brendan_foo
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Hi there,
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} :
\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}}
So this is the question at hand :
Evaluate the surface integral : \iint_S G(x,y,z) dS
where S is the portion of the plane x + y + z = 1 in the first octant, and G(x,y,z) = z
So, finding the partial derivatives etc.. i arrive at :
\sqrt{3}\int_0^1 \int_0^1 z dx dy which is equivalent to : \sqrt{3}\int_0^1 \int_0^1 (1 - x - y) dx dy...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 :
\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 if I am not mistaken..
The official solution however is \frac{\sqrt{3}}{6}
Help :( :(
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} :
\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}}
So this is the question at hand :
Evaluate the surface integral : \iint_S G(x,y,z) dS
where S is the portion of the plane x + y + z = 1 in the first octant, and G(x,y,z) = z
So, finding the partial derivatives etc.. i arrive at :
\sqrt{3}\int_0^1 \int_0^1 z dx dy which is equivalent to : \sqrt{3}\int_0^1 \int_0^1 (1 - x - y) dx dy...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 :
\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 if I am not mistaken..
The official solution however is \frac{\sqrt{3}}{6}
Help :( :(