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Evaluating the triple integral $\int^1_0 \int^{1-z}_0 \int^{1-y-z}_0 \text{dxdydz}$, I get $\frac 16$.

Evaluating the triple integral $\int^1_0 \int^1_0 \int^1_0 \text{dxdydz}$, I get $1$.

So I subtract them like: $$\int^1_0 \int^1_0 \int^1_0 \text{dxdydz} - \int^1_0 \int^{1-z}_0 \int^{1-y-z}_0 \text{dxdydz} = \int^1_0 \left(\int^1_0 \int^1_0 \text{dxdy} + \int^0_{1-z} \int^0_{1-y-z} \text{dxdy} \right)\text{dz}= \int^1_0 \int^1_{1-z} \int^1_{1-y-z} \text{dxdydz}$$ and I get $\frac 23$, not $1-\frac 16 = \frac 56$. So I must be combining the integrals wrong. Why can't I do it this way?

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You cannot do it because there is no law of integration that you are applying. There is a law which says $$-\int_a^b BLAH = \int_b^a BLAH$$ And there is a law which says $$\int_a^b BLAH + \int_b^c BLAH = \int_a^c BLAH$$ These laws work with just a single integral sign at a time. You seem to be trying to stretch these laws to apply in improper ways to iterated integrals. But you've found a counterexample, proving that your stretched laws are false.

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  • $\begingroup$ So then how could I combine these 2 integrals into one integral? Is there some interated integral combination law I can apply? $\endgroup$
    – user184709
    Oct 16 '14 at 1:30
  • $\begingroup$ You will have to look at the regions of space over which you are integrating, and find a parametrization for the region which is their difference. $\endgroup$ Oct 16 '14 at 1:46

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