Assume we have a function $f(r,\theta)=e^{-r}$ in polar coordinates.
The value at the origin point is $f(0,0)=1$. But actually $f(0,0)$, $f(0,1)$, ... , $f(0,\theta)$ are at the same location/point. So the total value at the origin should be the result of an integral:$\int_{0}^{2\pi}f(r,\theta)d\theta$, with $r=0$, which equals to $2\pi$.
Now we have two values for the origin point, could someone tell me which one is right?
You've noticed that you have a function that is constant in one of its variables. To simplify the setting, if $f:\mathbb R\to\mathbb R$ is a constant function with value $f(x)=C$ for all $x$, then $$\int_a^bf(x)\,dx=(b-a)C$$ In other words, you can integrate $f$ to find out its constant value, but the answer will be sensitive to the limits of integration. Perhaps you "just want to get rid" of the variable $x$, but to formulate the integral over $x$, you have to choose bounds that the variable $x$ will take.
We can rearrange the equation: $$\frac{\int_a^bf(x)\,dx}{\int_a^bdx}=\frac{\int_a^bf(x)\,dx}{b-a}=C$$ This equation says that you can recover the constant value of $x$ by taking an average value over some interval; and when you do that, the answer is not sensitive to the bounds of integration.
This is all a long way of explaining herb steinberg's comment. If you want to get an average value out of an integral like $\int_0^{2\pi}$, then you have to divide by $2\pi$.
So far, I haven't even had to say anything about polar coordinates. One relevant generalization of this principle is the mean value property for harmonic functions, which states that the value of such a function at any point is equal to the average value of the function over a sphere centered at that point. In two dimensions, that means taking an integral like $\frac{1}{2\pi}\int_{0}^{2\pi}f(r,\theta)\,d\theta$. This example provides some evidence that average values are useful, natural things to consider. (But it isn't directly relevant to your question, because your function isn't harmonic over the plane.)