Rewriting Triple Integral in different order How can I rewrite, say:
$$\int^1_0\int^{z^2}_0\int^y_0{f}dxdydz$$
In the orders $dzdxdy$ and $dydzdx$?
Would I have to solve this triple integral all the way or is there a simple way to do this?
 A: Hint
Your region is defined by the following inequalities, written directly from the integral limits:
$$\begin{split}
&0 \le x \le y \\
&0 \le y \le z^2 \\
&0 \le z \le 1 \\
\end{split}$$
Notice the first integral eliminates $x$ (so only $y,z$ remain) and the second eliminates $y$ (so only $z$ remains) and the last one eliminates $z$
You need to change the inequalities to be in the order defined by desired integration - $dzdxdy$ means you want to eliminate $z$ first by integrating over it, and then $x$ and finally, $y$.
Drawing a picture of the region will help you a lot.
UPDATE Let's do one simple example to give you an idea. Suppose I want to integrate $dydxdz$ instead. The last integral is the same, but the first two inequalities will need to be changed.
Assuming $z^2$ is some constant, the first two inequalities describe a triangle in the $x,y$ plane with vertices $(0,0), (0,z^2)$ and $(z^2,z^2)$. Going $x$-first, we bound it $0\le x \le y$ and $0 \le y \le z^2$.
Now we go $y$ first. $y$ changes from the curve $y=x$ to the horizontal line $y=z^2$, so the inequality would be $x \le y \le z^2$ and now $x$ varies from $0$ to $z^2$ to cover the same region. So the ineqialities become
$$
x \le y \le z^2\\
0 \le x \le z^2\\
0 \le z \le 1
$$
