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Math Help - Surface Area Calc 3

  1. #1
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    Surface Area Calc 3

    My problem says find the surface area of the part of the surface
    z=y^2-x^2 that is inside the cylinder x^2+y^2=2

    So im thinking of polar coordinates
    saying f(x,y)=y^2-x^2
    and fx=-2x, fy=2y
    and the area=
    double integral of (sqrt(1+4x^2+4y^2))

    but my problem is how to go about finding the limits
    Im not sure exactly, can i get direction so i can learn
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by begin View Post
    My problem says find the surface area of the part of the surface
    z=y^2-x^2 that is inside the cylinder x^2+y^2=2

    So im thinking of polar coordinates
    saying f(x,y)=y^2-x^2
    and fx=-2x, fy=2y
    and the area=
    double integral of (sqrt(1+4x^2+4y^2))

    but my problem is how to go about finding the limits
    Im not sure exactly, can i get direction so i can learn
    The part of the surfact that is it inside the cylindar is above the disk

    x^2+y^2 \le 2 and yes polar coordinates will be the best system.

    \displaysytle \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}f(r,\theta)rdrd\  theta
    Last edited by TheEmptySet; December 10th 2010 at 09:06 AM. Reason: error in post
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  3. #3
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    If your integrating from 0-2pie first,
    would the limits of integration be the other way?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by begin View Post
    If your integrating from 0-2pie first,
    would the limits of integration be the other way?
    You are correct. I have fixed the above post.
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  5. #5
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    okk all done, thanks for the help...
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