VC.07 Transforming 2D Integrals
TUTORIALS

\(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" = 
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    \(Perpend::"usage" = 
      "Perpend[a] returns a unit vector which is perpendicular to the vector a."; \)\)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" = 
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length of the shaft of a vector."; HeadRatio::"usage" = 
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length of the shaft of a vector."; HeadSize::"usage" = 
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    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[{
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              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[{
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                    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]; \)

T.1)  Transforming (∫_^)(∫_( (Subscript[R, xy]))^) 
          f[x, y] dx dy when the boundary of (Subscript[R, xy]) is given by parametric formulas

When the boundary of a region (Subscript[R, xy]) is plotted with parametric formulas, you often have a good shot at using the parametric formulas to come up with uv-paper on which (Subscript[R, uv]) is a rectangle. This is especially good because calculating 2D integrals over rectangles is usually very easy.

T.1.a.i)

The region (Subscript[R, xy]) is everything inside and on the circle
       (((x -  1))^2) +  (((y +  2))^2) = 5:

\(Clear[x, y, t]; 
{x[t_], y[t_]} = {1, \(-2\)} + \@5\ {Cos[t], Sin[t]}; 
    
\(Rxyplot = 
      ParametricPlot[{x[t], y[t]}, {t, 0, 2\ π}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AspectRatio -> Automatic, AxesLabel -> {"x", "y"}, 
        Epilog -> 
          Text[StyleForm["\!\(R\_\( x y\)\)", FontSize -> 16], {1, 
              \(-0.5\)}]]; \)\)

Calculate 
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          (3  (x^2) +  5  (y^4)) dx 
            dy

Look at the functions used to plot the boundary of (Subscript[R, xy]) :

{x[t], y[t]}

Put

\(Clear[x, y, u, t]; 
    
{x[u_, t_], y[u_, t_]} = {1, \(-2\)} + u\ {Cos[t], Sin[t]}\)

Realize this:
When you run u from 0 to (Sqrt[5]) , and you run t from 0 to 2  π, then {x[u, t], y[u, t]} runs through all of (Subscript[R, xy]):

{{ulow = 0, uhigh = (Sqrt[5])}, {tlow = 0, thigh = 2 π}}

The upshot:
(Subscript[R, uv]) is the rectangle 
       ulow ≤ u ≤ uhigh and tlow ≤ t ≤ thigh
on ut-paper.
This, and the fact that  
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          f[x, y] dx dy
       
        = 
        (∫_^)(∫_( (Subscript[R, ut]))^) 
            f[x[u, t], y[u, t]] (Subscript[A, xy])[u, t] du 
              dt
       
        = 
        (∫_tlow^thigh)(∫_ulow^uhigh) 
            f[x[u, t], y[u, t]] (Subscript[A, xy])[u, t] du 
              dt,
tell you that you can turn everything over to the machine,
right now:

\(Clear[gradx, grady, Axy]; 

    gradx[u_, t_] = {D[x[u, t], u], D[x[u, t], t]}; 

    grady[u_, t_] = {D[y[u, t], u], D[y[u, t], t]}; 

    Axy[u_, t_] = TrigExpand[Det[{gradx[u, t], grady[u, t]}]]\)

Here comes the calculation of (∫_^)(∫_( (Subscript[R, xy]))^) 
          (3  (x^2) +  5  (y^4)) dx 
            dy:

calculation = 
      Simplify[(∫_tlow^thigh)
            (∫_ulow^uhigh)
              (3 x[u, t]^2 + 5 y[u, t]^4) Axy[u, t] 
              \[DifferentialD]u \[DifferentialD]t]

Nasty answer, but it wasn't hard to get.

T.1.a.ii)

Could you have used the Gauss-Green formula to calculate the integral in part i)?

