Refer to this MSE example where:
\begin{eqnarray*} \int_{0}^{8}\int_{\sqrt[3]{\vphantom{\large a}y\,}}^{2}{\rm f}\left(x, y\right)\,{\rm d}x\,{\rm d}y & = & \int_{0}^{8}\left\lbrack\int_{0}^{2}\Theta\left(x - \sqrt[3]{\vphantom{\large a}y\,} \right) {\rm f}\left(x, y\right)\,{\rm d}x\right\rbrack{\rm d}y \\ & = & \int_{0}^{2}\left\lbrack\int_{0}^{8}\Theta\left(x^{3} - y\right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \\ & = & \int_{0}^{2}\left\lbrack\int_{0}^{x^{3}} {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \end{eqnarray*}
Is given as $\int\int_{R}f(x,y)\; dxdy$. I'm curious to know how order reversal would be done if it were $\int\int_{R}f(x,y)\; dydx$ instead - particularly when the Heaviside function is given in terms of $x$: $$\Theta(x) = \left\{\begin{array}{lr} 1 & x \geq 0 \\ 0 & x < 0 \end{array}\right.$$
I suspect you can make a similar definition in terms of $y$, by direct substitution:
$$\Theta(y) = \left\{\begin{array}{lr} 1 & y \geq 0 \\ 0 & y < 0 \end{array}\right.$$
and proceeding with similar manipulation as given in the example above, when given:
$$\int\int_{R}f(x,y)\; dydx$$
Can someone help me out? thanks
Edit: I suspect that when the reversal is $dydx\rightarrow\;dxdy$ it may require manipulation (taking the inverse) of an upper bound.