# Changing the order of integration for $\int_{-t}^t \int_{z-a}^z \int_z^{y+a} f(x,y,z) d x d y d z$

I'd like to change the order of integration in the following triple integral \begin{equation*} \int_{-t}^t \int_{z-a}^z \int_z^{y+a} f(x,y,z) d x d y d z \end{equation*} where $a > 0$ and $t > 0$. I'd like to integrate over $z$ first. The region of integration appears to be a prism; however, I'm having a hard time getting the correct limits.

• Have you tried drawing it out carefully on graph paper? – The Count Mar 27 '18 at 4:01
• what is the relation between $a$ and $t$ ? ( $a\ge t$ or $a\le t$) – освящение Mar 27 '18 at 4:57
• I suppose if I had to choose, I would take $t \leq a$. – John Mar 27 '18 at 13:59

## 1 Answer

Not a terribly good idea but you can do "blindly" via the inequalities $$z\le x\le y+a,$$ $$z-a\le y\le z,$$ $$-t\le z\le t.$$ The $z$ limits are obvious: $$y\le z\le x.$$ The upper limit for $x$ is also obviously $x\le y.$ Now, using $-t\le z$ and $z\le x$: $$-t\le x\le y.$$ Finally, $$-(t + a)\le y\le t.$$

• Thanks for the idea. I was also attempting to do so along these lines, but wasn't positive my limits were correct. For example, I think the limits for $z$ should actually be $\max \left\{y,-t \right\} \leq z \leq x$. I also think you flipped the relation for $x,y$, I think it should be $y \leq x$. I believe the limits for $x$ should be something like $\max\left\{-t,y\right\} \leq x \leq t$. – John Mar 27 '18 at 13:55
• This is what I believed the limits should be: $\int_{-t-a}^t \int_{\max \left\{ -t, y \right\} }^{y+a} \int_{\max \left\{-t,y \right\}}^{\min \left\{ x,t \right\}} dz dx dy$. I'd really like it if someone could confirm it though. – John Mar 27 '18 at 14:17
• @John, both wrong. Do the calculations with $f = 1$. I will re-check the answer. – Martín-Blas Pérez Pinilla Mar 28 '18 at 8:14