I am trying to solve an integration like; $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(a-\sqrt{x^2+y^2)}f(x,y)\,dx\,dy$. I am wondering what will be the limit of integration! There are methods to handel this kind of problems by simply doing a coordinate transformation in plane polar, but I don't want to do that. For $f(x,y)=1$, the integration should give the perimeter of the circle, i.e. $2\pi a$.
When we try to solve the problems like $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta(a-\sqrt{x^2+y^2)}f(x,y)\,dx\,dy$, we consider the limit of $x$ and $y$ in such a way that $\sqrt{x^2+y^2}\leq r$, i.e. $x^2+y^2=r^2$, which leads to $\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\Theta(a-\sqrt{x^2+y^2)}f(x,y)\,dx\,dy=\pi a^2$ for $f(x,y)=1$.
Your assistance in determining and understanding the limits of the integration for the integration containing Dirac delta function will be appreciated.
The $\delta$ 'function' is an odd mathematical construct in that it lowers the dimensional degree of an integral by $1$. For example
$$\int_{-\infty}^\infty \delta(x-a)f(x)dx=f(a)$$
That is, we have taken a $1$d integral over a single parameter and turned it into a $0$d integral (ie a constant. The same holds true for areas
$$\int_{-\infty}^\infty\int_{-\infty}^\infty \delta(x-a)f(x,y)dx=\int_{-\infty}^\infty f(a,y)dy$$
We have taken an integral over a $2$d area and turned it into a $1$d integral over a line.
Polar vs $2$d cartesian coordinates only make sense when talking about $2$d integrals over areas just as spherical, cylindrical, and $3$d only make sense over $3$d volumes. In a certain sense, every integral over a single parameter is using $1$d cartesian coordinates but you won't see them called this since that is the only option. In your case, you are taking a $2$d area and the $\delta$ function turns it into a $1$d integral over a line. Thus, it doesn't make any sense to talk about polar vs cartesian coordinates since the integral is no longer over an area. The Heaviside function is different. It DOES NOT change a $2$d integral over an area into a $1$d integral of a single parameter. Rather, it just tells you to ignore a portion of the $2$d plain and still leaves an integral over a $2$d area. In fact, you can see this difference if you consider the units of both integrals for $f=1$. In the $\delta$ case you get $2\pi a$ so the units will length. In the Heaviside case, you get $\pi a^2$ so the units will be length squared (an area).
Now, suppose you want to go through the process of solving the integral in $2$d cartesian coordinates. By process here I mean the transfer from $2$d down to $1$d. In polar, this is easy: assuming $a>0$
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(a-\sqrt{x^2+y^2})f(x,y)dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}\delta(a-r)r f(r\cos(\theta),r\sin(\theta))drd\theta$$
But the inner integral is
$$\int_{0}^{\infty}\delta(a-r)r f(r\cos(\theta),r\sin(\theta))dr=af(a\cos(\theta),a\sin(\theta))$$
so the whole integral is
$$=a\int_{0}^{2\pi}f(a\cos(\theta),a\sin(\theta))d\theta$$
Doing this in cartesian is more difficult, but doable by the following method
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(a-\sqrt{x^2+y^2})f(x,y)dxdy$$
$$=\lim_{\epsilon\to 0^+}\frac{1}{\epsilon}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left[\Theta(a+\epsilon-\sqrt{x^2+y^2})-\Theta(a-\sqrt{x^2+y^2}\right]f(x,y)dxdy$$
Basically, integrate over a thin band near the radius $a$, and then divide by the width of that thin band. The division by the width (represented by the $\frac{1}{\epsilon}$ out front) serves to change the units from length squared to length. In the case $f(x,y)=1$ this becomes
$$\lim_{\epsilon\to 0^{+}}\frac{\pi(a+\epsilon)^2-\pi a^2}{\epsilon}=\frac{d}{da}\pi a^2=2\pi a$$
You'll notice that this turned into a derivative at the end of the problem. In a certain sense, one can call the $\delta$ function the derivative of the Heaviside function although I am using the terms loosely here and care has to be taken to define what we mean in this context. That is, if we are loose with our notation, it is not incorrect to say
$$\delta(x)=\lim_{\epsilon\to 0^+}\frac{\Theta(x+\epsilon)-\Theta(x)}{\epsilon}$$
which is what I replaced the $\delta$ function in the integral with. Again, care must be taken when using the word 'function' to describe the $\delta$... thing... but it will suffice to solve the problem in $2$d cartesian coordinates.