Rewrite the triple integral $$ \int_0^1\int_0^x\int_0^y f(x,y,z)\,dz\,dy\,dx $$ as $$ \int_a^b\int_{g_1(z)}^{g_2(z)}\int_{h_1(yz)}^{h_2(y,z)}\,dx\,dy\,dz $$ I found $a=0$ , $b=1$, $g_2(z)=1$, $h_2(y,z)=1$, so I need to find $g_1(z)$ and $h_1(y,z)$.

  • $\begingroup$ In addition to my answer below, you might find my answer to a related problem about finding the limits of integration for double integrals after changing the order of integration helpful. $\endgroup$ – David H Mar 14 '14 at 2:44

The domain of integration of the first triple integral can be represented by the following three inequalities given in the order of integration where the bounds are obtained from the corresponding limits of integration:

$$0 \leq z \leq y,\\ 0 \leq y \leq x,\\ 0 \leq x \leq 1.$$

Equivalently, the above information contained in the above triple of inequalities can be concisely expressed in the single line chain of inequalities,

$$0 \leq z \leq y \leq x \leq 1.$$

This representation of the domain of integration has the advantage of describing the domain of integration in a more natural geometric manner that doesn't presuppose any order of integration. After choosing a different order of integration, it is ideal for deriving the appropriate bounds for a new equivalent triple of inequalities analogous to the triple above. Since the new order of integration is $dxdydz$, we find:

$$y \leq x \leq 1,\\ z \leq y \leq 1,\\ 0 \leq z \leq 1.$$


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