The question asks for the value of [tex]I=\int\int_Sx+y\textrm{ }dS[/tex] where [tex]S=\{(x,y,z)\mid2x+3z+y=6,x\ge0,y\ge0,z\ge0\}[/tex].
First let's look at what that surface looks like.
Letting [tex]y=z=0[/tex] yields [tex]x=3[/tex]
Letting [tex]x=z=0[/tex] yields [tex]y=2[/tex]
Letting [tex]x=y=0[/tex] yields [tex]z=6[/tex]
Therefore [tex]S[/tex] is the area of the triangle defined by the three points [tex](3,0,0),(0,2,0),(0,0,6)[/tex].
We can thus reformulate the integral as [tex]I=\int_{z=0}^6\int_{x=0}^{6-z}x+ydxdz[/tex].
By definition on the plane [tex]y=\dfrac{6-2x-z}3[/tex] thus [tex]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[/tex]
[tex]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[/tex]
[tex]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[/tex]
Hence [tex]\boxed{I=\int\int_Sx+y\textrm{ }dS=36}[/tex]
