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Notebook[{

Cell[CellGroupData[{
Cell["Means and Moments and Exploring New Plotting Techniques", "Title"],

Cell[TextData[StyleBox["Chapter 12, Sections 2 & 5",
  FontFamily->"Arial",
  FontSize->16,
  FontWeight->"Bold"]], "Text"],

Cell[CellGroupData[{

Cell["Introduction", "Section"],

Cell[TextData[StyleBox[
"OBJECTIVE: Extend the concept of the moments of a density function of a \
solid to applications in probability and engineering."]], "Text"],

Cell[TextData[StyleBox[
"What do means and multiple integrals have to do with probabilities? How can \
the method of moments be used to help determine whether or not an object will \
float in an upright position? These questions will be answered as you explore \
this module. You will also get to practice new and interesting ways to plot \
functions."]], "Text"],

Cell[CellGroupData[{

Cell["Technology Guidelines", "Subsection",
  CellDingbat->"\[LightBulb]"],

Cell[TextData[{
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    Background->None],
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    Background->None],
  "\nTO STOP AN EXECUTION\n\tSelect the ",
  StyleBox["Kernel",
    FontSlant->"Italic"],
  " pull-down menu and click on ",
  StyleBox["Abort Evaluation.\n",
    FontSlant->"Italic"],
  "ORDER OF EXECUTION\n\tExecute cells in the order given. Do not skip any \
Input cells within a given notebook.\nSAVING NOTEBOOKS\n\tYou can save \
anytime to any directory you choose, and it is wise to save often.\n\t\
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the \n\t ",
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    FontSlant->"Italic"],
  " selection under the ",
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  " pull-down menu.\nEXPERIENCING MAJOR PROBLEMS\n\tSave if appropriate, and \
then shut down ",
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    FontSlant->"Italic"],
  " and start it up again."
}], "Text"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Part I: Masses and Moments Applied to Probability Functions", "Section"],

Cell[TextData[{
  "In multivariable probability, probability density functions are defined \
over specified regions (domains) in the same way that density functions are \
defined over regions of a solid. These probability functions are designed so \
that the total \"mass\" is always 1. Probabilities are then defined to be the \
integrals of the density functions over a particular subregion of the domain. \
Consider the following example.\n\nSuppose that a national fast-food outlet \
is interested in the joint behavior of the random variables ",
  StyleBox["x",
    FontSlant->"Italic"],
  ", defined as the total time between a customer's arrival at the store and \
departing from the service window, and ",
  StyleBox["y",
    FontSlant->"Italic"],
  ", the time that a customer waits in line before reaching the service \
window. Since ",
  StyleBox["x",
    FontSlant->"Italic"],
  " contains the time a customer waits in line, ",
  StyleBox["x",
    FontSlant->"Italic"],
  " must be greater than ",
  StyleBox["y",
    FontSlant->"Italic"],
  ". The relative frequency distribution of observed values of ",
  StyleBox["x",
    FontSlant->"Italic"],
  " and ",
  StyleBox["y",
    FontSlant->"Italic"],
  " can be modeled by the probability density function:  \n",
  Cell[BoxData[
      FormBox[
        RowBox[{\(f(x, y)\), " ", "=", " ", 
          FormBox[\( .04  e\^\(\(- .1\) x -  .3  y\ \)\),
            "TraditionalForm"]}], TraditionalForm]]],
  "  for  ",
  Cell[BoxData[
      \(TraditionalForm
      \`0\  \[LessEqual] \ y\  \[LessEqual] \ x\  < \ \[Infinity]\)]],
  "  and 0 elsewhere, where ",
  StyleBox["x",
    FontSlant->"Italic"],
  " and ",
  StyleBox["y",
    FontSlant->"Italic"],
  " are measured in minutes."
}], "Text"],

Cell[BoxData[
    \(f[x_, y_] :=  .04 
         E\^\(\(- .1\) x - \(\(.3\)  \(y\)\(\ \)\)\)\)], "Input"],

Cell["\<\
We examine the domain of this function. First we must load a package to \
assist us in graphing.\
\>", "Text"],

Cell[BoxData[
    \(<< \ Graphics`FilledPlot`\)], "Input"],

Cell[BoxData[
    \(\(FilledPlot[x, {x, 0, 50}, 
        AxesLabel -> {"\<time in system\>", "\<time in line\>"}];\)\)], \
"Input"],

Cell["\<\
We begin by verifying that the integral of the density function over the \
entire domain is 1.\
\>", "Text"],

Cell[BoxData[{
    \(Clear[x, y]\), "\n", 
    \(Integrate[f[x, y], {x, 0, \[Infinity]}, {y, 0, x}]\)}], "Input"],

