VC.07 Transforming 2D Integrals
GIVE IT A TRY!

\(Off[General::spell]; 
Off[General::spell1]; 

    \(Off[ParametricPlot3D::ppcom]; \)\)Off[ParametricPlot::ppcom]; Off[Plot::plnr]; SetOptions[Limit, Analytic -> True];

CMView = {2.7, 1.6, 1.2};

Vector Drawer

BeginPackage["Vector3D`", "Graphics`Colors`"];

\(Axes3D::"usage" = 
      "Axes3D[a, b] creates a Graphics3D object of cartesian axes with x, y, and z running from -a/3 to a, and with axes labels b units beyond the tips of the axes.  Axes3D[a] is Axes3D[a, a/8]."; 

    \(Perpend::"usage" = 
      "Perpend[a] returns a unit vector which is perpendicular to the vector a."; \)\)Arrow::"usage" = 
      "Arrow[a] produces a 2 or 3 dimensional vector from the origin to a. \
Arrow[a, Tail -> tail] gives a vector from tail to a."; Vector::"usage" = 
      "Vector[a, b] gives a 2D or 3D vector from a to b."; ArrowHead::"usage" = 
      "ArrowHead[a, vec] produces an arrowhead with its tip placed at the \
point a, pointing in the direction of the vector vec."; Tail::"usage" = 
      "Tail -> point puts the tail of the vector at point."; Norm::"usage" = 
      "Norm[vec] returns the Euclidean length of vec."; Normalize::"usage" = 
      "Normalize[vec] returns a unit vector in the direction of vec, \
provided vec is nonzero."; Aperture::"usage" = 
      "Aperture is the ratio of the radius of the base to the length of the \
head of a vector."; TipSize::"usage" = 
      "TipSize specifies an absolute size for the inner tip length of the \
head of a vector."; TipRatio::"usage" = 
      "TipRatio is the ratio of the inner length of the arrowhead to the \
length of the shaft of a vector."; HeadRatio::"usage" = 
      "HeadRatio is the ratio of the outer length of the arrowhead to the \
length of the shaft of a vector."; HeadSize::"usage" = 
      "HeadSize specifies an absolute size for the head of a vector."; 
    ScaleFactor::"usage" = 
      "ScaleFactor specifies the amount to scale a vector in length and may \
be either a number or a pure function."; ZeroVectorPointSize::"usage" = 
      "ZeroVectorPointSize is the size of the point used to represent a \
zero vector."; Options[Vector] = 
      Options[Arrow] = 
        Options[ArrowHead] = {HeadRatio -> 0.18, 
            TipRatio -> 0.14, Aperture -> 0.3, 
            ScaleFactor -> 1, HeadType -> Polygon, 
            ShaftQ -> True, EdgesQ -> True, 
            VectorColor -> RGBColor[0, 0, 1], ShaftWidth -> 0.005, 
            ZeroVectorPointSize -> 0.01}; Begin["`Private`"]; SetOptions[ParametricPlot, AspectRatio -> Automatic]; SetOptions[Plot, AspectRatio -> Automatic]; SetOptions[Graphics, AspectRatio -> Automatic]; Axes3D[u_, v_] := 
      Graphics3D[{{Blue, Line[{{-(u/3), 0, 0}, {u, 0, 0}}]}, 
          Text["x", {u + v, 0, 0}], {Blue, 
            Line[{{0, -(u/3), 0}, {0, u, 0}}]}, 
          Text["y", {0, u + v, 0}], {Blue, 
            Line[{{0, 0, -(u/3)}, {0, 0, u}}]}, 
          Text["z", {0, 0, u + v}]}]; Axes3D[u_] := Axes3D[u, (u/8)]; Norm[a_?VectorQ] := N[(Sqrt[a . a])]; Normalize[a_?VectorQ] := If[a . a == 0, a, (a/Norm)[a]]; Perpend[a_?VectorQ] := 
      Normalize[{a[[2]], 
            -a[[1]]}] /; 
        Length[a] == 2Perpend[a_?VectorQ] := 
      Normalize[
          If[a[[1]] == 0, {1, 0, 0}, {0, 
              a[[3]], 
              -a[[2]]}]] /; 
        Length[a] == 3Base3D = 
      Table[{Cos[((2 k π)/8.) ], Sin[((2 k π)/8.) ]}, {k, 0, 
          8}]; Vector[from_?VectorQ, to_?VectorQ, opts___] := 
      Module[{sq, sw, ht, hs, hr, tr, ap, sf, vc, zvps, diff = to - from, 
            tip}, {sq, sw, ht, hs, hr, ts, tr, ap, sf, vc, zvps} = 
            {ShaftQ, ShaftWidth, HeadType, HeadSize, HeadRatio, TipSize, 
                  TipRatio, Aperture, ScaleFactor, VectorColor, 
                  ZeroVectorPointSize} /. {opts} /. Options[Vector]; 
          If[from == to, 
            Return[Graphics[{PointSize[zvps], vc, Point[from]}]]]; 
          diff = If[NumberQ[N[sf]], sf diff, sf[diff]]; tip = from + diff; 
          If[NumberQ[hs], hr = (hs/Norm)[diff]; tr = 0.75 hr]; 
          If[NumberQ[ts], tr = (ts/Norm)[diff]]; 
          Graphics[{
              If[sq, {vc, Thickness[sw], Line[{from, tip - tr diff}]}, {}], 
              With[{trans = ap hr Norm[diff] Perpend[diff]}, {vc, 
                  Thickness[sw], 
                  ht[{tip, tip - hr diff + trans, tip - tr diff, 
                      tip - hr diff - trans, tip}]}]}]] /; 
        Length[from] == Length[to] == 2Vector[from_?VectorQ, to_?VectorQ, opts___] := 
      Module[{sq, eq, sw, hs, hr, tr, ap, sf, vc, zvps, diff = to - from, 
            tip}, {sq, eq, sw, hs, hr, ap, sf, vc, zvps} = 
            {ShaftQ, EdgesQ, ShaftWidth, HeadSize, HeadRatio, Aperture, 
                  ScaleFactor, VectorColor, ZeroVectorPointSize} /. {opts}
                   /. Options[Vector]; 
          If[from == to, 
            Return[Graphics3D[{PointSize[zvps], vc, Point[from]}]]]; 
          diff = If[NumberQ[N[sf]], sf diff, sf[diff]]; tip = from + diff; 
          If[NumberQ[hs], hr = (hs/Norm)[diff]]; tr = (hr/2); 
          Graphics3D[{
              If[sq, {vc, Thickness[sw], Line[{from, tip - tr diff}]}, {}], 
              If[eq, {}, EdgeForm[]], {SurfaceColor[vc], 
                With[{temp = Perpend[diff]}, 
                  (Polygon[Append[#1, tip]]&)/@
                    Partition[
                      (tip - hr diff + #1&)/@
                        ((ap hr Norm[diff] Base3D) . {temp, 
                              Normalize[temp\[Cross]diff]}), 2, 1]]}}]] /; 
        Length[from] == Length[to] == 3Arrow[a_, Tail -> b_, opts___] := Vector[b, a + b, opts]Arrow[a_, opts___] := Vector[Table[0, {Length[a]}], a, opts]\(ArrowHead[a_, b_, opts___] := 
      Vector[a - b, a, ShaftQ -> False, opts]; 

End[]; 

    \(EndPackage[]; \)\)

Graphics`Colors`GosiaGreen = RGBColor[0,  0.392187,  0];

Spherical Setup

\(Clear[polar]; 

    polar[{x_, y_, z_}] := {\@\(x\^2 + y\^2 + z\^2\), ArcTan[y\/x], 
        ArcCos[z\/\@\(x\^2 + y\^2 + z\^2\)]}; 
Clear[X, length]; 

    length[X_] := \@\(X . X\); 
\(rr = 3; \)\)R = 5; scale = 0.8; red = RGBColor[1, 0, 0]; blue = RGBColor[0, 0, 1]; color1 = RGBColor[0.8, 0.2, 0.8]; color2 = RGBColor[0.2, 0.8, 0.2]; black = RGBColor[0, 0, 0]; \(Clear[axes]; 

    axes := Graphics3D[{Line[{X, {1.5\ rr, 0, 0}}], 
          Line[{X, {0, 1.5\ rr, 0}}], Line[{X, {0, 0, 2\ rr}}]}]; 

