Finding the limits of integration for a triple integral.

Question

I am trying to compute a triple integral of the form $\int_Vf(x,y,z)dV$ where $f(x,y,z)=8xyz$ for where our limits are $0<x<y<z<1$. Does this mean we have limits of integration at: $dz: \ y<z<1$

$dy:\ x<y<z$

$dx: \ 0<x<1$

Is this on the right lines or have I made a mistake in my thought process?

Solving the integral shouldn't be a problem, it is just defining the limits that I am struggling with.

Answer 1

Let's work from outside to inside, first for the limit of $x$, it is between $0$ and $1$.

For the limit of $y$, we have not considered $z$ yet. We have $x<y<1$.

For the limit of $z$, we have $y<z<1.$