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Math Help - Surface area problem

  1. #1
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    Surface area problem

    Find the area of the surface.

    z = f(x,y) = 9 - x^2 over the region R: Square with vertices (0,0), (3,0), (0,3), (3,3).

    can anyone show me how to solve this problem? I seriously have no clue how to start.
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  2. #2
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    The function f(x,y)=9-x^2 defines a parabolic sheet of constant altitude extending over the y-axis. The cross-sections of the portion of the surface that lies over [0,3]\times[0,3] are the same for all values of y.

    Hint: We may find the surface area by first calculating the length of the curve defined by f(x)=9-x^2 from x=0 to x=3.
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  3. #3
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    I'm lost

    Quote Originally Posted by Scott H View Post
    The function f(x,y)=9-x^2 defines a parabolic sheet of constant altitude extending over the y-axis. The cross-sections of the portion of the surface that lies over [0,3]\times[0,3] are the same for all values of y.

    Hint: We may find the surface area by first calculating the length of the curve defined by f(x)=9-x^2 from x=0 to x=3.

    I can't get it. I just started surface integrals this week and I'm totally lost. I need to see some resolved problems so I can back engineer what's going on..
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  4. #4
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    The surface you are looking can be devided in a part where the graph has not entered the square (x=0 to sqrt(6)) and a part where the graph has entered the square (x=sqrt(6) to 3).

    The surface of the first part is easy. sqrt(6)x3


    The surface of the second part can be written as an integral.

    -x^2+9 integrated from sqrt(6) to 3 =
    27-27/3-9*sqrt(6)-(sqrt(6)^3)/3=0.85

    which leads to an surface of 8.20

    Hope this will help you!
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  5. #5
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    shouldn't there be 2 integrals to solve this problem?
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