0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you tried drawing it out carefully on graph paper? $\endgroup$ – The Count Mar 27 '18 at 4:01
  • $\begingroup$ what is the relation between $a$ and $t$ ? ( $ a\ge t$ or $a\le t$) $\endgroup$ – освящение Mar 27 '18 at 4:57
  • $\begingroup$ I suppose if I had to choose, I would take $t \leq a$. $\endgroup$ – John Mar 27 '18 at 13:59
0
$\begingroup$

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, $$-(t + a)\le y\le t.$$

$\endgroup$
  • $\begingroup$ 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 $\max\left\{-t,y\right\} \leq x \leq t$. $\endgroup$ – John Mar 27 '18 at 13:55
  • $\begingroup$ 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. $\endgroup$ – John Mar 27 '18 at 14:17
  • $\begingroup$ @John, both wrong. Do the calculations with $f = 1$. I will re-check the answer. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 28 '18 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.