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Polar Double Integral

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the integral: [tex]\int_1^2\int_{\frac{3}{\sqrt{x}}}^{\sqrt{3}x}{{(x^2+y^2)}^{\frac{3}{2}}}dy \; dx[/tex]. Convert to polar and evaluate.

    2. Relevant equations
    [tex]r=\sqrt{(x^2+y^2)}[/tex]


    3. The attempt at a solution
    Ok, I've gotten bounds on [tex]\theta[/tex], [tex]\frac{\pi}{6} \le \theta \le \frac{\pi}{3}[/tex]. I'm not sure what the bounds for r should be, otherwise I have the integral:
    [tex]\int_\frac{\pi}{6}^\frac{\pi}{3}\int_{?}^{?}{r^3}dr \; d\theta[/tex]
     
  2. jcsd
  3. Nov 7, 2008 #2

    Dick

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    Do you have that x-y integral right? Did you draw a sketch of the region? At x=1 the lower limit of y is 3 and the upper limit is sqrt(3). At x=2 the lower is 3/sqrt(2) and the upper is 2*sqrt(3). The two limits are equal at theta=pi/3. This sounds really funny. At the very least, I don't believe your theta limits.
     
  4. Nov 7, 2008 #3
    That is the rectangular integral given in the book. I think the region is a quadrilateral-like thing that is enclosed by x=1,2 and y=rt(3)x, x/rt(3). I found the limits for theta by setting tan(theta)=rt(3), 1/rt(3) because then the thetas produce the necessary lines.
     
  5. Nov 7, 2008 #4

    Dick

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    That helps. In the problem statement you wrote 3/sqrt(x), not x/sqrt(3). Yes, the region is a trapezoid (a quadrilateral-like thing). Now I agree with your theta limits. That means your r limits will be a function of theta, right? Find them by intersecting a line at an angle theta through the origin with the two vertical lines in your quadrilateral-like thing.
     
  6. Nov 7, 2008 #5
    Wow, I'm sorry for that typo. That would change things. Would the correct limits for r be [tex]\frac{1}{cos(\theta)}\text{and}\frac{2}{cos(\theta)}[/tex]?
     
  7. Nov 7, 2008 #6

    Dick

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    They sure would. Don't forget dx*dy=r*dr*dtheta.
     
  8. Nov 7, 2008 #7
    Hmm, to check it, I plugged both expressions into my calculator and they're spitting out different numbers. The rectangular one gives me around 28.4 and the polar one gives me around 10.6.
     
  9. Nov 7, 2008 #8

    Dick

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    You've got a better calculator than I do. Did you catch "Don't forget dx*dy=r*dr*dtheta." I edited that into my last post.
     
  10. Nov 7, 2008 #9

    Dick

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    Ok, I've mastered my 'calculator'. I get 28.4 for both. Not exactly, of course.
     
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