Rewriting triple integrals e.g. dz dy dx -> dx dy dz

[eeme]

I'm having a tough time rewriting integrals from one form to another when the first integrand is not a function of two variables.

As an example, when writing the integral to find the volume of a tetrahedron, I can easily write all 6 versions of the integral based on z = 1 - x - y or some variant of that. When the first integral is a function of only one variable e.g. z = 1 - y² things start to get hairy.

Here's an example of something I'm having problems with:

from dz dy dx to dx dy dz

My best guess so far is:

dx is x=0 -> y/2
dy is y=0 -> \sqrt{z-1}
dz is z=0 -> 1

I'm not looking for just an answer to that problem...I'd like to know how to approach these. I drew the graph for the above equation, but didn't know what my limits for dx needed to be as the graph was hard to analyze. Is there an easy way in the case of tetrahedrons where you can take the starting equation of z=1-x-y and just remove the previous variable and solve for the one your integrating for?

 

[fresh_42]

The formula is
$$
\left\{\int_{0}^{2} 1\,dx\, , \,\int_{2x}^{4} 1\,dy\, , \, \int_{0}^{1-y^2} 1\,dz\right\}
$$
and so it's not quite clear, whether it is a triplet of integrals or one a triple integral.
In both interpretations, the boundaries are considered constants.

triplet of integrals:
$$
\int_{0}^{2} 1\,dx = [x]_0^2 = 2\, , \,\int_{2x}^{4} 1\,dy = [y]_{2x}^4=4-2x\, , \, \int_{0}^{1-y^2} 1\,dz=[z]_{0}^{1-y^2}=1-y^2
$$
triple integral:
You solve it from inside out.
\begin{align*}
\int_{0}^{2} dx \int_{2x}^{4} dy \int_{0}^{1-y^2} \,dz &= \int_{0}^{2} dx \int_{2x}^{4} (1-y^2) \,dy\\
&= \int_{0}^{2} \left[y-\dfrac{1}{3}y^3\right]_{2x}^4\,dx =\int_{0}^2 \left( -\dfrac{61}{3} - 2x + \dfrac{8}{3}x^3 \right)\,dx \\
&= \left[-\dfrac{61}{3}x-x^2+\dfrac{8}{12}x^4 \right]_{0}^{2} = -\dfrac{122}{3} -4 +\dfrac{2}{3}2^4=-34
\end{align*}