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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.2012-08-12

2 Answers 2

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This is a community-wiki answer trying to remove this question from the unanswered queue.

$\displaystyle{\iint_\sigma} (x-y-z)d\sigma = \sqrt{2} {\iint_R} \big(x-(1-x)-z\big)dxdz $, where $R$ is the projection of the given region $\sigma$ on the $xz$ plane.

This is correct.


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

No, the projection of the plane on the $xz$-plane is a rectangle: $R = \{0\leq x\leq 1,\;\text{and }\;0\leq z\leq 1\}.$ There is no direct relation between $x$ and $z$.

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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.

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    Could you please something fixed? I didn't edit the question- it was a surface integral.2012-08-13