Question

What is if g(x,y,z) = x + y and S is the first octant portion of the plane 2x + 3y + z = 6 ?

Answer 1

The question asks for the value of
`I=\int\int_Sx+y\textrm{ }dS` where
`S=\{(x,y,z)\mid2x+3z+y=6,x\ge0,y\ge0,z\ge0\}`.

First let's look at what that surface looks like.

Letting
`y=z=0` yields
`x=3`
Letting
`x=z=0` yields
`y=2`
Letting
`x=y=0` yields
`z=6`

Therefore
`S` is the area of the triangle defined by the three points
`(3,0,0),(0,2,0),(0,0,6)`.

We can thus reformulate the integral as
`I=\int_(z=0)^6\int_(x=0)^(6-z)x+ydxdz`.

By definition on the plane
`y=\frac{6-2x-z}3` thus
`I=\int_(z=0)^6\int_(x=0)^(6-z)x+\frac{6-2x-z}3dxdz=\int_(z=0)^6\int_(x=0)^(6-z)2+\frac x3-\frac z3 dxdz`


`I=\int_(z=0)^6\left[2x+\frac{x^2}6-\frac{zx}3\right]_(x=0)^(6-z)dz=\int_(z=0)^62(6-z)+\frac{(6-z)^2}6-\frac{z(6-z)}3\right]dz`


`I=\int_(z=0)^6\frac{z^2}2-6z+18=\left[\frac{z^ 3}6-3z^2+18z\right]_(z=0)^6=36-108+108`

Hence
`\boxed{I=\int\int_Sx+y\textrm{ }dS=36}`