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Homework Help: Double Integral Question

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    See Figure.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Simplifying the double integral,

    [tex]\int \int_{R} \sqrt{1 + 4x^2 + 4y^2} dA[/tex]

    Am I suppose to put in the bounds for part a, as part of simplifying the integral?

    This brings me to part b along with a few questions.

    The whole point of this problem is to compute the area between the two circles mentioned above correct? I'm confused as to how this is done with double integrals.

    If I wanted to find the area between the two circles I would simply find the area of the outer circle, [tex]x^2 + y^2 = 4[/tex] and subtract the area of the inner circle, [tex]x^2 + y^2 = 1[/tex].

    That being said, wouldn't the natural geometrical interpretation of this simply be something like a doughnut?

    I haven't gotten to part C yet, but I'd like to work out the misunderstandings I currently have and tackle that after.

    Thanks again for the help!
     

    Attached Files:

  2. jcsd
  3. Aug 24, 2010 #2

    Dick

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    Your integral isn't the area of the region between the two circles. It's the area of a surface z=y^2-x^2 defined over the area between the two circles, isn't it?
     
  4. Aug 25, 2010 #3
    That seems to make more sense, I think I got a little confused with the wording of the question.

    However, I still have a few more questions. Where exactly did you get z = y^2 - x^2?

    And how can I go about setting my boundaries for the given region?
     
  5. Aug 25, 2010 #4

    hunt_mat

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    You find the boundaries by turning the integral from cartesian to polar, so:
    [tex]
    dxdy=rdrd\theta
    [/tex]
    I think Dick is lookinf at the equation of a sphere:
    [tex]
    x^{2}+y^{2}+z^{2}=1
    [/tex]
    then
    [tex]
    z=\sqrt{1-x^{2}-y^{2}}
    [/tex]
     
  6. Aug 25, 2010 #5

    Dick

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    I'm looking at http://mathworld.wolfram.com/SurfaceArea.html equation 3. The surface must be z=y^2-x^2. It's not a sphere.
     
  7. Aug 25, 2010 #6
    Is what I wrote in the original post enough to satisfy part a? In other words, do I need to have the boundaries on my integral in order to have "simplified" it?

    Also, I'm still confused as to what the natural geometric interpretation of this integral is, is it simply a sphere?
     
  8. Aug 25, 2010 #7

    Dick

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    Are you asking what the surface z=y^2-x^2 is called? It's not a sphere. It's a hyperbolic paraboloid or something. It doesn't matter. The 'natural' interpretation is just that the integral is the area of that surface over the region between the two circles. Your dxdy form of the integral is fine without the bounds. Convert it to polar coordinates, then put in explicit boundaries.
     
  9. Aug 26, 2010 #8
    Pardon my calculus (which is pretty basic) if I'm wrong, but doesn't the double integral represent volume under the surface?
     
  10. Aug 26, 2010 #9

    Dick

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    The double integral of z*dx*dy would be volume. The double integral of sqrt(1+(dz/dx)^2+(dz/dy)^2)*dx*dy is surface area.
     
  11. Aug 26, 2010 #10
    You have definitely got to go out of your way to find a good Calculus book and review the section on Area of a Surface. Mine is Leithold. I guarantee if you find Leithold in the Library and spend no more than one hour on the section on surface area and then interpret your integral from that perspective, you'll see exactly what's going on: the integral is simply the area of the surface of [itex]z=y^2-x^2[/itex] over the annular region between the two circles with center at the origin and radii of 1 and 2.
     
    Last edited: Aug 26, 2010
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