Integrating Heaviside Step Function of two Variables Suppose we have a definite integral like $$ I=\int_0^\infty dx \int_0^\infty dy \,  Θ(α-x-y) $$
where $a \in R_+^*$ and $Θ$ is the Heaviside step function. Of course this is easy in that we can find the answer without working with integrals,  since it's simply the area of a triangle with vertices $O(0,0)$ , $A(a,0)$, $B(0,a)$, which is $\frac{a^2}{2}$. 
However trying to work the integral out doesn't seem so trivial (to me at least). One could, naively, do for example $I=\int_0^\infty dx \int_0^{a-x} dy = -\infty$ (since $Θ=0$ for $y \ge a-x$) which is obviously wrong- somehow $Θ$ should affect both integral limits, but I can't see how this can happen. Probably I'm misunderstanding the definition of $Θ(α-x-y)$.  
EDIT : As @John Polcari pointed out, in the above exaple, we should limit $x$ such that $a-x \ge 0$, so we should have $I= \int_0^\infty dx \int_0^{a-x}Θ(a-x) \, dy $ which indeed gives the right result. However this seems to me like adding the $Θ(α-x)$ by hand, which is no different from the first argument with the triangle's area. Is there a prettier-more strict- way of doing it?
 A: The formally correct result is $$\int\limits_0^\infty  {dz\,\Theta \left( {a - z} \right)}  = a\,\Theta \left( a \right)$$
In case it is unclear to see how to apply this:
By implication from above
$$\int\limits_0^\infty  {dz\,z} \,\Theta \left( {a - z} \right) = \frac{{{a^2}}}{2}\Theta \left( a \right)$$
Then
$$\begin{array}{l}
\int\limits_0^\infty  {dx\int\limits_0^\infty  {dy\,\Theta \left( {a - x - y} \right)} }  = \int\limits_0^\infty  {dx\left( {a - x} \right)\Theta \left( {a - x} \right)} \\
 = a\int\limits_0^\infty  {dx\,\Theta \left( {a - x} \right)}  - \int\limits_0^\infty  {dx\,x\;\Theta \left( {a - x} \right)} \\
 = {a^2}\Theta \left( a \right) - \frac{{{a^2}}}{2}\Theta \left( a \right) = \frac{{{a^2}}}{2}\Theta \left( a \right)
\end{array}$$
The final step function is appropriate because the if $a<0$, the answer really is zero.
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A: Let $I$ be the integral defined by 
$$I=\int_0^\infty \int_0^\infty \Theta(a-x-y)\,dy\,dx\tag1$$
Enforcing the substitution $u=a-x-y$ in $(1)$ reveals
$$I=\int_0^\infty \int_{-\infty}^{a-x} \Theta(u)\,du\,dx\tag2$$
Changing the order of integration in $(2)$ we find that 
$$\begin{align}
I&=\int_{-\infty}^a \int_0^{a-u} \Theta(u)\,dx\,du\\\\
&=\int_{-\infty}^a \Theta(u) \int_0^{a-u} (1)\,dx\,du\\\\
&=\int_{-\infty}^a \Theta(u) (a-u)\,du\\\\
&=\int_0^a (a-u)\,du\\\\
&=\frac12a^2
\end{align}$$
