Integrate G(x, y,z)=yz over the surface of x+y+z=1 in the first octant.​

Parameterize the surface (I'll call it S) by

r(u, v) = (1 - u) (1 - v) i + u (1 - v) j + v k

with 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.

Take the normal vector to this surface to be

n = ∂r/∂u × ∂r/∂v = ((v - 1) i + (1 - v) j) × ((u - 1) i - u j + k) = (1 - v) (i + j + k)

with magnitude

||n|| = √3 (1 - v)

Then in the integral, we have

Alternatively, if you're not familiar with parameterizing surfaces, you can use the "projection" formula:

where I write to mean the projection of the surface onto the (x, y)-plane, and z = f(x, y). We would then use

x + y + z = 1   ==>   z = f(x, y) = 1 - x - y

and is the triangle,

{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x}

Then the integral becomes

the answer is ! ?