    Clear[xyplane]; 

    xyplane := 
      Graphics3D[
        Polygon[{{\(-2\)\ rr, \(-2\)\ rr, 0}, {\(-2\)\ rr, 2\ rr, 0}, {2\ rr, 
              2\ rr, 0}, {2\ rr, \(-2\)\ rr, 0}}]]; 
\(X = {0, 0, 0}; \)\)origin = Graphics3D[{black, Point[X]}]; Z = {2, 3, 5}; \(Clear[point]; 

    point := Graphics3D[{red, PointSize[0.03], Point[scale\ Z]}]; 

    \(Z0 = {0, 0, length[0.8\ Z]}; \)\)\(Clear[zpoint]; 
zpoint = Graphics3D[{black, Point[Z0]}]; 

    \(XY = scale\ Z\ {1, 1, 0}; \)\)\(Clear[xypoint]; 
xypoint := Graphics3D[{black, Point[XY]}]; 

    \(X0 = {length[scale\ Z\ {1, 1, 0}], 0, 0}; \)\)xpoint = Graphics3D[{black, Point[X0]}]; Y0 = {0, length[scale Z {1, 1, 0}], 0}; ypoint = Graphics3D[{black, Point[Y0]}]; \(Clear[ray]; 
ray := Graphics3D[{blue, Line[{X, scale\ Z}]}]; 

    Clear[ray1]; 
ray1 := Graphics3D[{color1, Line[{scale\ Z, Z}]}]; 

    Clear[xyray]; 
xyray := Graphics3D[{red, Line[{X, Z\ {1, 1, 0}}]}]; 

    \(rays := {ray, xyray}; \)\)xyaux = 
      (length[0.8 Z] (0.8 Z {1, 1, 0}))/
        length[0.8 Z {1, 1, 0}]; \(Clear[zarc]; 

    zarc = Graphics3D[{blue, 
          Line[Table[
              \(Cos[t]\ \) {0, 0, length[0.8\ Z]} + Sin[t]\ xyaux, {t, 0, 
                \(polar[0.8\ Z]\)[[3
                  ]],  .1}]]}]; 
Clear[zarc1]; 

    zarc1 = Graphics3D[{color1, 
          Line[Table[
              \(Cos[t]\ \) {0, 0, length[0.8\ Z]} + Sin[t]\ xyaux, {t, 
                \(polar[0.8\ Z]\)[[3]], 
                1,  .1}]]}]; 
Clear[zarc2]; 

    zarc2 = Graphics3D[{color1, 
          Line[Table[
              \(Cos[t]\ \) {0, 0, length[0.8\ Z]} + Sin[t]\ xyaux, {t, 
                \(-0.8\), 0,  .1}]]}]; 
Clear[xyarc]; 

    xyarc = Graphics3D[{blue, 
          Line[Table[
              Cos[t]\ X0 + Sin[t]\ Y0, {t, 0, 
                \(polar[XY]\)[[2
                  ]],  .1}]]}]; 
Clear[xyarc1]; 

    xyarc1 = Graphics3D[{color1, 
          Line[Table[Cos[t]\ X0 + Sin[t]\ Y0, {t, \(- .5\), 0,  .1}]]}]; 

    Clear[xyarc2]; 

    xyarc2 = Graphics3D[{color1, 
          Line[Table[
              Cos[t]\ X0 + Sin[t]\ Y0, {t, 
                \(polar[XY]\)[[2]], 
                \(polar[XY]\)[[2]] + 
                  0.3,  .1}]]}]; 
Clear[stick]; 

    stick = Graphics3D[{red, Line[{scale\ Z, XY}]}]; 

    \(xaxislabel = Text["x", {0.363, 0.26}]; \)\)yaxislabel = Text["y", {0.73, 0.3}]; zaxislabel = Text["z", { .5, 0.67}]; stickplot = 
      Show[xyplane, axes, ray, point, Boxed -> False, 
        ViewPoint -> {2.747, 1.594, 1.168}, 
        Epilog -> {xaxislabel, yaxislabel, zaxislabel}, 
        DisplayFunction -> Identity]; rho = Text["r", {0.53, 0.45}]; rlabel = 
      Show[xyplane, axes, ray, point, Boxed -> False, 
        ViewPoint -> {2.747, 1.594, 1.168}, 
        Epilog -> {rho, xaxislabel, yaxislabel, zaxislabel}, 
        DisplayFunction -> Identity]; phi = Text["s", {0.55, 0.57}]; slabel = 
      Show[xyplane, zpoint, axes, ray, point, zarc, zarc1, zarc2, 
        Boxed -> False, ViewPoint -> {2.747, 1.594, 1.168}, 
        Epilog -> {phi, xaxislabel, yaxislabel, zaxislabel}, 
        DisplayFunction -> Identity]; theta = Text["t", { .5, 0.3}]; tlabel = 
      Show[xyplane, axes, xypoint, xpoint, ray, stick, point, xyarc, xyarc1, 
        xyarc2, Boxed -> False, ViewPoint -> {2.747, 1.594, 1.168}, 
        Epilog -> {theta, xaxislabel, yaxislabel, zaxislabel}, 
        DisplayFunction -> Identity]; all = 
      Show[xyplane, axes, ray, stick, point, zpoint, xpoint, xypoint, xyarc, 
        xyarc1, xyarc2, zarc, zarc1, zarc2, Boxed -> False, 
        ViewPoint -> {2.747, 1.594, 1.168}, 
        Epilog -> {theta, phi, rho, xaxislabel, yaxislabel, zaxislabel}, 
        DisplayFunction -> Identity];

Crisscross

\(Clear[x]; 

    crisscross = 
      Plot[{Cos[x], Sin[x]}, {x, π\/8, \(3\ π\)\/8}, 
        PlotStyle -> {{RGBColor[1, 0, 0], Thickness[0.02]}}, 
        Axes -> None, AspectRatio -> 1\/3, 
        DisplayFunction -> Identity]; \)

G.1)  Transforming 2D integrals*

G.1.a)

Here's a plot of the part of the surface
       z = (e^(-((x^2) +  4  (y^2))))
above the ellipse 
       ((((x/2)))^2)+ (y^2)=1 
in the xy-plane:

\(Clear[x, y, r, t]; 
    
{x[r_, t_], y[r_, t_]} = {2\ r\ Cos[t], r\ Sin[t]}; 

    \({{rlow, rhigh}, {tlow, thigh}} = {{0, 1}, {0, 2\ π}}; \)\)ParametricPlot3D[{x[r, t], y[r, t], 
        (E^(-(x[r, t]^2 + 4 y[r, t]^2)))}, {r, rlow, rhigh}, {t, 
        tlow, thigh}, AxesLabel -> {"x", "y", "z"}, 
      ViewPoint -> CMView];

Measure the volume of the solid whose top skin is the surface plotted above and whose base is everything on xy-plane directly below this surface.

G.1.b)

Here's a plot of a region (Subscript[R, xy]) on xy-paper:

\(Clear[x, y, r, t]; 
    
{x[r_, t_], y[r_, t_]} = {5\ r\ Cos[t], 3\ r\ Sin[t]}; 

    \({rlow, rhigh} = {1, 2}; \)\){tlow, thigh} = {-2, 2}; twosides = 
      ParametricPlot[{{x[rlow, t], y[rlow, t]}, {x[rhigh, t], y[rhigh, t]}}, {
          t, tlow, thigh}, PlotStyle -> {{Thickness[0.01], Red}}, 
        DisplayFunction -> Identity]; twomoresides = 
      ParametricPlot[{{x[r, tlow], y[r, tlow]}, {x[r, thigh], y[r, thigh]}}, {
          r, rlow, rhigh}, PlotStyle -> {{Thickness[0.01], Red}}, 
        DisplayFunction -> Identity]; \(nametag = 
      Graphics[Text[
          StyleForm["\!\(R\_xy\)", FontSize -> 16], {6, 3}]]; 


    Show[twosides, twomoresides, nametag, AspectRatio -> Automatic, 
      AxesLabel -> {"x", "y"}, 
      DisplayFunction -> $DisplayFunction]; \)

(Subscript[R, xy]) is everything inside and on this boundary.
Use a transformation to favorable uv-paper to measure the area of (Subscript[R, xy]) with little thought and with Mathematica doing the work.

