Skip to main content

calculus and analysis - Can Mathematica help me evaluate an integral over disjoint disks I=intD1intD2log|x−y|dydx?


I want to evaluate an integral that involves two disjoint unit disks D1 and D2. D1 is centered at (−2,0) and D2 is centered at (0,2). The integral I want to compute is



I=∫D1∫D2log|x−y|dydx.


I looked at the in-built Python integration methods and also the quadpy library but although they have lots of options for integration over a single disk, I couldn't find anything that can help me with integrating over two disjoint disks.


Is it possible to evaluate this integral in Mathematica? I don't need an optimum method, I just need to obtain the value of this integral.



Answer



NIntegrate[
 Log[Norm[{x1, x2} - {y1, y2}]], 
 {x1, x2} ∈ Disk[{-2, 0}, 1], 
 {y1, y2} ∈ Disk[{2, 0}, 1]
 ]



13.6822



(the actual result being 13.682176919165677),


In order to enter ∈, just type Esc e l Esc.


In order to increase the precision, use the option PrecisionGoal.


Edit


Another possibility that relieves Mathematica from the need to discretize the disks and that allows her to use higher-order quadrature formulas is to employ polar coordinates on each of the disks:


NIntegrate[
 Log[Norm[{r1 Cos[θ1] - 2,r1 Sin[θ1]} - {r2 Cos[θ2] + 2,r2 Sin[θ2]}]] r1 r2,

 {r1, 0, 1}, {θ1, -Pi, Pi},
 {r2, 0, 1}, {θ2, -Pi, Pi},
 Method -> "LocalAdaptive"
 ]

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more