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Surface area of sqrt(x^2+y^2)

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    What's the surface area of the following 3D curve over the restricted range:
    z=f(x,y)=[tex]\sqrt{x^2+y^2}[/tex]
    0[tex]\leq[/tex]f(x,y)[tex]\leq[/tex]8

    2. Relevant equations
    **The answer is [tex]\sqrt{2}\pi[/tex]**

    The surface area equation (with partials)
    [tex]\sqrt{1+(Fx)^2+(Fy)^2}[/tex]

    Reduces to
    [tex]\sqrt{2}[/tex]

    So, for an as yet unknown integration range, we have
    [tex]\int\int\sqrt{2}dydx[/tex]

    3. The attempt at a solution
    Since the Z is restricted to [0,8] it would seem x and y should both be limited to [-8,8] but that integration range doesn't compute the the correct answer (listed above).

    What's the range of integral for both dy and dx?
     
  2. jcsd
  3. Nov 15, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    sqrt(x^2+y^2)=8 is a circle of radius 8, isn't it? What does that tell you about the domain? But I don't see how you are going to get sqrt(2)*pi out of that.
     
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