As it is not possible to show mathematical expressions here I am attaching link to the question.
link: https://math.stackexchange.com/q/4700487/1136237
Your expertise in determining and comprehending the boundaries of integration within the Delta function's tantalizing grip will be treasured beyond measure.
Use δ(a-√x2+y2)=(a/√a2-x2)(δ(y-√a2-x2)+δ(y+√a2-x2)) to do the integral over y. Then the integral over x remains and its integration interval is [-a,a].
The general recipe is to transform the δ function δ(a-g(y)) into a sum of δ functions δ(y-yk), where yk are the zeros of g(y)-a. Each term acquires a denominator |g'(yk)| in the process.
To find the limits of integration for a double integral involving a delta function, you need to consider the properties and behavior of the delta function. The delta function is a distribution that is zero everywhere except at a single point, where it is infinite. Its integral over the entire real line is equal to 1.
When dealing with a delta function in a double integral, it is often associated with a constraint or condition on one or both variables. Let's consider an example where we have a double integral with a delta function:
I = ∫∫ f(x, y) δ(g(x, y)) dx dy
Here, f(x, y) is the function being integrated, g(x, y) is the function inside the delta function, and δ represents the delta function.
To determine the limits of integration for this double integral, you need to identify the region in the (x, y) plane where the delta function is non-zero. This region is determined by the condition g(x, y) = 0.
1. Solve g(x, y) = 0 to find the equation(s) that define the region of interest.
2. Use these equations to determine the boundaries of the region in the (x, y) plane. This may involve solving for x or y in terms of the other variable or using additional conditions.
3. Once you have the boundaries of the region, set up the limits of integration accordingly.
For example, let's say we have the following double integral:
I = ∫∫ f(x, y) δ(x - y^2) dx dy
To find the limits of integration, we need to solve the equation x - y^2 = 0. In this case, the equation represents a parabolic curve.
1. Solve x - y^2 = 0 for x:
x = y^2
2. Determine the boundaries of the region:
The parabolic curve x = y^2 represents the region of interest. Depending on the specific problem, you may have additional conditions that further restrict the region.
3. Set up the limits of integration:
Based on the determined region, you need to set the appropriate limits for x and y in the double integral. For example, if the region is bounded by y = -1 and y = 1, and x = y^2, the limits of integration would be:
∫[-1, 1] ∫[y^2, y^2] f(x, y) dx dy
Remember to adjust the limits of integration based on the specific problem and the conditions associated with the delta function.
The delta function practically lowers the dimensional degree of an integral by 1. For example
∫∞−∞𝛿(𝑥−𝑎)𝑓(𝑥)𝑑𝑥=𝑓(𝑎), in which the 1d integral is turned into a 0d integral and thus practically simplified.
As a colleague pointed out:
Polar vs 2d cartesian coordinates only make sense when talking about 2d integrals over areas just as spherical, cylindrical, and 3d only make sense over 3d volumes. In a certain sense, every integral over a single parameter is using 1d cartesian coordinates but you won't see them called this since that is the only option. In your case, you are taking a 2d area and the 𝛿 function turns it into a 1d 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 step function is, indeed, different because it does not change a 2d integral over an area into a 1d integral of a single parameter. It simply allows you to ignore a portion of the 2d plain and still leaves an integral over a 2d area, kind of like a partial derivative, if you understand what I mean.
As another fellow colleague has pointed out, however, under certain circumstances you can turn the delta-"function":
δ(b-g(y)) into a sum of delta-"functions":
δ(y-yk),
where yk are the zeros of g(y)-b. You will find this in the image I attached. Each term then gets a denominator the likes of g'(yk) in the process. I hope my answer has gotten you a bit further along in the process of solving this problem.
The best explanation I could find was on the website
https://www.mathpages.com/home/kmath663/kmath663.htm
Let me know if this answer seemed helpful!
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