Yes. 
Here is how it goes:
You want to calculate (∫_^)(∫_( (Subscript[R, xy]))^) 
          (3  (x^2) +  5  (y^4)) dx 
            dy,
and you have a counterclockwise parameterization of the boundary of (Subscript[R, xy]):

\(Clear[x, y, f, t]; 
f[x_, y_] = 3\ x\^2 + 5\ y\^4; 

    \({x[t_], y[t_]} = {1, \(-2\)} + \@5\ {Cos[t], Sin[t]}; \)\){a, b} = {0, 2 π}

With a ≤ t ≤ b.
The Gauss-Green formula says
       (∫_^)(∫_( (Subscript[R, xy]))) 
          (D[n[x, y], x] -  D[m[x, y], y]) dx 
            dy
        = 
        (∫_a^b) m[x[t], y[t]] x'[t] +  
          n[x[t], y[t]] y'[t] dt,
where a = 0 and b = 2  π. 
To calculate 
       (∫_^)(∫_( (Subscript[R, xy]))) 
          f[x, y] dx dy,
You just say 
       m[x, y] = 0, and 
       n[x, y] = (∫_0^x) f[s, y] ds:

\(Clear[n, m, s, y]; 
m[x_, y_] = 0; 

    n[x_, y_] = ∫\_0\%x f[s, y] \[DifferentialD]s\)

This gives you f[x, y] = D[n[x, y], x] -  D[m[x, y], y]:

{D[n[x, y], x] - D[m[x, y], y], f[x, y]}

Now you know that
       (∫_^)(∫_( (Subscript[R, xy]))) 
          f[x, y] dx dy
       (∫_^)(∫_( (Subscript[R, xy]))) 
          (D[n[x, y], x] -  D[m[x, y], y]) dx 
            dy
        = 
        (∫_a^b) m[x[t], y[t]] x'[t] +  
          n[x[t], y[t]] y'[t] dt:

{x[t_], y[t_]} = {1, -2} + (Sqrt[5]) {Cos[t], Sin[t]}; GGcalculation=(∫_a^b)(m[x[t], y[t]] x'[t]+n[x[t], y[t]] y'[t])\[DifferentialD]t

As expected, this is the same nasty answer you got in part i).

T.1.a.iii)

How do you decide whether to go with transformations as in part i), or with the Gauss-Green approach as in part ii)?

From the scientific point of view, it's a toss-up.  Both methods work, so the decision is really a matter of personal choice. 
Most folks prefer the approach using transformations in part i).  But when you're putting on the ritz, you might want to go with the Gauss-Green approach.
Because you're already familiar with the Gauss-Green formula, this lesson will concentrate on the approach using transformations.  You should do that too, because the approach using transformations has a clear extension to three dimensions.  Versions of Gauss-Green in three dimensions are complicated.

T.1.b.i)

Here's a plot of the part of the surface
       z = x y +  5
above the ellipse 
     ((((x/2)))^2)+ ((((y/3)))^2)=1 
in the xy-plane:

\(Clear[x, y, r, t]; 
    
{x[r_, t_], y[r_, t_]} = {2\ r\ Cos[t], 3\ r\ Sin[t]}; 

    \({{rlow, rhigh}, {tlow, thigh}} = {{0, 1}, {0, 2\ π}}; \)\)top = 
      ParametricPlot3D[{x[r, t], y[r, t], x[r, t] y[r, t] + 5}, {r, rlow, 
          rhigh}, {t, tlow, thigh}, DisplayFunction -> Identity]; \(base = 
      ParametricPlot3D[{x[r, t], y[r, t], 0}, {r, rlow, rhigh}, {t, tlow, 
          thigh}, PlotPoints -> {2, Automatic}, 
        DisplayFunction -> Identity]; 


    Show[top, base, PlotRange -> All, 
      AxesLabel -> {"x", "y", "z"}, 
      ViewPoint -> CMView, DisplayFunction -> $DisplayFunction]; 
    \)

Measure the volume under the plotted surface and above its base in the xy-plane.