Cell[TextData[{
  "To compute the average or expected values of ",
  StyleBox["x",
    FontSlant->"Italic"],
  " and ",
  StyleBox["y",
    FontSlant->"Italic"],
  " over the domain, we find the first moments of the density function with \
respect to the",
  StyleBox[" x",
    FontSlant->"Italic"],
  " and ",
  StyleBox["y",
    FontSlant->"Italic"],
  " axes. The second integral is difficult to evaluate as ",
  StyleBox["x",
    FontSlant->"Italic"],
  " approaches \[Infinity], so we substitute a large value of ",
  StyleBox["x",
    FontSlant->"Italic"],
  " for the upper bound."
}], "Text"],

Cell[BoxData[{
    \(Print[\ "\<Expected time in system = \>", 
      Integrate[
        x\ f[x, y], {x, 0, \[Infinity]}, {y, 0, 
          x}], "\< minutes.\>"]\), "\n", 
    \(Print[\ "\<Expected time waiting in line = \>", 
      Integrate[
        y\ f[x, y] // Simplify, {x, 0, 1000000}, {y, 0, 
          x}], "\< minutes.\>"]\)}], "Input"],

Cell[TextData[{
  "Now we compute the probability that ",
  StyleBox["y",
    FontSlant->"Italic"],
  " is at least one-half as large as ",
  StyleBox["x",
    FontSlant->"Italic"],
  ". First we can draw the region over which we will integrate (the part for \
",
  StyleBox["x < 5",
    FontSlant->"Italic"],
  "). "
}], "Text"],

Cell[BoxData[
    \(\(FilledPlot[{x, 1/2  x}, {x, 0, 5}, 
        AxesLabel -> {"\<time in system\>", "\<time in line\>"}];\)\)], \
"Input"],

Cell["\<\
The probability of this event is the integral of the density function over \
this region.\
\>", "Text"],

Cell[BoxData[
    \(Print["\<The probability that the time in line is at least half as long \
as the time in the system is \>", 
      Integrate[f[x, y], {x, 0, \[Infinity]}, {y, 1/2  x, x}]]\)], "Input"],

Cell[TextData[{
  "If we want to know the probability that a customer spends no more than six \
minutes at the service window itself, we represent that quantity ",
  Cell[BoxData[
      \(TraditionalForm\`x\  - \ y\)]],
  ", and we integrate over the portion of the domain where ",
  Cell[BoxData[
      \(TraditionalForm\`x\  - \ y\  \[LessEqual] \ 6. \)]]
}], "Text"],

Cell[BoxData[{
    \(\(p1 = 
        FilledPlot[\n\t\t{x, x - 6}, {x, 6, 20}, Fills\  -> \ GrayLevel[ .8], 
          DisplayFunction -> Identity];\)\), "\n", 
    \(\(p2 = 
        FilledPlot[x, {x, 0, 6}, \ Fills\  -> GrayLevel[ .8], 
          DisplayFunction -> Identity];\)\), "\n", 
    \(\(Show[p1, p2, DisplayFunction -> $DisplayFunction, 
        AxesLabel -> {"\<time in system\>", "\<time in line\>"}];\)\)}], \
"Input"],

Cell[TextData[{
  StyleBox[
  "Looking at the domain, you can see that it is easier here to integrate \
with respect to "],
  StyleBox["x",
    FontSlant->"Italic"],
  StyleBox[" first."]
}], "Text"],

Cell[BoxData[
    \(Print["\<The probability that the time spent at the service desk is no \
more than 6 minutes is \>", 
      Integrate[f[x, y], {y, 0, \[Infinity]}, {x, y, y + 6}]]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["You Try It: Part I", "Section"],

Cell["\<\
Try your hand at computing the probability that the time at the service desk \
will be at least 8 minutes. Change the entries in red to the appropriate \
terms.\
\>", "Text"],

Cell[BoxData[
    RowBox[{"Print", "[", 
      RowBox[{"\"\<The probability is \>\"", ",", 
        RowBox[{"Integrate", "[", 
          RowBox[{\(f[x, y]\), ",", \({y, 0, \[Infinity]}\), ",", 
            RowBox[{"{", 
              RowBox[{"x", ",", 
                StyleBox["y",
                  FontColor->RGBColor[1, 0, 0]], ",", 
                StyleBox[\(y + 6\),
                  FontColor->RGBColor[1, 0, 0]]}], "}"}]}], "]"}]}], 
      "]"}]], "Input"],

Cell["\<\
Compute the probability that the time in line is no more than 30% of the time \
in the system. Change the entries in red to the appropriate terms.\
\>", "Text"],

