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 \(y=\dfrac{6-2x-z}3\) 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 \(\boxed{I=\int\int_Sx+y\textrm{ }dS=36}\)
