral, the +/-sqrt(4-y^2) are just the x-values. The square root function is always greater than or equal to zero for real numbers. So from that I infer that the 'range' of x-values would then just be [-2,2]. Continuing with my observation of the original integral the 'range' of y-values is [0,2]. A semi-circle in the upper half of the x-y plane satisifies these conditions but a full circle doesn't seem to as it would include negative y-values. So I think that I only need to worry about the upper part of the circle.
     
  5. Jun 17, 2005 #4

    OlderDan

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    It is only the upper part of the circle. If it were the whole circle the original y integral limits would have been from -2 to +2 instead of 0 to 2.
     
  6. Jun 17, 2005 #5

    HallsofIvy

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    Now, the right end of the base diameter of the circle is at (2, 0). That corresponds to θ= 0 (θ= arctan(y/x)= 0).
    The left end of the base diameter of the circle is at (-2, 0). That corresponds to θ= π.
     
  7. Jun 17, 2005 #6

    OlderDan

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    I just read all this again, and it occurs to me that maybe what you are not clear on are these x limits. The range of x is a function of y on the left hand side integral because the x integral corresponds to a horizontal chord whose length gets shorter as y increases. The x interval of integration is [-2,2] only when y = 0 and shrinks as y increases until it becomes zero at y = 2.

    You can imagine "painting" the area of the semicircle with a brush of width dy starting on the x axis with a horizontal stroke of length 4, followed by shorter horizontal strokes (corresponding to integrals over x) as you move upward. Successive strokes correspond to integrating over y. Alternatvely, you could paint the area using semi-circular strokes of width dr and length r*pi (corresponding to integrals over theta) starting at the origin and moving outward. Successive strokes correspond to integrating over r. Either technique will cover the entire area.
     
  8. Jun 17, 2005 #7
    Gah, sorry about that, it was really late. I should've seen it.
     
  9. Jun 18, 2005 #8
    Thanks for the help.
     
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