It's duck soup.
The volume is measured by 
       (∫_^)(∫_( 
            (Subscript[R, xy]))^)(x y +  5) dx 
            dy, 
where (Subscript[R, xy]) is everything inside and on the ellipse 
     ((((x/2)))^2)+ ((((y/3)))^2)=1 
plotted in the xy-plane.
Go with rt-paper coming from:

{x[r, t], y[r, t]}

As you run r from rlow to rhigh, and as you run t from tlow to thigh, {x[r, t], y[r, t]} sweeps out everything inside and on (Subscript[R, xy]). 
So (Subscript[R, rt]) is the rectangle rlow ≤ r ≤ rhigh and tlow ≤ t ≤ thigh, and
       volume = 
        (∫_^)(∫_( 
              (Subscript[R, xy]))^)(x y +  5) dx 
              dy  
        = 
        (∫_^)(∫_( 
              (Subscript[R,  r t]))^)(x[r, t] y[r, t] +  5) 
              (Subscript[A, xy])[r, t] dr dt
        = 
        (∫_tlow^thigh)(∫_rlow^rhigh) 
            (x[r, t] y[r, t] +  5) (Subscript[A, xy])[r, t] dr 
              dt:

\(Clear[gradx, grady, Axy]; 

    gradx[r_, t_] = {D[x[r, t], r], D[x[r, t], t]}; 

    grady[r_, t_] = {D[y[r, t], r], D[y[r, t], t]}; 

    Axy[r_, t_] = TrigExpand[Det[{gradx[r, t], grady[r, t]}]]\)

Measure the volume:

(∫_tlow^thigh)
        (∫_rlow^rhigh)(x[r, t] y[r, t] + 5) Axy[r, t] 
          \[DifferentialD]r \[DifferentialD]t

Nice answer. 
But if you hadn't gone to rt-paper to calculate the integral, getting this nice answer wouldn't have been so simple.

T.1.b.ii)

Could you have measured the same volume using the Gauss-Green formula?

Yes.

T.1.c.i)

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

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

    \({rlow, rhigh} = {1, 3}; \)\){tlow, thigh} = {(π/4), ((7 π)/4)}; 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], {\(-6\), 1}]]; 

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

(Subscript[R, xy]) is everything inside and on this boundary.
Calculate 
       (∫_^)(∫_( (Subscript[R, xy]))^) x 
            dx dy 
with little thought, and with Mathematica doing the work.

Look at:

{x[r, t], y[r, t]}

When you look at the plotting instructions above, then you see that when you run r from rlow to rhigh, and you run t from tlow to thigh, then {x[r, t], y[r, t]} describes (Subscript[R, xy]).  This tells you that on rt-paper (Subscript[R, rt]) is the rectangle 
       rlow ≤ r ≤ rhigh and tlow ≤ t ≤ thigh.
And your thinking is almost done.
Remembering that
       (∫_^)(∫_( (Subscript[R, xy]))^) x 
              dx dy = 
        (∫_^)(∫_( 
              (Subscript[R,  r t]))^) x[r, t] (Subscript[A, xy])[r, t] dr 
              dt, 
turn Mathematica loose:

\(Clear[gradx, grady, Axy]; 

    gradx[r_, t_] = {D[x[r, t], r], D[x[r, t], t]}; 

    grady[r_, t_] = {D[y[r, t], r], D[y[r, t], t]}; 

    Axy[r_, t_] = Factor[TrigExpand[Det[{gradx[r, t], grady[r, t]}]]]\)

You can see that (Subscript[A, xy])[r, t] is never negative because r never goes negative. Now your thinking is done.
Now calculate
       (∫_^)(∫_( (Subscript[R, xy]))^) x 
              dx dy = 
        (∫_^)(∫_( 
              (Subscript[R,  r t]))^) x[r, t] (Subscript[A, xy])[r, t] dr 
              dt
                      = 
        (∫_tlow^thigh)(∫_rlow^rhigh) 
            x[r, t] (Subscript[A, xy])[r, t] dr dt:

(∫_tlow^thigh)
        (∫_rlow^rhigh) x[r, t] Axy[r, t] \[DifferentialD]r 
          \[DifferentialD]t

And you're out of here.

T.1.c.ii)

Could you have calculated the same integral using the Gauss-Green formula?

In theory, yes.
In practice, it would have been a very tedious job, because you would have to do a lot of bureaucratic work to come up with the required counterclockwise parameterization of the boundary. This would involve separate parameterizations for each of the four boundary segments.
Ugh.

T.1.d.i) A constant-width ribbon.