Cell[BoxData[
    RowBox[{"Print", "[", 
      RowBox[{"\"\<The probability is \>\"", ",", 
        RowBox[{"Integrate", "[", 
          RowBox[{\(f[x, y]\), ",", \({x, 0, \[Infinity]}\), ",", 
            RowBox[{"{", 
              RowBox[{"y", ",", 
                StyleBox[\(1/2  x\),
                  FontColor->RGBColor[1, 0, 0]], ",", 
                StyleBox["x",
                  FontColor->RGBColor[1, 0, 0]]}], "}"}]}], "]"}]}], 
      "]"}]], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Part II: Using Moments to Determine Acceptable Heights", "Section"],

Cell[TextData[StyleBox["Section 12.2, Extension of Exercise 43",
  FontWeight->"Bold"]], "Text"],

Cell["\<\
Suppose that you have a buoy in the form of a solid hemisphere of radius 40 \
centimeters and uniform density that is topped with a solid cylinder of the \
same material and radius 40 centimeters. This buoy floats with the center of \
the hemisphere above the water surface. Determine the maximum height that the \
cylinder may attain that permits the buoy to continue floating upright. To \
picture this, we load two packages and plot the shape of the body.\
\>", "Text"],

Cell[BoxData[{
    \(<< Graphics`ParametricPlot3D`\), "\n", 
    \(<< Graphics`Graphics3D`\)}], "Input"],

Cell[BoxData[{
    \(Clear[x, y, z, u, v, t, r, \[Theta]]\), "\n", 
    \(\(hemisplot = 
        CylindricalPlot3D[\(-\@\(1600 - r\^2\)\), {r, 0, 40}, {\[Theta], 0, 
            2  \[Pi]}, AxesLabel -> {x, y, z}, 
          DisplayFunction -> Identity];\)\), "\n", 
    \(\(cylpts\  = \ 
        Table[{40  Cos[t]\ , 40\ Sin[t], \n\ \ u}, \ {t, \ 0, \ 2  Pi, \ 
            Pi/12}, \n\ \ {u, \ 0, 40}];\)\), "\n", 
    \(\(cylplot = 
        ListSurfacePlot3D[cylpts, DisplayFunction -> Identity];\)\), "\n", 
    \(\(topplot = 
        CylindricalPlot3D[40, {r, 0, 40}, {\[Theta], 0, 2  \[Pi]}, 
          AxesLabel -> {x, y, z}, DisplayFunction -> Identity];\)\), "\n", 
    \(\(flagpolepts = Table[{0, 0, t}, {t, 35, 80}];\)\), "\n", 
    \(\(flagpts = Table[{0, u, v}, {u, 0, 20}, {v, 60, 80}];\)\), "\n", 
    \(\(fp1 = 
        ScatterPlot3D[flagpolepts, DisplayFunction -> Identity];\)\), "\n", 
    \(\(fp2 = 
        ListSurfacePlot3D[flagpts, DisplayFunction -> Identity];\)\), "\n", 
    \(\(Show[hemisplot, cylplot, topplot, fp1, fp2, 
        DisplayFunction :> $DisplayFunction];\)\)}], "Input"],

Cell[TextData[{
  "You may assume that the flag on top is weightless. \n\nThe method of \
moments can help us. It can be determined that the buoyancy force exerted by \
the body of water due to the amount of water displaced will be in a direction \
through the point that would be the midpoint of the sphere, as long as the \
object is upright or tilted so that no part of the cylinder is immersed. \
Hence, if the center of gravity of the solid is at that point or closer to to \
the bottom of the hemisphere, the object will stay upright. If the center of \
gravity is in the region of the cylinder, the object will tilt. \n\nWe first \
determine the ",
  StyleBox["z",
    FontSlant->"Italic"],
  "-coordinate of the center of gravity of the hemisphere. Recall that the \
volume of a hemisphere with radius of ",
  StyleBox["r",
    FontSlant->"Italic"],
  " is 2/3 \[Pi] ",
  Cell[BoxData[
      \(TraditionalForm\`r\^3\)]],
  "."
}], "Text"],

Cell[BoxData[{
    \(\(volhemis = 2/3\ \[Pi]\ 40\^3;\)\), "\n", 
    \(\(hemis = \(-\@\(1600 - x\^2 - y\^2\)\);\)\), "\n", 
    \(zbar = 
      Integrate[
          z\ , {x, \(-40\), 
            40}, {y, \(-\@\(1600 - x\^2\)\), \@\(1600 - x\^2\)}, {z, hemis, 
            0}]/volhemis\)}], "Input"],

Cell[TextData[{
  "The ",
  StyleBox["z",
    FontSlant->"Italic"],
  "-coordinate of the center of mass of the cylinder is ",
  StyleBox["h/2",
    FontSlant->"Italic"],
  ", so if we equate the first moment of the hemisphere to the first moment \
of the cylinder, we can solve for ",
  StyleBox["h",
    FontSlant->"Italic"],
  ".  Since the radius is 40 here, the volume of the cylinder is ",
  "1600",
  StyleBox[" ",
    FontSlant->"Italic"],
  "\[Pi] ",
  StyleBox["h",
    FontSlant->"Italic"],
  "."
}], "Text"],

