1. ## Triple integral equality

Well, Hi. I wanted to know if I did this exercise right. It gives a triple integral and ask me for the other five triple integrals equal to it.

This is it: $\displaystyle\int_{0}^{1}\displaystyle\int_{\sqrt[ ]{x}}^{1}\displaystyle\int_{0}^{1-y}dzdydx$

And this is what I did:

$z=1-y\rightarrow{y=1-z}$, $\sqrt[ ]{x}=1-z\longrightarrow{z=1-\sqrt[ ]{x}}$
$x=(1-z)^2=(z-1)^2$

And the equalities.

$\displaystyle\int_{0}^{1}\displaystyle\int_{\sqrt[ ]{x}}^{1}\displaystyle\int_{0}^{1-y}dzdydx=\displaystyle\int_{0}^{1}\displaystyle\in t_{0}^{1-\sqrt[ ]{x}}\displaystyle\int_{0}^{1-z}dydzdx=\displaystyle\int_{0}^{1}\displaystyle\in t_{0}^{(z-1)^2}\displaystyle\int_{0}^{1-z}dydxdz=\displaystyle\int_{0}^{1}\displaystyle\in t_{0}^{1-z}\displaystyle\int_{0}^{y^2}dxdydx=$
$\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-y}\displaystyle\int_{0}^{(z-1)^2}dxdzdy=\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{y^2}\displaystyle\int_{0}^{1-\sqrt[ ]{x}}dzdydx$

Is this right?

Bye and thanks!

2. I haven't checked all the math, but looking at the differentials I believe you're right.