How to find out how namespace got polluted?

var arra = [];
for (var i = 0; i < 2003000; i++) {
 arra.push(i * i + i);
}

(function(){
 var arra = [];
 for (var i = 0; i < 2003000; i++) {
  arra.push(i * i + i);
 }
})();

var x1 = 5;
var x2 = 20;
var y1 = 3
var y2 = 16;

var rise = y2 - y1;
var run = x2 - x1;

var slope = rise / run;

var risesquared = rise * rise;
var runsquared = run * run;

var distancesquared = risesquared + runsquared;

var distance = Math.sqrt(dinstancesquared);

//Calculate is the only exposed global variable
var Calculate = function () {
 //all defintions in this closure are local, and will not be exposed to the global namespace
 var Coordinates = [];//array for coordinates
 var Coordinate = function (xcoord, ycoord) {//definition for type Coordinate
   this.x = xcoord;//assign values similar to a constructor
   this.y = ycoord;
  };

  return {//these methods will be exposed through the Calculate object
   AddCoordinate: function (x, y) {
   Coordinates.push(new Coordinate(x, y));//Add a new coordinate
  },

  Slope: function () {//Calculates slope and returns the value
   var c1 = Coordinates[0];
   var c2 = Coordinates[1];
   return c2.y - c1.y / c2.x - c1.x;//calculates rise over run and returns result
  },

  Distance: function () {
   //even with an excessive amount of variables declared, these are all still local
   var c1 = Coordinates[0];
   var c2 = Coordinates[1];

   var rise = c2.y - c1.y;
   var run = c2.x - c1.x;

   var risesquared = rise * rise;
   var runsquared = run * run;

   var distancesquared = risesquared + runsquared;

   var distance = Math.sqrt(distancesquared);

   return distance;
  }
 };
};

//this is a "self executing closure" and is used because these variables will be
//scoped to the function, and will not be available globally nor will they collide
//with any variable names in the global namespace
(function () {
 var calc = Calculate();
 calc.AddCoordinate(5, 20);
 calc.AddCoordinate(3, 16);
 console.log(calc.Slope());
 console.log(calc.Distance());
})();

from .baz import do_something
del baz