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Double Integrals in polar coordinates: Calculus 3

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Given [itex]\int^{\sqrt{6}}_{0}[/itex][itex]\int^{x}_{-x}[/itex]dydx, convert to ploar coordinates and evaluate.



    2. Relevant equations
    We know that x=rcos[itex]\theta[/itex] and y=rsin[itex]\theta[/itex] and r =x^2+y^2


    3. The attempt at a solution

    First, I defined the region of the original integral: R = 0 [itex]\leq[/itex]x[itex]\leq[/itex] [itex]\sqrt{6}[/itex] and -x[itex]\leq[/itex]y[itex]\leq[/itex]x.

    In other words, we know that y is bounded on the lower limit by the line -x and the upper limit x. Because the line x produces a 45 degree angle, -[itex]\pi[/itex]/4 [itex]\leq[/itex] [itex]\theta[/itex] [itex]\leq[/itex] [itex]\pi[/itex]/4.

    However, I do not understand how to find domain of r. The region the original integral covers is not circular. We can prove this because at ([itex]\sqrt{6}[/itex],0), r = [itex]\sqrt{6}[/itex] and at ([itex]\sqrt{6}[/itex], [itex]\sqrt{6}[/itex]), which is still in the region defined by the original integral, r = [itex]\sqrt{12}[/itex] ([itex]\sqrt{\sqrt{6}^2+\sqrt{6}^2}[/itex])

    So that is where I am stuck. The answer to the problem according to the book is 6 but I do not know how to get there. All the examples we worked were circular regions.

    Additionally, assuming the region is 0 [itex]\leq[/itex] r [itex]\leq[/itex] [itex]\sqrt{6}[/itex] does not work either. [itex]\int^{pi/4}_{-pi/4}[/itex] [itex]\int^{sqrt{6}}_{0}[/itex] r dr d[itex]\theta[/itex] = (3[itex]\pi[/itex])/2
     
    Last edited: Feb 27, 2013
  2. jcsd
  3. Feb 27, 2013 #2
    3pi/4 is 135 degrees. According the limits of integration, x only varies from 0 to [itex]\sqrt{6}[/itex]. [itex]\theta[/itex] cannot go past the limits set by x.

    If [itex]\theta[/itex] varied from pi/4 to to 3pi/4 wouldn't the limits of x be different?

    However, I agree it is a triangle but I think the triangle is in positive x region, and the lower side of the triangle is the line -x.

    I am fairly confident on the bounds of [itex]\theta[/itex], but very confused by the bounds of r.
     
  4. Feb 27, 2013 #3

    Mark44

    Staff: Mentor

    You're right. I'm trying to multitask, and not doing to well at it. θ
    is exactly as you say.
    The limit for r is the vertical line x = √6. Going to polar coordinates, this is r cos(θ) = √6. You can solve this equation for r. The lower limit for r is, of course, 0.
     
  5. Feb 27, 2013 #4
    No problem, multitasking makes breathing hard, haha.

    That makes perfect sense though. That means the upper limit of integration depends on θ. So many problems in polar coordinates have static bounds, I completely forgot about that.

    Thanks!

    P.S. I just calculated the new integral, it is indeed correct.
     
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