G.1.c.i)

Here's a parallelogram plotted on xy-paper

Plot[{0.5 x - 1, 0.5 x + 2, -0.2 x, -0.2 x + 4}, {x, 
        -3.5, 7.5}, PlotStyle -> {{Magenta, Thickness[0.01]}}, 
      PlotRange -> {-0.5, 3.5}, AspectRatio -> Automatic, 
      AxesLabel -> {"x", "y"}];

Call R everything inside and on this parallelogram, and use a favorable transformation to help calculate  
       (∫_^)(∫_( R)^) 
          (e^(y -  x)) dx dy.

G.1.c.ii)

If a ≠ b, c < d, and r < s, then you are guaranteed that the lines
       y = a x +  c, y = a x +  d, 
       y = b x +  r,  and y = b x +  s 
define a parallelogram on xy-paper.
Reason:
Non-parallel lines cross each other.
Assume b < a, c < d, and r < s, and come up with a formula that measures the area of this parallelogram in terms of a, b, c, d, r, and s.

After you have your formula, make it look pretty by applying the Mathematica instruction Factor[Together[ ]].
After you do that, then you should get
       (((d  -  c) (s  -  r))/(b  -  a)).

G.1.d)

Use a transformation to favorable uv-paper to calculate 
       (∫_^)(∫_( 
            (Subscript[R, xy]))^)(x +  y) dx 
            dy
where (Subscript[R, xy]) is the region with x ≥ 0 and y ≥ 0 bounded by the curves
       (x^2) -  (y^2) = 1, 
       (x^2) -  (y^2) = 4,
       (x^2) +  (y^2) = 4, and
       (x^2) +  (y^2) = 9.

G.1.e)

Calculate
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          (e^(-(x^2) -  (y^2))) dx 
            dy
where (Subscript[R, xy]) is the region on xy-paper consisting of everything within and on the circle (x^2) +  (y^2) = 2.

G.2)  Ribbons*

G.2.a)

Here's  a curve on xy-paper:

\(Clear[x, y, r, t]; 
{x[t_], y[t_]} = {2\ t\ Cos[t], 2\ t\ Sin[t]}; 

    \({tlow, thigh} = {π, 3\ π}; \)\)curveplot = 
      ParametricPlot[{x[t], y[t]}, {t, tlow, thigh}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AspectRatio -> Automatic, PlotRange -> All, 
        AxesLabel -> {"x", "y"}];

Here's the boundary of a ribbon of constant width 1 centered on this curve:

Clear[unitnormal];
unitnormal[t_]=TrigExpand[(({y'[t],-x'[t]})/(Sqrt[((x'[t])^2)+((y'[t])^2)]))];
Clear[r, outerribbon, innerribbon, oneend, otherend];
outerribbon[t_] = {x[t], y[t]} + 0.5 unitnormal[t];
innerribbon[t_] = {x[t], y[t]} - 0.5 unitnormal[t]; oneend[r_] = {x[tlow], y[tlow]} + r unitnormal[tlow]; otherend[r_] = {x[thigh], y[thigh]} + r unitnormal[thigh]; {rlow = -0.5, rhigh = 0.5}; twosides = 
      ParametricPlot[{outerribbon[t], innerribbon[t]}, {t, tlow, thigh}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; \(twomoresides = 
      ParametricPlot[{oneend[r], otherend[r]}, {r, rlow, rhigh}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; 


    Show[curveplot, twosides, twomoresides, AspectRatio -> Automatic, 
      AxesLabel -> {"x", "y"}, 
      DisplayFunction -> $DisplayFunction]; \)

Change this ribbon to a new ribbon with constant width 2 centered on the given curve, and measure the area of the new ribbon.

G.2.b) A variable width ribbon

Go with the same base curve as in part i).

\(Clear[x, y, r, t]; 
{x[t_], y[t_]} = {2\ t\ Cos[t], 2\ t\ Sin[t]}; 

    \({tlow, thigh} = {π, 3\ π}; \)\)curveplot = 
      ParametricPlot[{x[t], y[t]}, {t, tlow, thigh}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AspectRatio -> Automatic, PlotRange -> All, 
        AxesLabel -> {"x", "y"}];

But this time make the width of the ribbon 
       2 +  4  Sin[2  t]^2 at {x[t], y[t]}.
Plot the resulting ribbon, and measure its area.

This ribbon is wider than the ribbon in part i).  As a result, your area measurement of this ribbon should be bigger than your area measurment of the ribbon in part i).

G.2.c)

Do something artistic with ribbons.
How about a real eye-catcher coming from your own mind?

G.3)  Flow measurements*

G.3.a)

To calculate the net flow of a vector field 
       Field[x, y] = {m[x, y], n[x, y]} 
across the boundary C of a region R, you have your choice:
-> You can go to the labor of parameterizing C, and then calculate 
       (∮_( C)) -n[x, y] dx  +  
        m[x, y] dy.
-> Or if the field has no singularities inside R, you can put 
       divField[x, y] = D[m[x, y], x] +  D[n[x, y], y] 
and calculate the 2D integral
       (∫_^)(∫_( R)) 
          divField[x, y] dx dy.
Here's a vector field:

\(Clear[x, y, m, n, Field]; 
    
{m[x_, y_], n[x_, y_]} = {x\^3 + y, x + y\^2}\)

R is everything inside the parallelogram you see below:

Plot[{0.6 x - 2, 0.6 x + 5, -0.3 x, -0.3 x + 4}, {x, 
        -5.6, 6.8}, PlotStyle -> {{Coral, Thickness[0.01]}}, 
      PlotRange -> {-0.8, 4.5}, AspectRatio -> Automatic, 
      AxesLabel -> {"x", "y"}];

Transform the 2D integral
       (∫_^)(∫_( R)) 
          divField[x, y] dx dy
to favorable uv-paper to measure the net flow of this vector field across the parallelogram.  Is the net flow of this vector field across this parallelogram from outside to inside or inside to outside?

G.3.b)

To calculate the net flow of a vector field 
       Field[x, y] = {m[x, y], n[x, y]} 
along the boundary C of a region R, you have your choice:
-> You can go to the labor of parameterizing C, and then calculate 
       (∮_( C)) m[x, y] dx  +  
        n[x, y] dy.
-> Or if the field has no singularities inside R, you can put 
       rotField[x, y] = D[n[x, y], x] -  D[m[x, y], y] 
and calculate the 2D integral
       (∫_^)(∫_( R)) rotField[x, y]dxdy.
Here's a vector field:

\(Clear[x, y, m, n, Field]; 
    
{m[x_, y_], n[x_, y_]} = {2\ Sin[y] + x, x + y}; \)

R is everything inside the parallelogram plotted in part a).
Transform the 2D integral
       (∫_^)(∫_( R)) rotField[x, y]dxdy
to favorable uv-paper to measure the net flow of this vector field along the parallelogram.  Is the net flow of this vector field along the parallelogram clockwise or counterclockwise?

G.4)  Interpret the plots

G.4.a)

All the plots below give information concerning the same phenomenon about what happens when you plot the region within the uv-paper square with corners at 
       {(1/2), -2}, {2, -2}, {2, 2} and {(1/2), 2} 
on xy-paper coming from 
       x[u, v] = Log[u] and y[u, v] = ArcTan[v].
Interpret the information conveyed by each plot.

