# Help to change the orders of this triple integral

In the page 1950 of the Stewart's multivariable calculus book (section 15.7) he asks to rewrite an integral in the other orders, I'm having troubles particularly with the order $dxdydz$:

My solution is

$$\int^1_0\int^1_0\int^{\sqrt{1-z}}_0f(x,y,z)dxdydz$$

The solutions manual gives me other solution:

Following my reasoning, the graph is a rectangle in the $zy-$ plane, that's why I chose the limits $1$ and $0$ in the integral in the middle. I didn't understand why I'm wrong and why the limits he has chosen are the correct ones.

• I’m voting to close this question because OP did not respond to the answer. Commented Aug 28, 2023 at 11:24

The region of integration is given by the inequalities $0\le x\le1$, $0\le y\le1-x$ and $0\le z\le1-x^2$. The latter two are equivalent to $x\le 1-y\le 1$ and $x^2\le1-z\le1$ (which is equivalent to $x\le\sqrt{1-z}\le1$).
Your integral is over the region defined by $0\le y\le1$, $0\le z\le1$ and $0\le x \le\sqrt{1-z}$. This is not the same as the original region. There are points like $(x,y,z)=(3/5,1,16/25)$ in your region which are not in the original region. You have neglected the condition that $0\le y\le 1-x$.
As you say, we must have $0\le y\le 1$ and $0\le z\le1$. Then the condition on $x$ is that $0\le x\le\min(1-y,\sqrt{1-z})$. When $0\le y\le1-\sqrt{1-z}$ that amounts to $0\le x\le \sqrt{1-z}$ and when $1-\sqrt{1-z}\le y\le 1$ it amounts to to $0\le x\le1-y$. So we get Stewart's limits.