Question

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

Answer 1

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

`\displaystyle\iint_SG(x,y,z)\,\mathrm ds = \int_0^1\int_0^1 G((1-u)(1-v),u(1-v),v) \|\mathbf n\| \,\mathrm du\,\mathrm dv \\\\= \sqrt3 \int_0^1\int_0^1uv(1-v)^2\,\mathrm du\,\mathrm dv \\\\= \boxed{\frac1{8\sqrt3}}`

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

`\displaystyle\iint_S G(x,y,z)\,\mathrm ds = \int_{S_(xy)}G(x,y,z)\sqrt{1+\left((\partial f)/(\partial x)\right)^2+\left((\partial f)/(\partial y)\right)^2}\,\mathrm dx\,\mathrm dy`

where I write
`S_(xy)` 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
`S_(xy)` is the triangle,

{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x}

Then the integral becomes

`\displaystyle\int_0^1\int_0^(1-x)y(1-x-y)√(1+(-1)^2+(-1)^2)\,\mathrm dy\,\mathrm dx \\\\= \sqrt3\int_0^1\int_0^(1-x) y(1-x-y)\,\mathrm dy\,\mathrm dx \\\\= (\sqrt3)/(24) \\\\= \boxed{\frac1{8\sqrt3}}`