\(Clear[x, y, u, v, gradx, grady, Axy]; 
x[u_, v_] = Log[u]; 

    \(y[u_, v_] = ArcTan[v]; \)\)\(gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 

    grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 

    Axy[u_, v_] = Det[{gradx[u, v], grady[u, v]}]\)

Plot3D[Axy[u, v], {u, (1/2), 2}, {v, -2, 2}, 
      AxesLabel -> {"u", "v", "Axy[u,v]"}, ViewPoint -> CMView];

\(Clear[gradAxy]; 
gradAxy[u_, v_] = {D[Axy[u, v], u], D[Axy[u, v], v]}; 
    
scalefactor = 0.3; 

    \(Show[Table[
        Arrow[gradAxy[u, v], Tail -> {u, v}, VectorColor -> Red, 
          ScaleFactor -> scalefactor], {u, 1\/2, 2, 1\/4}, {v, \(-2\), 
          2, 1\/4}], Axes -> Automatic]; \)\)

\(Clear[sizer]; 
scalefactor = 0.08; 

    \(sizer[u_, v_] = scalefactor\ \@Axy[u, v]; \)\)\(Clear[k]; 

    uvpoints = 
      Table[{Random[Real, {1\/2, 2}], Random[Real, {\(-2\), 2}]}, {k, 1, 
          150}]; 
\(Show[
      Table[Graphics[{
            PointSize[
              sizer@@uvpoints[[k]]], 
            Red, Point[
              uvpoints[[k]]]}], {k, 1, 
          150}], AspectRatio -> 1, Axes -> Automatic, 
      AxesLabel -> {"u", "v"}]; \)\)

ContourPlot[Axy[u, v], {u, (1/2), 2}, {v, -2, 2}];

The lighter the shading, the bigger (Subscript[A, xy])[u, v] is.

G.5)  Semi-log paper and log-log paper

Semi-log and log paper are friends of every scientist because they make analysis of exponential and power functions very easy. Back at the beginning of Calculus&Mathematica, you used semi-log paper to some advantage.
Here are the lines 
       y = (e^(-2)), y = (e^(-1)), y = (e^0), y = (e^1), y = (e^2), and y = (e^3)
plotted on  xy-paper for -5 ≤ x ≤ 15:

\(Clear[t]; 

    xylines = 
      ParametricPlot[{{t, 1\/E\^2}, {t, 1\/E}, {t, E\^0}, {t, E\^1}, {t, 
            E\^2}, {t, E\^3}}, {t, \(-5\), 15}, PlotStyle -> {{Blue}}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic]; 
    \)

Semi-log paper is uv-paper for 
       u[x, y] = x, and
       v[x, y] = Log[y].
Here are the same lines plotted on semi-log paper:

\(Clear[u, v, x, y]; 
u[x_, y_] = x; 
\(v[x_, y_] = Log[y]; \)\)uvlines = 
      ParametricPlot[{{u[t, (1/(E^2))], v[t, (1/(E^2))]}, {u[t, (1/E)], 
            v[t, (1/E)]}, {u[t, (E^0)], v[t, (E^0)]}, {u[t, (E^1)], v[t, (E^1)]}, {
            u[t, (E^2)], v[t, (E^2)]}, {u[t, (E^3)], v[t, (E^3)]}}, {t, -5, 
          15}, PlotStyle -> {Blue}, 
        AxesLabel -> {"u", "v"}, AspectRatio -> Automatic];

G.5.a)

Plot the xy-curves y = 3  (e^(-0.75 x)) and y = 2  (e^(1.5  x)) on semilog paper.
Describe what you see, and try to explain why you see it.
Why is it a good idea to plot xy-data on semi-log paper to reveal exponential relationships between the x-coordinate and the y-coordinate?

G.5.b) Log-log paper

Log-log paper is uv-paper for 
       u[x, y] = Log[x] and 
       v[x, y] = Log[y].
Plot the power curves y = 4  (x^(-0.6)) and y = 5  (x^1.4) on xy-paper, and then plot them on log-log paper.  
Why is the result not terribly surprising?
Why is it a good idea to plot xy-data on log-log paper to reveal power relationships between the x-coordinate and the y-coordinate?

G.6)  What can happen when (Subscript[A, xy])[u, v] is 0:
What does the sign of Det[{gradx[u, v], grady[u, v]}] tell you?

G.6.a)

Ever wonder what can happen when (Subscript[A, xy])[u, v] = 0?
The truth is that a uv-paper point {(Subscript[u, 0]), (Subscript[v, 0])} with 
       (Subscript[A, xy])[(Subscript[u, 0]), (Subscript[v, 0])] = 0
suffers the ultimate insult when it is plotted on xy-paper. 
Here's why: 
If (Subscript[A, xy])[(Subscript[u, 0]), (Subscript[v, 0])] = 0, then the uv-paper point {(Subscript[u, 0]), (Subscript[v, 0])} effectively disappears when it is plotted on xy-paper.
This sets up a situation in which crazy things can happen, because the xy-paper plot in the vicinity of one of these unlucky points is compressed so much.
Here's a sample:

\(Clear[u, v, x, y]; 
x[u_, v_] = u\^2 - v\^2\)

y[u_, v_] = u v + 10

\(Clear[gradx, grady, Axy]; 

    gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 

    grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 

    Axy[u_, v_] = Det[{gradx[u, v], grady[u, v]}]\)

At {(Subscript[u, 0]), (Subscript[v, 0])} = {0, 0}, the uv-paper to xy-paper area conversion factor 
       (Subscript[A, xy])[u, v] 
is 0, so {0, 0} is rubbed out on xy-paper.  On xy-paper strange things should happen near the plot of the uv-paper point {0, 0}. 
The uv-paper circle 
       (((u -  3))^2) +  (((v -  4))^2) = (5^2) 
goes right through {0, 0}:

\(Clear[u, v, t]; 
{u[t_], v[t_]} = {3, 4} + 5\ {Cos[t], Sin[t]}; 

    \(ParametricPlot[{u[t], v[t]}, {t, 0, 2\ π}, 
      PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, AxesLabel -> {"u", "v"}]; 
    \)\)

Now look at the plot of this circle on xy-paper:

ParametricPlot[{x[u[t], v[t]], y[u[t], v[t]]}, {t, 0, 2 π}, 
      PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, AxesLabel -> {"x", "y"}];

Sure enough; a dimple right at {0, 10}, which is the xy-plot of the uv-point {0, 0} because:

{x[0, 0], y[0, 0]}

Now you try one.

G.6.a.i)

Here's a parametric plot of the parabola v = (u^2) -  2  u on uv-paper:

\(Clear[t]; 

    ParametricPlot[{t, t\^2 - 2\ t}, {t, \(-1\), 2}, 
      PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, AxesLabel -> {"u", "v"}]; 
    \)

Go with:

\(Clear[u, v, x, y]; 
{x[u_, v_], y[u_, v_]} = {u\^2 - v\^2, u\ v}\)

Here is a plot of the same parabola on xy-paper:

ParametricPlot[{x[t, (t^2) - 2 t], y[t, (t^2) - 2 t]}, {t, -1, 
        2}, PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, PlotRange -> All, 
      AxesLabel -> {"x", "y"}];

Try to explain why you see what you see.

G.6.a.ii)

Here is a parametric plot of the ellipse 
       ((((u/4)))^2)+ (((((v  -  2)/2)))^2)=1 
on uv-paper:

\(Clear[t]; 

    ParametricPlot[{4\ Cos[t], 2 + 2\ Sin[t]}, {t, 0, 2\ π}, 
      PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, AxesLabel -> {"u", "v"}, 
      AxesOrigin -> {0, 0}]; \)

Go with polar paper:

\(Clear[u, v, x, y]; 
    
{x[u_, v_], y[u_, v_]} = {u\ Cos[v], u\ Sin[v]}\)

Here's a plot of the same ellipse on xy-paper:

\(Clear[t]; 

    ParametricPlot[{x[4\ Cos[t], 2 + 2\ Sin[t]], 
        y[4\ Cos[t], 2 + 2\ Sin[t]]}, {t, 0, 2\ π}, 
      PlotStyle -> {{Thickness[0.01], Red}}, 
      AspectRatio -> Automatic, AxesLabel -> {"u", "v"}]; 
    \)

Try to explain why you see what you see.

