Integrate $f(x,y,z) = x+ y+ z$ over the plane $2x + 2y + z = 8$ that lies in the first quadrant

Question

Integrate $f(x,y,z) = x+ y+ z$ over the plane $2x + 2y + z = 8$ that lies in the first quadrant

My answer:

I set up the following integral

$$\int\int_R\int_{z = 0}^{8-2(x+y)} (x+y+z) \:dz dR$$ where $R: x+ y = 4$ so we get

$$I = \int_{x=0}^4\int_{y=0}^{4-x}\int_{z = 0}^{8-2(x+y)} (x+y+z )\:dzdydx$$

Is this integral set up correctly?

Answer 1

Hint: You are asked to integrate $f(x,y,z)$ over the surface of the given plane (say $S$) in first octant, so it is a surface integral, i.e. $\iint_{S} f(x,y,z) dS.$

Now, on $xy$-plane, $dS=\sqrt{1+z_x^{2}+z_y^{2}}dxdy=3dxdy$. We use the equation of the plane in $f(x,y,z)$ which gives:

$$\iint_{S} f(x,y,z) dS=3\int_{x=0}^4 \int_{y=0}^{4-x} (8-x-y) dxdy.$$

I hope, you may take it from here.