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Math Help - Surface Area (2 Parts)

  1. #1
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    Surface Area (2 Parts)

    Find the surface area of that portion of the plane
    z = 1+2 x-3 y that lies above the triangular region in the xy -plane with vertices (0, 0 ), (3, 0) , and (0, 2) .

    I dont understand how to set up the integral or evaluate it....

    Using the same plane, z=1+2x-3y, find the surface area that lies above the xy-plane

    I am lost...can any one help????
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  2. #2
    Eater of Worlds
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    Find the equation of the line that passes from (3,0) to (0,2).

    We find it has equation y=\frac{-2}{3}x+2

    Now, z_{x}=2, \;\ z_{y}=-3, \;\ z_{x}^{2}+z_{y}^{2}+1=\sqrt{14}

    \int_{0}^{3}\int_{0}^{\frac{-2x}{3}+2}\sqrt{14}dydx
    Last edited by galactus; July 6th 2008 at 03:35 PM.
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  3. #3
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    ok i get that...i dont understand the xy-plane part???
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  4. #4
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    Quote Originally Posted by galactus View Post
    Find the equation of the line that passes from (3,0) to (0,2).

    We find it has equation y=\frac{-2}{3}x+2

    Now, z_{x}=2, \;\ z_{y}=-3, \;\ {\color{red}\sqrt{z_{x}^{2}+z_{y}^{2}+1}} =\sqrt{14}

    \int_{0}^{3}\int_{0}^{\frac{-2x}{3}+2}\sqrt{14}dydx
    Small typo (corrected in red) that has no affect on the final double integral given by the big G.

    Quote Originally Posted by jme44
    Is the answer 3 times the square root of 14 for the first question?
    Yes.
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