G.6.b.i) The sign of Det[{gradx[u, v], grady[u, v]}]

Here is a rather small circle on uv-paper parameterized in the counterclockwise direction:

\(Clear[u, v, t, uvpaperplotter]; 
r = 0.2; 

    \(uvpaperplotter[t_] = {r\ Cos[t], r\ Sin[t]}; \)\)uvpaperplot = 
      ParametricPlot[uvpaperplotter[t], {t, 0, 2 π}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; jump = (π/4); uvtangentvectors=Table[Arrow[uvpaperplotter'[t],Tail -> uvpaperplotter[t],VectorColor -> Blue],{t, 0, 2 π - jump, jump}];
Show[uvpaperplot, uvtangentvectors, AspectRatio -> Automatic, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> $DisplayFunction];

Now go to xy-paper with:

\(Clear[x, y, u, v]; 
    
{x[u_, v_], y[u_, v_]} = {3\ u - v + 3, u - 2\ v + 3}\)

Here's the xy-paper plot of the same circle with tangent vectors coming from its resulting xy-parameterization:

\(Clear[xypaperplotter]; 

    xypaperplotter[t_] = {x@@uvpaperplotter[t], y@@uvpaperplotter[t]}; 

    \(xypaperplot = 
      ParametricPlot[xypaperplotter[t], {t, 0, 2\ π}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; \)\)jump = (π/4); xytangentvectors=Table[Arrow[xypaperplotter'[t],Tail -> xypaperplotter[t],VectorColor -> Blue],{t, 0, 2 π - jump, jump}];
Show[xypaperplot, xytangentvectors, AspectRatio -> Automatic, 
        AxesLabel -> {"x", "y"}, 
        DisplayFunction -> $DisplayFunction];

Even though the curve was parameterized in the counterclockwise direction on the original uv-paper, it is parameterized in the clockwise direction on xy-paper.
Now look at (Subscript[A, xy])[u, v] with the absolute value dropped:

\(Clear[gradx, grady, Axy]; 

    gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 

    grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 

    Axy[u_, v_] = Det[{gradx[u, v], grady[u, v]}]\)

Negative.
Do you think this is an accident?

G.6.b.ii)

Here's what happens on xy-paper to the same uv-circle when you go with
       x[u, v] = 3  u -  v +  3, and 
       y[u, v] = u +  2  v:

\(Clear[x, y, u, v, xypaperplotter]; 
    
{x[u_, v_], y[u_, v_]} = {3\ u - v + 3, u + 2\ v}; 

    \(xypaperplotter[t_] = {x@@uvpaperplotter[t], y@@uvpaperplotter[t]}; 
    \)\)xypaperplot = 
      ParametricPlot[xypaperplotter[t], {t, 0, 2 π}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; jump = (π/4); xytangentvectors=Table[Arrow[xypaperplotter'[t],Tail -> xypaperplotter[t],VectorColor -> Blue],{t, 0, 2 π - jump, jump}];

Show[xypaperplot, xytangentvectors, AspectRatio -> Automatic, 
        AxesLabel -> {"x", "y"}, 
        DisplayFunction -> $DisplayFunction];

Counterclockwise on xy-paper.
Now look at (Subscript[A, xy])[u, v] with the absolute value dropped:

\(Clear[gradx, grady, Axy]; 

    gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 

    grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 

    Axy[u_, v_] = Det[{gradx[u, v], grady[u, v]}]\)

Are you surprised?
Why or why not?

G.6.b.iii)

Fill in numbers 
       a, b, c, d, e, and f 
of your own choice.  Go with
       x[u, v] = a u +  b v +  c, and 
       y[u, v] = d u +  e v +  f,
and investigate, as above, what happens to the same uv-circle on xy-paper, and how what happens is related to the sign of Det[{gradx[u, v], grady[u, v]}].

G.6.b.iv)

Here are two small circles on uv-paper parameterized in the counterclockwise direction:

\(Clear[u, v, t, uvpaperplotter1, uvpaperplotter2]; 
r = 0.5; 

    \(uvpaperplotter1[t_] = {\(-1\), 0} + {r\ Cos[t], r\ Sin[t]}; \)\)uvpaperplotter2[t_] = {1, 0} + {r Cos[t], r Sin[t]}; uvpaperplot = 
      ParametricPlot[{uvpaperplotter1[t], uvpaperplotter2[t]}, {t, 0, 
          2 π}, PlotStyle -> {{Thickness[0.01], Red}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; jump = (π/4); uvtangentvectors=Table[{Arrow[uvpaperplotter1'[t],Tail -> uvpaperplotter1[t],VectorColor -> Blue],Arrow[uvpaperplotter2'[t],Tail -> uvpaperplotter2[t],VectorColor -> Blue]},{t, 0, 2 π - jump, jump}];

Show[uvpaperplot, uvtangentvectors, AspectRatio -> Automatic, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> $DisplayFunction];

Here are the xy-paper plot of the same circles with tangent vectors coming from its own xy-parameterization when you go with
       x[u, v] = u Cos[v], and 
       y[u, v] = u Sin[v]:

\(Clear[x, y, u, v]; 
{x[u_, v_], y[u_, v_]} = {u\ Cos[v], u\ Sin[v]}; 

    Clear[xypaperplotter1, xypaperplotter2]; 

    xypaperplotter1[t_] = {x@@uvpaperplotter1[t], y@@uvpaperplotter1[t]}; 

    \(xypaperplotter2[t_] = {x@@uvpaperplotter2[t], y@@uvpaperplotter2[t]}; 
    \)\)xypaperplot = 
      ParametricPlot[{xypaperplotter1[t], xypaperplotter2[t]}, {t, 0, 
          2 π}, PlotStyle -> {{Thickness[0.01], Red}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; jump = (π/4); xytangentvectors=Table[{Arrow[xypaperplotter1'[t],Tail -> xypaperplotter1[t],VectorColor -> Blue],Arrow[xypaperplotter2'[t],Tail -> xypaperplotter2[t],VectorColor -> Blue]},{t, 0, 2 π - jump, jump}];

Show[xypaperplot, xytangentvectors, AspectRatio -> Automatic, 
        AxesLabel -> {"x", "y"}, 
        DisplayFunction -> $DisplayFunction];

Now look at:

\(Clear[gradx, grady, Axy]; 

    gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 

    grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 

    Axy[u_, v_] = TrigExpand[Det[{gradx[u, v], grady[u, v]}]]\)

Say why you think one of xy-parameterizations was reversed from counterclockwise to clockwise, but the other remained counterclockwise.

Look hard at the plots of the original circles on uv-paper:

Show[uvpaperplot, uvtangentvectors, AspectRatio -> Automatic, 
      AxesLabel -> {"u", "v"}, 
      DisplayFunction -> $DisplayFunction];

And look hard at:

Axy[u, v]

G.7)  Volume, mass, and density

G.7.a.i)

An object is made by forming a uniform substance that has density 1.71  (grams/(unit^3)) in the shape of of the surface  
       f[x, y]=(3.14/(1  +  (x^2)  +  (y^2))) 
over the region inside and on the circle (x^2) +  (y^2) = 8 on  xy-paper.
Use polar coordinate paper with 
       x[u, v] = u Cos[v], and 
       y[u, v] = u Sin[v]
to measure the total mass and volume of this object.

G.7.a.ii)

When you rubberized the xy-paper and stretched it out to make the uv-paper, you also deformed the original shape of the object in part i).
What does the resulting deformed object look like when it is plotted on uv-paper for the functions 
       x[u, v] = u Cos[v], and 
       y[u, v] = u Sin[v] 
used above in part i)?
What is the density of this deformed object at a point {u, v} inside the uv-paper rectangle 0 ≤ u ≤ Sqrt[8] and 0 ≤ v ≤ 2  π?

