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Math Help - surface area of portion of plane

  1. #1
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    surface area of portion of plane

    find the surface area of that portion of the plane z=1 + 2x - 3y that lies above the xy-plane

    i don't understand how to set up the integral for this and i think it is easier to do it one way either dxdy or dydx but i am so lost...can any one help me????
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  2. #2
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    Unless you are talking about the portion that lies above the 2nd quadrant in the xy plane the surface area is infinite.

    I think that this is what you are looking for.

    surface area of portion of plane-graph.jpg

    If so this will work

    z=1+2x-3y

    using the formula

    \iint_S dS=\iint_D\sqrt{1+\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}}dA=\iint_D\sqrt{1+(2)^2+(-3)^2}dA=\sqrt{14}\iint_D dA

    From here the calculation of the integral can be done by letting y go from y=0 to y=\frac{2}{3}x+\frac{1}{3} and x from x=-\frac{1}{2} to x=0

    Or you can notice the region in the 2nd quadrant is a triangle with a base of \frac{1}{2} and a height of \frac{1}{3}

    A=\frac{1}{2}bh=\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)\left( \frac{1}{3} \right)=\left( \frac{1}{12} \right)

    So finally we get

    \sqrt{14}\iint_D dA=\sqrt{14}\left( \frac{1}{12}\right)=\frac{\sqrt{15}}{12}
    Last edited by TheEmptySet; July 7th 2008 at 11:44 AM. Reason: error in addition Thanks Galactus
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  3. #3
    Eater of Worlds
    galactus's Avatar
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    Empty done a fine job, but if I may show you one other little thing about these.

    We can set it up using dydx, dzdy, or dzdx and get the same thing.

    For dzdy

    \int_{0}^{\frac{1}{3}}\int_{0}^{1-3y}\frac{\sqrt{14}}{2}dzdy=\frac{\sqrt{14}}{12}

    What we do is solve the original equation for x, thus getting

    x=\frac{z+3y-1}{2}. Now, if we set x and z = 0 and solve for z we get the limit for z. Which is 0 to z=1-3y

    Of course, to get the y limit, set x and z=0 and solve for y and we see y=1/3.

    To do dzdx, it works the same way and we get:

    \int_{-\frac{1}{2}}^{0}\int_{0}^{2x+1}\frac{\sqrt{14}}{3}  dzdx=\frac{\sqrt{14}}{12}

    See what I mean?.
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