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Math Help - integral

  1. #1
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    integral

    Hey guys.

    I'm in the middle of this crazy exercise and I got stuck in this ugly integral in the bottom of the page, any ideas on how to solve it?
    Attached Thumbnails Attached Thumbnails integral-scan0008.jpg  
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  2. #2
    Super Member wingless's Avatar
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    What's the original question?
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  3. #3
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    The question is to find the area of this surface (in the pic).
    What it means, is to find the surface area of the cylinder between z=0 and z=sqrt(x^2+y^2).
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  4. #4
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    (It is assume that a>0)

    Let A be the disk (x-a)^2+y^2 = a^2.
    Then what you are looking for is \iint_A \sqrt{1+(\partial_1 f)^2+(\partial_1 f)^2}.
    Where f(x,y) = \sqrt{x^2+y^2}.

    This means, \partial_1 f(x,y) = - \frac{x}{\sqrt{x^2+y^2}} and \partial_2 f(x,y) = -\frac{y}{\sqrt{x^2+y^2}}

    Thus, we end up with \iint_A \sqrt{2} = \sqrt{2} \cdot \text{area}(A) = \sqrt{2}\pi a^2
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