G.8)  Two recreational plots

G.8.a) Flow out of an open pipe

Here is a uv-paper plot of the line segments
       v = -π, v = -((2 π)/3), v = -(π/3), v = -(π/6),
       v = 0, v = (π/6), v = (π/3), v = ((2 π)/3), and v = π 
with -6 ≤ u ≤ 2:

\(Clear[t]; 

    uvplot = ParametricPlot[{{t, \(-π\)}, {t, \(-\(\(2\ π\)\/3\)\)}, {
            t, \(-\(π\/3\)\)}, {t, \(-\(π\/6\)\)}, {t, 0}, {t, 
            π\/6}, {t, π\/3}, {t, \(2\ π\)\/3}, {t, π}}, {t, 
          \(-6\), 2}, PlotStyle -> {{Thickness[0.01], Blue}}, 
        AxesLabel -> {"u", "v"}]; \)

Plot these curves on xy-paper for 
       x[u, v] = u +  (e^u) Cos[v], and 
       y[u, v] = v +  (e^u) Sin[v].
If your xy-paper plot is correct, then it will depict streamlines of water flow out of an open pipe.

G.8.b) Airfoils

Ever wonder how to plot some airfoils (airplane wings)? 
If so, then wait no longer.
Take 
       x[u, v] = (((u^3) +  u (v^2) +  u)/((u^2) +  (v^2))), and 
       y[u, v] = (((v^3) +  (u^2) v -  v)/((u^2) +  (v^2))),
and plot on xy-paper some uv-paper circles 
       (((u -  a))^2) +  (v^2) = (((1 +  a))^2) 
for several choices of a with 0 < a < 1.

G.9)  What went wrong?*

G.9.a)

Let (Subscript[R, xy]) be the region on xy-paper consisting of everything inside and on the circle
        (x^2) +  (y^2) = 4.  
Polar cordinate paper is handy for calculating
       (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy:
Go to polar coordinates via:

\(Clear[x, y, u, v, gradx, grady, Axy]; 
x[u_, v_] = u\ Cos[v]; 

    y[u_, v_] = u\ Sin[v]; 
gradx[u_, v_] = {D[x[u, v], u], D[x[u, v], v]}; 
    
grady[u_, v_] = {D[y[u, v], u], D[y[u, v], v]}; 


    Axy[u_, v_] = TrigExpand[Det[{gradx[u, v], grady[u, v]}]]\)

\(Clear[f]; 
f[x_, y_] = x\^2 + y\^2\)

Here are six attempts at calculations of 
       (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy.
Identify the correct calculations, and determine what went wrong in the incorrect calculations.  
There is always a possibility that a calculation produces a correct answer, but the method is wrong.  Identify these as well.

Calculation 1:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_0
        ^(2 π))(∫_0^4) f[x[u, v], y[u, v]] Axy[u, v] 
          \[DifferentialD]u \[DifferentialD]v

Calculation 2:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_0
        ^(2 π))(∫_0^2) f[x[u, v], y[u, v]] Axy[u, v] 
          \[DifferentialD]u \[DifferentialD]v

Calculation 3:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_0
        ^(4 π))(∫_0^2) f[x[u, v], y[u, v]] Axy[u, v] 
          \[DifferentialD]u \[DifferentialD]v

Calculation 4:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_(-2)^0)
        (∫_0^(2 π))f[x[u, v], y[u, v]] Axy[u, v] 
          \[DifferentialD]v \[DifferentialD]u

Calculation 5:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_0^2)
        (∫_(-π)^π) f[x[u, v], y[u, v]] Axy[u, v] 
          \[DifferentialD]v \[DifferentialD]u

Calculation 6:

The uv-paper rectangle 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2  π plots out on xy-paper as the circle (x^2) +  (y^2) = 4 and everything inside it.
So (∫_^)(∫_( 
            (Subscript[R, xy]))^)((x^2) +  (y^2)) dx 
            dy is given by:

(∫_0^2)
        (∫_(10 π)^(12 π))f[x[u, v], y[u, v]] 
            Axy[u, v] \[DifferentialD]v \[DifferentialD]u

G.10)  Linear equations and area conversion factors

When you go from xy-paper to uv-paper through formulas 
       u[x, y] = a x +  b y, and 
       v[x, y] = c x +  d y, 
for fixed numbers a, b, c, and d, things get pictorial.  For one thing, the uv-grid on xy-paper comes from straight lines:

{a, b, c, d} = {2, 5, -1, 3}; \(Clear[u, v, x, y]; 
u[x_, y_] = a\ x + b\ y; 

    \(v[x_, y_] = c\ x + d\ y; \)\){{xlow, xhigh}, {ylow, yhigh}} = {{-5, 5}, {-5, 5}}; ulevelcurves = 
      ContourPlot[Evaluate[u[x, y]], {x, xlow, xhigh}, {y, ylow, yhigh}, 
        Contours 
          -> {-30, -25, -20, -15, -10, -5, 0, 5, 
            10, 15, 20, 25, 30}, ContourShading -> False, 
        DisplayFunction -> Identity]; vlevelcurves = 
      ContourPlot[Evaluate[v[x, y]], {x, xlow, xhigh}, {y, ylow, yhigh}, 
        Contours 
          -> {-15, -10, -5, 0, 5, 10, 15, 20, 25, 30}, 
        ContourShading -> False, DisplayFunction -> Identity]; 
    \(uvGridonxyPaper = 
      Show[ulevelcurves, vlevelcurves, Frame -> False]; 


    Show[uvGridonxyPaper, Axes -> True, 
      AxesLabel -> {"x", "y"}, 
      DisplayFunction -> $DisplayFunction]; \)

This is one reason that lots of folks call the transformation from xy-paper to uv-paper coming from
       u[x, y] = a x +  b y, and 
       v[x, y] = c x +  d y
by the name linear transformation.  When you take a course called linear algebra, you will hear a lot about linear transformations.  In this problem, you get to see some stuff about linear transformations that you won't see in a standard linear algebra course.

G.10.a.i) Tripping

Go with:

\(Clear[x, y, u, v]; 
{a, b, c, d} = {2, \(-1\), 3, 2}; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

Make two vectors:

directionx = {a, c}

directiony = {b, d}

Look at these uv-paper plots of 
     x directionx with its tail at {0, 0} and 
     y directiony with its tail at the tip of x directionx 
and the point 
       {u[x, y], v[x, y]} for x = 0.4 and y = -0.3:

{x, y} = {0.4, -0.3}; uvpoint = 
      Graphics[{Red, PointSize[0.05], Point[{u[x, y], v[x, y]}]}]; \(trip = {Arrow[x\ directionx, Tail -> {0, 0}], 
        Arrow[y\ directiony, Tail -> x\ directionx]}; 

    labels = {Graphics[Text["x directionx", \(x\ directionx\)\/2]], 
        Graphics[Text["y directiony", x\ directionx + \(y\ directiony\)\/2]], 
        Graphics[Text["{u[x,y],v[x,y]", {u[x, y], v[x, y]}, {0, 4}]]}; 


    Show[uvpoint, labels, trip, PlotRange -> All, Axes -> True, 
      AxesLabel -> {"u", "v"}]; \)

Go again with the same u[x, y] and v[x, y] but this time take x = -2 and y = 4:

{x, y} = {-2, 4}; uvpoint = 
      Graphics[{Red, PointSize[0.05], Point[{u[x, y], v[x, y]}]}]; \(trip = {Arrow[x\ directionx, Tail -> {0, 0}], 
        Arrow[y\ directiony, Tail -> x\ directionx]}; 

    \(labels = {Graphics[Text["x directionx", \(x\ directionx\)\/2]], 
        Graphics[Text["y directiony", x\ directionx + \(y\ directiony\)\/2]], 
        Graphics[Text["{u[x,y],v[x,y]", {u[x, y], v[x, y]}, {0, 4}]]}; 


    Show[uvpoint, labels, trip, PlotRange -> All, Axes -> True, 
      AxesLabel -> {"u", "v"}]; \)\)

Read the code, edit it and and run a few more if you like. Once you are certain what's happening, explain the physical process of going with an xy-paper point {x, y} and the vectors directionx and directiony to take a trip starting at {0, 0} and ending at the uv-paper plot {u[x, y], v[x, y]} of the original xy-paper point {x, y}.