Cell[BoxData[{
    \(solh = 
      Solve[\(-zbar\)*volhemis == h/2*\((1600  \[Pi]\ h)\), h]\), "\n", 
    \(Print["\<The height can be at most \>", 
      solh[\([2, 1, 2]\)], "\< or \>", 
      N[solh[\([2, 1, 2]\)]], "\< centimeters.\>"]\)}], "Input"],

Cell[TextData[{
  "In general, if the radius were ",
  StyleBox["R",
    FontSlant->"Italic"],
  ", the maximum height of the cylinder could be at most  ",
  Cell[BoxData[
      \(TraditionalForm\`\(R \@ 2\)\/2\)]],
  "."
}], "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["You Try It: Part II", "Section"],

Cell[TextData[{
  "Experiment with some of the 3-D plot functions presented in this section. \
First, just practice with a cylindrical plot. It is understood in these plots \
that ",
  StyleBox["z",
    FontSlant->"Italic"],
  " is a function of ",
  StyleBox["r",
    FontSlant->"Italic"],
  " and \[Theta] . Try some of your own functions. The package ",
  StyleBox["ParametricPlot3D",
    FontWeight->"Bold"],
  " is required for executing the cylindrical or spherical plots. Change the \
terms in red."
}], "Text"],

Cell[BoxData[
    RowBox[{
      RowBox[{"z", "=", 
        StyleBox[\(r\ \ Cos[\[Theta]]\^2\),
          FontColor->RGBColor[1, 0, 0]]}], ";", "\n", 
      RowBox[{"newcyl", "=", 
        RowBox[{"CylindricalPlot3D", "[", 
          RowBox[{"z", ",", 
            RowBox[{"{", 
              RowBox[{"r", ",", "0", ",", 
                StyleBox["10",
                  FontColor->RGBColor[1, 0, 0]]}], "}"}], ",", 
            RowBox[{"{", 
              RowBox[{"\[Theta]", ",", "0", 
                StyleBox[",",
                  FontColor->GrayLevel[0]], \(2  \[Pi]\)}], "}"}], 
            ",", \(AxesLabel -> {x, y, z}\)}], "]"}]}], ";"}]], "Input"],

Cell["\<\
You can create spherical plots with the same package you used to create \
cylindrical plots. In these plots, it is assumed that \[Rho] is a function of \
\[Phi] and \[Theta]. Try some of your own by changing the terms in red. Not \
everything will give satisfying results.\
\>", "Text"],

Cell[BoxData[
    RowBox[{
      RowBox[{"\[Rho]", "=", 
        StyleBox[\(\[Theta]\ Cos[\[Phi]]\),
          FontColor->RGBColor[1, 0, 0]]}], ";", 
      "\n", \(SphericalPlot3D[\[Rho], {\[Phi], 0, \[Pi]}, {\[Theta], 0, 
          2  \[Pi]}, AxesLabel -> {x, y, z}]\), ";"}]], "Input"],

Cell["\<\
You may have noticed that to draw the flag we had to first generate points \
and then plot the points. The difference between the two was that the pole \
could be thought of as a line, whereas, the flag was more like a surface. \
Experiment with the following commands and draw something else by changing \
the terms in red. Remember to leave one parameter in the first expression and \
two parameters in the second experession.\
\>", "Text"],

Cell[BoxData[
    RowBox[{
      RowBox[{"newflagpolepts", "=", 
        RowBox[{"Table", "[", 
          RowBox[{
            RowBox[{"{", 
              StyleBox[\(0, 0, t\),
                FontColor->RGBColor[1, 0, 0]], "}"}], ",", 
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                FontColor->RGBColor[1, 0, 0]], "}"}]}], "]"}]}], ";", "\n", 
      RowBox[{"newflagpts", "=", 
        RowBox[{"Table", "[", 
          RowBox[{
            RowBox[{"{", 
              StyleBox[\(0, u, v\),
                FontColor->RGBColor[1, 0, 0]], "}"}], ",", 
            RowBox[{"{", 
              StyleBox[\(u, 0, 20\),
                FontColor->RGBColor[1, 0, 0]], "}"}], ",", 
            RowBox[{"{", 
              StyleBox[\(v, 15, 30\),
                FontColor->RGBColor[1, 0, 0]], "}"}]}], "]"}]}], ";", 
      "\n", \(fp1 = ScatterPlot3D[newflagpolepts]\), ";", 
      "\n", \(fp2 = ListSurfacePlot3D[newflagpts]\), ";", 
      "\n", \(Show[fp1, fp2, Boxed -> False, Axes -> False]\), ";"}]], "Input"]
}, Closed]]
}, Open  ]]
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