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