Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

2 Answers 2

up vote 2 down vote accepted

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 $z$.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.