# Surface integral - need a bit of explaining

I'm going through some old calculus, and I'm struggling a bit with surface integrals.

Here's the problem:

Compute the integral

$$\iint\limits_{\sigma} (x-y-z)d\sigma$$

where $\sigma$ is the plane $x+y=1$ in the first octant, limited by $z=0$ and $z=1$.

So, what I've done so far is to convert the equation for sigma as a function of $y$, i.e. $y = 1-x$

$\therefore\quad\frac{\partial y}{\partial x} = -1,\quad\frac{\partial y}{\partial z} = 0\\ \therefore\quad\sqrt{(\frac{\partial y}{\partial x})^2 + (\frac{\partial y}{\partial z})^2+1} = \sqrt{2}\\ \therefore\quad\displaystyle{\iint\limits_{\sigma}} (x-y-z)d\sigma = \sqrt{2} \displaystyle{\iint\limits_{R}} (x-(1-x)-z)dxdz$

where $R$ is the projection of the given region $\sigma$ on the $xz$ plane.

Simplifying:

$$\sqrt{2}\iint\limits_{R}(2x-1-z)dxdz$$

So, it seems that the projection is a right triangle, with vertices at $(0,0,0), (0,0,1), (1,0,0)$.

Am I on the right track, and, how do I proceed from here? I full worked out example would help me a lot.

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 Now you should go on with the double integral of $2x-1-z$ on the triangle. – enzotib Aug 12 '12 at 13:58

Limits: $$0\leq x\leq 1$$ $$0\leq y\leq 1-x$$ $$0\leq z\leq 1$$
So the integral is $$\int_0^1dx\int_0^1dz\int_0^{1-x}(x-y-z)dy=\int_0^1\int_0^1\left[(x-z)(1-x)-\frac{1}{2}(1-x)^2\right]dzdx=$$ $$=\int_0^1\left[(x-x^2)-\frac{1}{2}(1-x)-\frac{1}{2}(1-x)^2\right]dx=\int_0^1\left(-\frac{3}{2}x^2+\frac{5}{2}x-1\right)dx$$ and now you can finish the exercise.
 Is that a triple integral? I understood it was a surface integral. – enzotib Aug 12 '12 at 14:28 I'm thinking exactly what enzotib said. I thought the limits would be $0 \ge x \ge 1$, $0 \ge z \ge 1-x$. – Dhaivat Pandya Aug 12 '12 at 16:06 Rats, I misread the OP (or it was edited). I shall delete my answer and, perhaps, later post something fixed. – DonAntonio Aug 12 '12 at 16:48 Could you please something fixed? I didn't edit the question- it was a surface integral. – Dhaivat Pandya Aug 13 '12 at 9:07