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September 6th, 2017, 12:27 PM   #1
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Triple integral question

Hello folks. I have a question on this triple integral.

$$\iiint_S z \,dx\,dy\,dz$$ where S is bounded by $x+y+z=2$, $x=0$, $y=0$, and $z=0$

I have the work on my white board. I got an answer of $\frac{5}{4}$ ... I sketched out a 3D coordinate system and noticed that this looks like a tetrahedron with Vertices (2,0,0), (0,2,0), and (0,0,2). However, I did not like how the integral was set up in the question, so I decided to use Fubini's theorem to write it in the form of $$\iiint_S z \,dz\,dy\,dx$$ so I could eventually write the limits like ... $$\int\limits_{0}^{1}\int\limits_{0}^{(2-x)}\int\limits_{0}^{(2-x-y)} z \,dz\,dy\,dx$$

which lead me to the result of $\frac{5}{4}$. I just feel a little bit uncomfortable with switching around the differentials especially since the question had them set up in a particular way. Did I set this up correctly?

Thank you!

Jacob
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September 6th, 2017, 01:12 PM   #2
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You can switch the "differentials" (they aren't really differentials) as long as you get the limits right.

Why do you think $0 \le x \le 1$? I'd have guessed that it went up to 2. (Edit: now verified by a plot).
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Last edited by v8archie; September 6th, 2017 at 01:21 PM.
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September 6th, 2017, 03:05 PM   #3
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Originally Posted by v8archie View Post
You can switch the "differentials" (they aren't really differentials) as long as you get the limits right.

Why do you think $0 \le x \le 1$? I'd have guessed that it went up to 2. (Edit: now verified by a plot).
Oh my god... What a silly, silly mistake. Thank you.
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September 6th, 2017, 03:13 PM   #4
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Originally Posted by v8archie View Post
You can switch the "differentials" (they aren't really differentials) as long as you get the limits right.

Why do you think $0 \le x \le 1$? I'd have guessed that it went up to 2. (Edit: now verified by a plot).
Excuse me v8archie... I can move the dx, dy, and dz's around, but if my last integral still includes variables in the limits of integration, then I run into problems. So I suppose this is what you are saying by getting the limits right. I would need to reevaluate all the limits of integration?

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September 6th, 2017, 03:17 PM   #5
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I got an answer of $\frac{5}{4}$
How?
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September 6th, 2017, 04:51 PM   #6
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Originally Posted by SenatorArmstrong View Post
Excuse me v8archie... I can move the dx, dy, and dz's around, but if my last integral still includes variables in the limits of integration, then I run into problems. So I suppose this is what you are saying by getting the limits right. I would need to reevaluate all the limits of integration?
Typically, all of the limits can change, not just change position. There's quite a lot of symmetry in this space, so the changes aren't so radical. But if you swap the $x$ and $y$ integrals, the $x$ interval changes from $(0,2)$ to $(0,2-y)$ and the $y$ interval changes from $(0,2-x)$ to $(0,2)$.
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September 6th, 2017, 04:53 PM   #7
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Originally Posted by v8archie View Post
Typically, all of the limits can change, not just change position. There's quite a lot of symmetry in this space, so the changes aren't so radical. But if you swap the $x$ and $y$ integrals, the $x$ interval changes from $(0,2)$ to $(0,2-y)$ and the $y$ interval changes from $(0,2-x)$ to $(0,2)$.
Thank you for the clarification!
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September 6th, 2017, 04:53 PM   #8
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How?
It was a mistake with the final integral with respect to x. It should have be 0 -> 2, not 0 -> 1.
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September 6th, 2017, 07:05 PM   #9
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How?
$$\iiint_S z \,dx\,dy\,dz$$

$$ S = {(x,y,z) \; |\; 0 \leq x \leq 2, \; 0 \leq y \leq (2-x), \; 0 \leq z \leq (2-x-y)}$$

$$\int\limits_{0}^{2}\int\limits_{0}^{(2-x)}\int\limits_{0}^{(2-x-y)} z \,dz\,dy\,dx = \frac{2}{3}$$

Finally got something reasonable.
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September 7th, 2017, 03:34 AM   #10
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That looks okay. When you had 1 instead of 2, you shouldn't have got 5/4 (you missed a factor of 2 somewhere).
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