What you saw was the geometry of the trip.
Here's the algebra:

Clear[x, y, u, v, a, b, c, d]; {u[x_, y_], v[x_, y_]} = {a x + b y, c x + d y}

directionx = {a, c}

directiony = {b, d}

{u[x, y], v[x, y]} == x directionx + y directiony

G.10.a.ii) Tripping when directionx and directiony are either parallel or point in opposite directions

Go with:

\(Clear[x, y, u, v]; 
{a, b, c, d} = {4, 6, 2, 3}; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

Make two the vectors:

directionx = {a, c}

directiony = {b, d}

Note that the vectors directionx and directiony are parallel:

\(vectors = {
        Arrow[directionx, Tail -> {0, 0}, VectorColor -> Blue], 
        Arrow[directiony, Tail -> {0, 1}, VectorColor -> Blue]}; 

    labels = {Graphics[Text["directionx", directionx\/2]], 
        Graphics[Text["directiony", {0, 1} + directiony\/2]]}; 


    Show[vectors, labels, PlotRange -> All, Axes -> True, 
      AxesLabel -> {"u", "v"}]; \)

Look at these uv-paper plots of 
       x directionx with its tail at {0, 0}, and 
       y directiony with its tail at the tip of x directionx, 
and the point 
     {u[x, y], v[x, y]} for x = 1 and y = 2:

{x, y} = {1, 2}; uvpoint = 
      Graphics[{Red, PointSize[0.05], Point[{u[x, y], v[x, y]}]}]; \(trip = {
        Arrow[x\ directionx, Tail -> {0, 0}, VectorColor -> Blue], 
        Arrow[y\ directiony, Tail -> x\ directionx, 
          VectorColor -> Blue]}; 

    labels = {Graphics[Text["x directionx", \(x\ directionx\)\/2]], 
        Graphics[Text["y directiony", x\ directionx + \(y\ directiony\)\/2]], 
        Graphics[Text["{u[x,y],v[x,y]", {u[x, y], v[x, y]}, {0, 4}]]}; 


    Show[uvpoint, labels, trip, PlotRange -> All, Axes -> True, 
      AxesLabel -> {"u", "v"}]; \)

Go again with the same u[x, y] and v[x, y], but this time take x = -1 and y = -2:

{x, y} = {-1, -2}; uvpoint = 
      Graphics[{Red, PointSize[0.05], Point[{u[x, y], v[x, y]}]}]; \(trip = {
        Arrow[x\ directionx, Tail -> {0, 0}, VectorColor -> Blue], 
        Arrow[y\ directiony, Tail -> x\ directionx, 
          VectorColor -> Blue]}; 

    labels = {Graphics[Text["x directionx", \(x\ directionx\)\/2]], 
        Graphics[Text["y directiony", x\ directionx + \(y\ directiony\)\/2]], 
        Graphics[Text["{u[x,y],v[x,y]", {u[x, y], v[x, y]}, {0, 4}]]}; 


    Show[uvpoint, labels, trip, PlotRange -> All, Axes -> True, 
      AxesLabel -> {"u", "v"}]; \)

Edit and run a few more if you like. 
Once you smell the truth, explain where the points {u[x, y], v[x, y]} must plot out as x and y vary.

G.10.a.iii)

Continue to go with parallel vectors directionx and directiony as above.

\(Clear[x, y, u, v]; 
{a, b, c, d} = {4, 6, 2, 3}; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

Make two the vectors:

directionx = {a, c}

directiony = {b, d}

Remember that:

{u[x, y], v[x, y]} == x directionx + y directiony

Look at the following attempt to solve:

Solve[{0 == a x + b y, 2 == c x + d y}, {x, y}]

Use the fact that
       {u[x, y], v[x, y]} 
        = {a x +  b y, c x +  d y} 
        = x directionx +  y directiony 
to explain why there was never any hope of solving the simultaneous equations
       0 = a x +  b y,
       2 = c x +  d y
for x and y.  Then look at:

directionx

directiony

Say how the numbers u and v must be related to guarantee that the simultaneous equations
       u = a x +  b y
       v = c x +  d y
can be solved for x and y.

G.10.a.iv)

Continue to go with parallel vectors directionx and directiony as above.

\(Clear[x, y, u, v]; 
{a, b, c, d} = {4, 6, 2, 3}; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

This time, look at the xy-to-uv-area conversion factor:

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, v_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]\)

What happens to the xy-paper when it is stretched and compressed to make uv-paper?
How does this explain why when you go with  
       a = 4, b = 6, c = 2, and d = 3 
as above, then for all but very special choices of u and v, it will be impossible to solve
       u = a x +  b y
       v = c x +  d y
for x and y?

G.10.b)

This time go with cleared values of a, b, c, and d:

\(Clear[x, y, u, v, a, b, c, d]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

directionx = {a, c}

directiony = {b, d}

Look at the xy-to-uv-area conversion factor:

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, y_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]\)

Explain:
Saying that the vectors directionx and directiony are either parallel or point in opposite directions is the same as saying that the area conversion factor (Subscript[A, uv])[x, y] = 0.
And explain:
If you go with numbers a, b, c, and d that make
       (Subscript[A, uv])[x, y] = 0, 
then for all but very special choices of numbers u and v, it will be impossible to solve
       u = a x +  b y
       v = c x +  d y
for x and y.

G.10.b.ii)

Look at this:

Clear[x, y, u, v, a, b, c, d, t]; {u[x_, y_], v[x_, y_]} = {a x + b y, c x + d y}

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, y_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]\)

Now look at what most folks call the coefficient matrix:

coefficientmatrix = {{a, b}, {c, d}}; MatrixForm[coefficientmatrix]

And its determinant:

Det[coefficientmatrix]

Compare the xy-to-uv-area conversion factor to the determinant of the coefficient matrix:

Det[coefficientmatrix] == Auv[x, y]

Explain:
When you go with numbers a, b, c, and d that make
       determinant of the coefficient matrix = 0, 
then for all but very special choices of u and v, it will be impossible to solve
       u = a x +  b y
       v = c x +  d y
for x and y.
And explain: 
When you go with numbers a, b, c, and d that make
       determinant of the coefficient matrix ≠ 0,
then, no matter what choice of u and v you make, it will be possible to solve
       u = a x +  b y
       v = c x +  d y
for x and y.

G.11)  Eigenvalues and eigenvectors

One good way to extend your mathematical horizons is to look up some unfamiliar Mathematica instructions and try them out.  This is what you will do here with a little coaching from your friends at Calculus&Mathematica.
Take four numbers a, b, c, and d with c = b like this:

{a, b, d} = {3.5, 1.0, 4.2}; {{a, b}, {c, d}} = {{a, b}, {b, d}}; matrix = {{a, b}, {c, d}}; MatrixForm[matrix]

Go from xy-paper to uv-paper with

\(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}\)

Here is the xy-paper circle
       (x^2) +  (y^2) = 1 
plotted on xy-paper:

xyplot = 
      ParametricPlot[{Cos[t], Sin[t]}, {t, 0, 2 π}, 
        AspectRatio -> Automatic, 
        PlotStyle -> {{Blue, Thickness[0.01]}}, 
        AxesLabel -> {"x", "y"}, PlotLabel -> "xy paper plot"];

Here is how this circle plots out on uv-paper in true scale:

uvplot = 
      ParametricPlot[{u[Cos[t], Sin[t]], v[Cos[t], Sin[t]]}, {t, 0, 
          2 π}, AspectRatio -> Automatic, 
        PlotStyle -> {{Blue, Thickness[0.01]}}, 
        AxesLabel -> {"u", "v"}, PlotLabel -> "uv paper plot"];

You guessed right! 
This is a tilted ellipse centered at {0, 0}.
Here comes the unfamiliar Mathematica instruction:

eigens = Eigensystem[matrix]

The eigenvalues of the matrix are:

{eigenvalue1, eigenvalue2} = 
      eigens[[1]]

The eigenvectors of the matrix are:

{eigenvector1, eigenvector2} = 
      eigens[[2]]

The unit eigenvectors of the matrix are:

{uniteigen1, uniteigen2} = {
        (eigenvector1/(Sqrt[eigenvector1 . eigenvector1])), 
        (eigenvector2/(Sqrt[eigenvector2 . eigenvector2]))}

Here are the unit eigenvectors of this matrix together with a plot of the circle (x^2) +  (y^2) = 1:

Show[xyplot, 
      Arrow[uniteigen1, Tail -> {0, 0}, VectorColor -> Red], 
      Arrow[uniteigen2, Tail -> {0, 0}, VectorColor -> Red]];

Now put:

{amazing1, amazing2} = {eigenvalue1 uniteigen1, 
        eigenvalue2 uniteigen2}

And look at this plot:

Show[uvplot, 
      Arrow[amazing1, Tail -> {0, 0}, VectorColor -> Red], 
      Arrow[amazing2, Tail -> {0, 0}, VectorColor -> Red], 
      AspectRatio -> Automatic];

Bingo. 
Just for good measure, look at:

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, y_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]; 
    
{eigenvalue1\ eigenvalue2, Auv[x, y]}\)

Cowabunga.
The xy-to-uv-area conversion factor is the same as the product of the two eigenvalues.

