# Surface integral of a plane

I have this function

$$f(x,y,z) = x + y$$

and I want to find

$$\iint f(x,y,z) dS$$

Where S is the first octant part of the plane

$$x + y + z =1$$

Now, I know the method, and have paramatrized like this

$$z = 1 - x - y,\, x = x,\, y = y.$$ However, I thought the bounds of integration for x and y were between 0 and 1, but doing this gives me an answer of $3^{0.5}$, when the answer is $(3^{1/2})/3$, implying that the bounds that my textbook is using are different.

What are they?

• The condition of being in the octant is that $x, y, z \geq 0$. So lower limits of $0$ for $x$ and $y$ are correct, but you need to use that $z \geq 0$ for the upper limits. It may help to try to visualize projecting the part of the plane $x + y + z = 1$ onto the $x$-$y$ plane. In particular, you'll see that it is a familiar shape, but not a square. – Michael Joyce Dec 9 '12 at 22:44
• so, when z = 0? you get x + y = 1, or y = 1-x? Does that not give the same bounds? Obviously that's wrong, but for some reason I think you're trying to get me to say a 'triangle' – praks5432 Dec 9 '12 at 22:46
• oh wait a minute - that actually helped - mind posting that as an answer and I can accept it? – praks5432 Dec 9 '12 at 23:01

The region of integration is the part of the plane $x + y + z = 1$ that is cut out by the inequalities $x, y, z \geq 0$. The resulting surface is a triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. You can project the triangle onto the $x$-$y$ plane for your calculation, in which case your region of integration is given by $x, y \geq 0$ and $1 - x - y \geq 0$. If you integrate $y$ first, then the limits of integration are from $y = 0$ to $y = 1 - x$. Then $x$ has to go from $x = 0$ to $x = 1$. Then $dS = \sqrt{3} \, dx \,dy$, so this yields $$\iint f \,\, dS = \int_{x = 0}^{x = 1} \int_{y = 0}^{y = 1 - x} (x + y) \, \sqrt{3} \,\, dy \,\, dx.$$
• Edited because the original had the wrong $dS$ term.
$$\iint f(x,y,z) dS=\int_{0}^{1}\int_{0}^{1-x} (x+y) \sqrt{1+(\frac{dz}{dx})^{2}+(\frac{dz}{dy})^2}dy\,dx$$
$$= \sqrt{3} \int_{0}^{1}\int_{0}^{1-x} (x+y)dydx= \frac{\sqrt{3}}{3}.$$
Note that, S is the first octant part of the plane, then you will have the region in xy-plane bounded by the $x$-axis, $y$-axis and the $y=1-x$, $x\geq 0, y\geq 0.$