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Math Help - surface area problem..

  1. #1
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    surface area problem..

    been trying to solve the folowing area calc:
    <br />
S=\{(x,y,z) |  x^2 + z^2 = 1, y^2+z^2\leq 1. z\geq 0\}<br />

    so it should be the area of the surface z=\sqrt{1-x^2}-\sqrt{1-y^2}

    then to use the formula for area integrals over uneven surfaces
     \int \int \sqrt{ (\frac{df}{dx})^2 + (\frac {df}{dy})^2 + 1}

    i derive as follows:

     <br />
\frac{dz}{dx} =  \frac{-x}{ \sqrt{1-x^2}}<br />
     <br />
\frac{dz}{dy} = \frac {y}{\sqrt{1-y^2}}<br />

    and then in to the formula:

     \int \int \sqrt{ (\frac{-x}{ \sqrt(1-x^2)})^2 + (\frac {y}{\sqrt(1-y^2)})^2 + 1}dxdy

    but then i'm stucked, anyone got an idea how to solve this one?
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  2. #2
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    Okay you have,
    <br />
z=\sqrt{1-x^2}-\sqrt{1-y^2}<br />
    Then,
    \frac{\partial z}{\partial x}=-\frac{x}{\sqrt{1-x^2}}
    And,
    \frac{\partial z}{\partial y}=\frac{y}{\sqrt{1-y^2}}
    Then, square them,
    \frac{x^2}{1-x^2}
    And,
    \frac{y^2}{1-y^2}
    Add them,
    \frac{x^2}{1-x^2}+\frac{y^2}{1-y^2}
    Which is,
    \frac{x^2(1-y^2)+y^2(1-x^2)}{(1-x^2)(1-y^2)}
    Thus,
    \frac{x^2-x^2y^2+y^2-x^2y^2}{(1-x^2)(1-y^2)}
    Combine,
    \frac{x^2-2x^2y^2+y^2}{(1-x^2)(1-y^2)}
    Add 1,
    \frac{x^2-2x^2y^2+y^2}{(1-x^2)(1-y^2)}+1
    Add fractions,
    \frac{x^2-2x^2y^2+y^2+1-x^2-y^2+x^2y^2}{(1-x^2)(1-y^2)}=\frac{1-x^2y^2}{(1-x^2)(1-y^2)}
    So the surface area is,
    \int_A \int \sqrt{\frac{1-x^2y^2}{(1-x^2)(1-y^2)}} dA

    Now, I did not simplify this integration. I am thinking about it. I tried using a Jacobian but alas not success yet.
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  3. #3
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    yepp, have been going that way, been trying reducing it to
    <br />
\frac{x^2+y^2-2x^2y^2}{(1-(y^2+x^2)+x^2y^2)} + 1<br />

    and then switch to polar coordinates but it get's even mesier that way, but it doesn't seem to be a very suitable area for polar coords anyway.
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  4. #4
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    just a thought, since y^2+z^2 \leq 1 doesn't that mean that the half tube in the zy plane is filled? witch makes
    <br />
z=\sqrt{1-x^2}-\sqrt{1-y^2}<br />
    incorrect?
    since  z=\sqrt{1-x^2}-\sqrt{1-y^2} makes something that looks like a hyperbolic paraboloid..
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