G.11.a)

Remembering that the ellipse above is the uv-paper plot of the xy-paper circle (x^2) +  (y^2) = 1, write down the uv-paper measurement of the area enclosed by the ellipse plotted above.

G.11.b)

Do the it again for another choice of 
        a, b, c, and d with c = b:

{a, b, d} = {2.7, -5.3, 0.7}; {{a, b}, {c, d}} = {{a, b}, {b, d}}; matrix = {{a, b}, {c, d}}; \(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}; 

    \(uvplot = 
      ParametricPlot[{u[Cos[t], Sin[t]], v[Cos[t], Sin[t]]}, {t, 0, 
          2\ π}, AspectRatio -> Automatic, 
        PlotStyle -> {{Blue, Thickness[0.01]}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; \)\)eigens = Eigensystem[matrix]; {eigenvalue1, eigenvalue2} = 
      eigens[[1]]; {eigenvector1, eigenvector2} = 
      eigens[[2]]; {uniteigen1, uniteigen2} = {
        (eigenvector1/(Sqrt[eigenvector1 . eigenvector1])), 
        (eigenvector2/(Sqrt[eigenvector2 . eigenvector2]))}; \({amazing1, amazing2} = {eigenvalue1\ uniteigen1, 
        eigenvalue2\ uniteigen2}; 


    Show[uvplot, 
      Arrow[amazing1, Tail -> {0, 0}, VectorColor -> Red], 
      Arrow[amazing2, Tail -> {0, 0}, VectorColor -> Red], 
      DisplayFunction -> $DisplayFunction]; \)

Bingo again.
Again just for good measure, look at:

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, y_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]; 
    
{eigenvalue1\ eigenvalue2, Auv[x, y]}\)

Totally cool.
Play with some more choices of a, b, c, and d with b = c until you get enough experience to form your own opinions about the anwers to the questions:
-> What do you think unit eigenvectors are?
-> What do you think eigenvalues are?
-> Why are lots of folks delighted with eigenvectors and eigenvalues?

Don't look up the answer in a linear algebra book because most linear algebra books don't look at eigenvector and eigenvalues from this visual perspective. Instead, they make eigenvector and eigenvalues the final products of some dreary algebra.
How sad.

G.11.c)

Look at this one:

{a, b, d} = {2, 2, 2}; {{a, b}, {c, d}} = {{a, b}, {b, d}}; matrix = {{a, b}, {c, d}}; \(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}; 

    \(uvplot = 
      ParametricPlot[{u[Cos[t], Sin[t]], v[Cos[t], Sin[t]]}, {t, 0, 
          2\ π}, AspectRatio -> Automatic, 
        PlotStyle -> {{Blue, Thickness[0.01]}}, 
        AxesLabel -> {"u", "v"}, 
        DisplayFunction -> Identity]; \)\)eigens = Eigensystem[matrix]; {eigenvalue1, eigenvalue2} = 
      eigens[[1]]; {eigenvector1, eigenvector2} = 
      eigens[[2]]; {uniteigen1, uniteigen2} = {
        (eigenvector1/(Sqrt[eigenvector1 . eigenvector1])), 
        (eigenvector2/(Sqrt[eigenvector2 . eigenvector2]))}; {amazing1, amazing2} = {eigenvalue1 uniteigen1, 
        eigenvalue2 uniteigen2}; Show[uvplot, DisplayFunction -> $DisplayFunction];

Show[uvplot, 
      Arrow[amazing1, Tail -> {0, 0}, VectorColor -> Red], 
      Arrow[amazing2, Tail -> {0, 0}, VectorColor -> Red], 
      DisplayFunction -> $DisplayFunction];

{eigenvalue1, eigenvalue2}

\(Clear[gradu, gradv, Auv]; 

    gradu[x_, y_] = {D[u[x, y], x], D[u[x, y], y]}; 

    gradv[x_, y_] = {D[v[x, y], x], D[v[x, y], y]}; 

    Auv[x_, y_] = Det[{gradu[x, y], gradv[x, y]}]; 
    
{eigenvalue1\ eigenvalue2, Auv[x, y]}\)

What do you think happened to the ellipse?

G.11.d)

If what you've seen so far hasn't blown your mind, look at these:

{a, b, d} = {9, -4, 2}; {{a, b}, {c, d}} = {{a, b}, {b, d}}; matrix = {{a, b}, {c, d}}; \(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}; 

    \(eigens = Eigensystem[matrix]; \)\){eigenvalue1, eigenvalue2} = 
      eigens[[1]]; {eigenvector1, eigenvector2} = 
      eigens[[2]]; {uniteigen1, uniteigen2} = {
        (eigenvector1/(Sqrt[eigenvector1 . eigenvector1])), 
        (eigenvector2/(Sqrt[eigenvector2 . eigenvector2]))}; {amazing1, amazing2} = {eigenvalue1 uniteigen1, 
        eigenvalue2 uniteigen2}; {u[x, y], v[x, y]} == 
      ExpandAll[
        Together[{x, y} . uniteigen1 amazing1 + {x, y} . uniteigen2 
              amazing2]]

It works for any a, b, c, and d you feed in as long as b = c:

{a, b, d} = {Random[Real, {-10, 10}], 
        Random[Real, {-10, 10}], Random[Real, {-10, 10}]}; {{a, b}, {c, d}} = {{a, b}, {b, d}}; matrix = {{a, b}, {c, d}}; \(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {a\ x + b\ y, c\ x + d\ y}; 

    \(eigens = Eigensystem[matrix]; \)\){eigenvalue1, eigenvalue2} = 
      eigens[[1]]; {eigenvector1, eigenvector2} = 
      eigens[[2]]; {uniteigen1, uniteigen2} = {
        (eigenvector1/(Sqrt[eigenvector1 . eigenvector1])), 
        (eigenvector2/(Sqrt[eigenvector2 . eigenvector2]))}; {amazing1, amazing2} = {eigenvalue1 uniteigen1, 
        eigenvalue2 uniteigen2}; {u[x, y], v[x, y]} == 
      ExpandAll[
        Together[{x, y} . uniteigen1 amazing1 + {x, y} . uniteigen2 
              amazing2]]

In other words when you go with any numbers a, b, c, and d with b = c, and you put 
       u[x, y] = a x +  b y, and 
       v[x, y] = c x +  d y, 
then you can count on having
       {u[x, y], v[x, y]}
        = 
        ({x, y} . uniteigen1) amazing1 +  
          ({x, y} . uniteigen2) amazing2.
This resolves {u[x, y], v[x, y]} into two perpendicular components in the directions of the two eigenvectors. 
Fancy folks call this the "spectral theorem" and make this the centerpiece of a good linear algebra course.  If you did the problem G.10) above, then some interesting plots might come to your mind.
But you are not asked to write anything about what all this means; so sit back and enjoy.
On the other hand, if you've got something to say, then go ahead and indulge yourself.