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Convert the integral into polar coordinates

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Convert the following integrals into polar coordinates and then calculate them.

    a) int(0 , 2^(1/2)) int(y, [(4-y^2)^1/2]) xydxdy .


    2. Relevant equations
    x = rcostheta
    y = rsintheta
    r = (x^2 + y^2)^(1/2)


    3. The attempt at a solution

    Would it simply be:

    int(0 , 2^(1/2)) int(rsintheta, [(4-(rsintheta)^2)^1/2]) r^2sinthetacostheta r dr dtheta

    Is that all I have to do to convert it? I know how to integrate..I just need a bit of help with converting it into polar coordinates.

    Thanks.
     
  2. jcsd
  3. Oct 3, 2007 #2

    dynamicsolo

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    It looks to me like almost everything is converted. If I'm reading your statement correctly, your integration limits on y are 0 to sqrt(2). Won't that need to be converted also? (It might be a good idea to draw a picture of the integration region to see how to express it in polar coordinates. That may also make it clearer how to deal with the other set of integration limits.)
     
    Last edited: Oct 3, 2007
  4. Oct 3, 2007 #3

    HallsofIvy

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    No, no, no! You have to change the limits of integration to polar coordinates as well. Draw a picture. x= y is a straight line. [itex]x= (1-y^2)^{1/2}= \sqrt{1- y^2}[/itex] is the right half of the unit circle [itex]x^2+ y^2= 4[/itex]. You are integrating over the eighth of a circle between the x-axis and the line y= x. What are the upper and lower limits of r in that region? What are the upper and lower limits of [itex]\theta[/itex]? You should find that the limits of integration are all constants!

    The "No, no, no" was a response to inner08, not to dynamicsolo, who got in just before me.
     
  5. Oct 3, 2007 #4
    Ok. I drew a graph. From looking at it, r would be from 0 to 1 and theta from 0 to pi/2?

    By the way, shouldn't x=y and x=(4-y^2)^1/2 instead of x=(1-y^2)^1/2? If that is the case, r would be between 0 and 2 I think.
     
    Last edited: Oct 3, 2007
  6. Oct 3, 2007 #5

    D H

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    How did you get r from 0 to 1 and theta from 0 to pi/2? Regarding the latter, pi/2 corresponds to an angle of 900, or the y-axis. The line y=x represents a different angle.

    Regarding the former, you have y alone going from 0 to [itex]\surd2[/itex]. In other words, y just by itself goes outside the unit circle. BTW, the unit circle is [itex]x^2+y^2equiv1[/itex]. You need to find some larger circle than the unit circle.
     
  7. Oct 3, 2007 #6

    HallsofIvy

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    What is the radius of the circle? Yes, r starts at 0 but it has to go all the way out to the circle.

    Now look at [itex]\theta[/itex]. [itex]\theta= 0[/itex] to [itex]\pi/2[/itex] goes from the x-axis to the y-axis. Yes, you want to go up to the y-axis but you don't want to start at the x-axis, you want to start at the line y= x. What is [itex]\theta[/itex] for that?
     
  8. Oct 3, 2007 #7

    D H

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    You do not want to go the the y-axis. That corresponds to starting from x=0. The x integration starts at the line y=x and goes to the point on the circle that is at a height y above the x axis. Think of the x integration as following a line segment parallel to the x axis.
     
  9. Oct 3, 2007 #8
    Ok, so theta is pi/4 so its from theta is from (0,pi/4). As for r, i'm still a bit confused. You said it has to go out of the circle so can I pick whatever I want like 3 or 4?..or how do I find it?

    Thanks for being patient with me.
     
  10. Oct 3, 2007 #9

    D H

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    Method 1: Sketch out the integration region on some graph paper. The circle's radius should be obvious.

    Method 2: You already know its a circle, and you already know that it is the upper range if the x integration that hits the circle. You know the range of the y integration. What does this tell you about the circle's radius?
     
  11. Oct 4, 2007 #10

    HallsofIvy

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    I misread the problem! Didn't notice that it was integrating with respect to x first, then y. Guess I am too used to it being the other way around.
     
  12. Oct 4, 2007 #11

    HallsofIvy

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    No, you cannot just "pick" whatever number you like!
    The original problem said that the upper limit on the x-integral was x= [(4-y^2)^1/2]
    Squaring both sides, [itex]x^2= 4- y^2[/itex] so [itex]x^2+ y^2= 4[/itex]. That's the equation of a circle. What is the radius of that circle?
     
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