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Calculate triple integral(zdV) over the area $E$, where $E$ is bounded by the planes $y=0 ~ z=0$ and $x+y=2$ and the cylinder $y^2 + z^2 = 1$ in the first octant.

I have seen the correct answer already, however I have difficulty in understanding why the limits in the triple integral are taken in that way. Could anyone give me advice on what strategy I should use to determine the boundries of the integrals in such relatively complex cases?

Here is the answer is the correct for the sake of completeness:

$$\displaystyle \int\limits_0^1 \int\limits_0^{\sqrt{1-y^2}} \int\limits_0^{2-y} \ z~dx~dz~dy$$

Thanks in advance

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@Amzoti Thanks for your edit but you left out a 'z' in front of the dx –  Dreamer Feb 19 '13 at 18:42
Thank you @Amzoti –  Dreamer Feb 19 '13 at 18:45
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1 Answer

up vote 2 down vote accepted

You want it in the first octant so we have $$x\ge0, y\ge0,z\geq0$$ The lower limit of $y$ is zero but another limit of it is ruled by $y^2+z^2=1$. Since $z=0$ is one of our boundary , so $y^2=1$ and so $y=1$. Clearly, the latter equation gives us the restriction for $z$, that is: $$z\big|_0^{\sqrt{-y^2+1}}$$ Note that here again we have $z\ge 0$. For $x$ is the same, that is $$x\big|_0^{2-y}$$ since we already know that $x+y=2$.

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WOW! Nice graphic! You make it? +1 –  amWhy Feb 20 '13 at 0:23
Thank you. Your explenation and the answer make sense to me, however what I dont really understand is how for instance you could find the restriction for y by setting z=0 in y^2 + z^2 =1 but why this would not work to find the restriction for z. Since y=>0, couldnt you apply the same argument to reason that a restriction for z would be z=1 (by setting y=0 in y^2 + z^2 =1)? And couldnt you also reason in this way that a restriction for x would be x=2 (by setting y=0 in x+y=2)? Why could you do this in your cases and not here? –  Dreamer Feb 20 '13 at 17:30
@user48288: The point is to see that two equations has $y$ in shared, so we'd better give it the constant ranges and so what you thought about $z$, for example, would not be a proper one. –  B. S. Feb 20 '13 at 22:25
Sorry, Im still not entirely sure what you mean. Do you mean that since y appears in 2 equations it should have constant ranges? Why is that? –  Dreamer Feb 21 '13 at 14:41
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