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I'm still reading on about surface integrals and stuff, and I've arrived at a formula that goes like this {based on plane projection methods} :

[tex] \iint_S G(x,y,z) dS = \iint_R G(x,y,z) \sqrt{1 + \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}}[/tex]

So this is the question at hand :

Evaluate the surface integral : [itex] \iint_S G(x,y,z) dS [/itex]

where S is the portion of the plane [itex] x + y + z = 1 [/itex] in the first octant, and [itex] G(x,y,z) = z [/itex]

So, finding the partial derivatives etc.. i arrive at :

[tex]\sqrt{3}\int_0^1 \int_0^1 z dx dy[/tex] which is equivalent to : [tex]\sqrt{3}\int_0^1 \int_0^1 (1 - x - y) dx dy[/tex]...I chose the limits for dx and dy as so.....

x + y + z = 1.... in the x plane... y+z = 0 thus x = 1, y = 0, and the same reasoning for y...

Computing this i get :

[tex]\sqrt{3}\int_0^1 (1 - \frac{1}{2} - y) dy = \sqrt{3}\left[ y - \frac{y}{2} - \frac{y^2}{2} \right]_{0}^{1} = 0[/tex] if im not mistaken..

The official solution however is [tex] \frac{\sqrt{3}}{6}[/tex]

Help :( :(

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# Simple surface integration blues {again}

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