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Cartesian Integral to Polar Integral

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Change the Cartesian integral to an equivalent polar integral and evaluate

    ∫∫dydx

    The bounds of the first integral (The outermost) are -5 to 5, and the bounds of the second (inner) are 0 to [itex]\sqrt{ 25-x^{2}}[/itex]


    2. Relevant equations

    ∫∫dydx == ∫∫r(dr)(d[itex]\Theta[/itex])

    [itex]x^{2}[/itex]+[itex]y^{2}[/itex]=[itex]r^{2}[/itex]

    x = rCos([itex]\Theta[/itex])

    y = rSin([itex]\Theta[/itex])

    3. The attempt at a solution

    I solved for r by setting y equal to [itex]\sqrt{ 25-x^{2}}[/itex]

    after doing this I found r to be 5

    I know that typically the next step is to find the bounds for \Theta, but I have no clue as to how to do this, I know how to set up the integral and evaluate it, but I do not know how to determine the bounds for [itex]\Theta[/itex] can anyone please explain that to me?

    Thanks for your time.
     
  2. jcsd
  3. Nov 24, 2012 #2

    LCKurtz

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    Draw a picture of the region bounded by ##y=\sqrt{25-x^2}## and use it to figure the ##\theta## limits.
     
  4. Nov 24, 2012 #3
    Ok, so it would be 0 to 5[itex]\Theta[/itex]?
     
    Last edited: Nov 24, 2012
  5. Nov 24, 2012 #4

    LCKurtz

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    No. The outer limits must be constant. What does your picture look like?
     
    Last edited: Nov 24, 2012
  6. Nov 24, 2012 #5
    This is what I got when I plotted it using wolfram, I assumed giving the max on the Y axis that it would be 5theta, I apologize if this is something simple, I am having a really difficult time understanding it

    http://www.wolframalpha.com/input/?i=y+=+(25-(x^2))^(1/2)
     
    Last edited: Nov 24, 2012
  7. Nov 24, 2012 #6

    LCKurtz

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    I wasn't looking for you to show me a picture of what Wolfram gives. I want you do describe the given region in words so I know you know what you are looking at. You need to understand the region to figure out the limits.
     
  8. Nov 24, 2012 #7
    It's an arc in the first quadrant, with a y maximum of 5 and an x maximum of 5. I don't really know what you're asking, I apologize. This is holding me up on every problem I try, I have no clue how you get these values for theta...
     
  9. Nov 24, 2012 #8

    HallsofIvy

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    What are the polar coordianates of the point (5, 0) in Cartesian coordinates? What are the polar coordinates of the point (-5, 0) in Cartesian coordinates?
     
  10. Nov 24, 2012 #9

    LCKurtz

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    What kind of arc? What is it? An arch of a sine curve? A piece of a parabola? A chunk of an ellipse?
     
  11. Nov 24, 2012 #10
    Well, for (5,0) the Polar Coordinates (r,theta) would be r = [itex]\sqrt{(5^2)+0^2}[/itex] = 5
    then the theta value would be Tan[itex]^{-1}[/itex](0)

    or (-5,0) the Polar Coordinates (r,theta) would be r = [itex]\sqrt{(-5^2)+0^2}[/itex] = 5
    then the theta value would be Tan[itex]^{-1}[/itex](0)

    for (0,5) the Polar Coordinates (r,theta) would be r = [itex]\sqrt{(0^2)+5^2}[/itex] = 5
    then the theta value would be Tan[itex]^{-1}[/itex](5/0)
     
    Last edited: Nov 24, 2012
  12. Nov 24, 2012 #11

    LCKurtz

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    What angle is that, in radians?

    The same angle???
    5/0 is undefined. Anyway, just look at the picture and tell us the angles. Do you understand what polar coordinates measure? And you haven't answered what kind of graph that is.
     
  13. Nov 24, 2012 #12
    I don't really know what you mean by explaining the graph...It looks like an elipse, and I believe that polar coordinate measure real and imaginary parts of a number - That may be completely wrong. I am also fully aware that 5/0 is undefined I was just using the definitions presented in my book.

    I'm assuming that the measurement is in radians, I'm very lost with this topic and I apologize
     
    Last edited: Nov 24, 2012
  14. Nov 25, 2012 #13
    Ok, so I understand this a lot better now after reading the following link...feeling pretty stupid at the moment...

    http://www.mathsisfun.com/polar-cartesian-coordinates.html

    I now understand that the measurement is in Degrees after taking Tangeant inverse of y/x.

    My question now is, how do we determine the bounds over which to use this formula

    For example: from (5,0) to (0,5); or (5,0) to (-5,0) etc
     
  15. Nov 25, 2012 #14

    haruspex

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    It will help you enormously to acquire a grasp of what's going on rather than just reaching for formulae.
    When we do an integral in Cartesian, dxdy, we're breaking the region of interest up into little rectangles. One such rectangle is from x to x+dx in the x-direction and y to y+dy in y-direction. That's fine when you have two parallel straight line boundaries (like, x≥a, x≤b), but can get messy otherwise.
    In polar co-ordinates, we break it into little annular sectors, r going from r to r+dr and theta from θ to θ+dθ. (The area of such a sector becomes, in the limit, rdrdθ, so part of the conversion turns dxdy into that.)
    If r is the inner integral, that means you are first of all summing these little sectors from the origin out towards the value of r which bounds the area. Think of this as being done separately for each θ, so θ is constant during this summation. So the range for r in this case is... what?
    Having done that integral, we know the value for sectors of angular width dθ, running out from the origin. We now have to sum those as theta runs from its minimum value for the region out to its maximum value for the region. Having drawn the region, what are those minimum and maximum angles?
     
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