Changing the order of integration without sketching? When changing the order of double integrals, I have always relied on sketching the region. I have recently come across this example on MSE by @FelixMartin which seems to avoid visual-based reasoning, and a comment below it says that it's "very nice and neat". But I have no idea what's going on with the $\Theta()$. Can someone please give me a hint?
\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*}

PS: sorry for the verbatim copying, I just reckoned it would save people some time.
 A: $\Theta$ here refers to the Heaviside Step Function (wiki: http://en.wikipedia.org/wiki/Heaviside_step_function). It is defined very simply as:
$$\Theta(x) = \left\{\begin{array}{lr} 1 & x \geq 0 \\ 0 & x < 0 \end{array}\right.$$
The usefulness of $\Theta$ is that it allows one to express 'cutoffs' using function notation, and you can use this to change the bounds of integrals without changing their values.
So, for instance, you could write:
$$\int_1^{10}f(x)\,dx = \int_0^{10}\Theta(x - 1)f(x)\,dx$$
The reason they are the same is that $\Theta(x-1)$ is equal to zero on the interval from 0 to 1.
So this explains the first line of the manipulation:
$$\int_{\sqrt[3]{y}}^{2}f(x,y)\, dx = \int_0^{2}\Theta(x - \sqrt[3]{y})f(x,y)\, dx$$
The second line of the manipulation is Fubini's theorem, along with the observation that $x - \sqrt[3]{y} \geq 0$ if and only if $x^3 - y \geq 0$, so $\Theta(x - \sqrt[3]{y}) = \Theta(x^3 - y)$.
Note that you can express TWO cutoffs with two instances of the theta function:
$$\int_0^{1}\int_{y^2}^{y}f(x,y)\,dx\,dy = \int_0^1\int_0^1 \Theta(x - y^2) \Theta(y - x)f(x,y)\, dx\,dy$$