Here's a curve on xy-paper:

\(Clear[x, y, r, t]; 
{x[t_], y[t_]} = {4\ Cos[t], 3\ Sin[t]}; 

    \({tlow, thigh} = {1, 5}; 


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

The outer normal to the curve at {x[t], y[t]} is:

Clear[unitnormal];
unitnormal[t_]=TrigExpand[(({y'[t],-x'[t]})/(Sqrt[((x'[t])^2)+((y'[t])^2)]))]

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

\(Clear[r, outeribbon, 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]; \)

Measure the area of this ribbon.

Call inside of the ribbon (Subscript[R, xy]). The integral 
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          dx dy
measures the area of the ribbon. 
Go to rt-paper with:

{rlow, rhigh} = {-0.5, 0.5}; \(Clear[x, y, r, t]; 
    
{x[r_, t_], y[r_, t_]} = {4\ Cos[t], 3\ Sin[t]} + r\ unitnormal[t]\)

When you look at the plotting instructions above, then you see that when you run r from rlow to rhigh, and you run t from tlow to thigh, then {x[r, t], y[r, t]} describes the whole ribbon (Subscript[R, xy]).  This tells you that on rt-paper (Subscript[R, rt]) is the rectangle 
       rlow ≤ r ≤ rhigh and tlow ≤ t ≤ thigh.
Remembering that
       area of ribbon = 
        (∫_^)(∫_( (Subscript[R, xy]))^) 
              dx dy = 
          (∫_^)(∫_( 
                (Subscript[R,  r t]))^) 
              (Subscript[A, xy])[r, t] dr dt, 
turn Mathematica loose:

\(Clear[gradx, grady, Axy]; 

    gradx[r_, t_] = {D[x[r, t], r], D[x[r, t], t]}; 

    grady[r_, t_] = {D[y[r, t], r], D[y[r, t], t]}; 

    Axy[r_, t_] = TrigExpand[Det[{gradx[r, t], grady[r, t]}]]\)

Analyzing this,it's hard to see whether this mess can ever go negative. Take the easy way out and integrate its absolute value using NIntegrate.

NIntegrate[
      Evaluate[Abs[Axy[r, t]]], {t, tlow, thigh}, {r, rlow, rhigh}, 
      AccuracyGoal -> 2]

About 14.4 square units.

T.1.d.ii) A variable width ribbon

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

\(Clear[x, y, r, t]; 
{x[t_], y[t_]} = {4\ Cos[t], 3\ Sin[t]}; 

    \({tlow, thigh} = {1, 5}; \)\)curveplot = 
      ParametricPlot[{x[t], y[t]}, {t, tlow, thigh}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        AspectRatio -> Automatic, AxesLabel -> {"x", "y"}];

But this time make the width of the ribbon 
       0.4  t at {x[t], y[t]}.
Here's what you get:

Clear[unitnormal];
unitnormal[t_]=TrigExpand[(({y'[t],-x'[t]})/(Sqrt[((x'[t])^2)+((y'[t])^2)]))];
Clear[r, outeribbon, innerribbon, oneend, otherend, halfwidth];
halfwidth[t_] = 0.2 t;
outerribbon[t_] = {x[t], y[t]} + halfwidth[t] unitnormal[t]; innerribbon[t_] = {x[t], y[t]} - halfwidth[t] unitnormal[t]; tlowend[r_] = {x[tlow], y[tlow]} + r unitnormal[tlow]; thighend[r_] = {x[thigh], y[thigh]} + r unitnormal[thigh]; twosides = 
      ParametricPlot[{outerribbon[t], innerribbon[t]}, {t, tlow, thigh}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; onemoreside = 
      ParametricPlot[tlowend[r], {r, -halfwidth[tlow], halfwidth[tlow]}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; \(andonemoreside = 
      ParametricPlot[
        thighend[r], {r, \(-halfwidth[thigh]\), halfwidth[thigh]}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; 


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

Measure the area of this ribbon.

Call the plot of the ribbon above (Subscript[R, xy]). The integral 
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          dx dy
measures the area of the ribbon. 
Go to rt-paper with:

\(Clear[x, y, r, t]; 
    
{x[r_, t_], y[r_, t_]} = {4\ Cos[t], 3\ Sin[t]} + 
        r\ halfwidth[t]\ unitnormal[t]\)

When you look at the plotting instructions above, then you see that when you run r from -1 to 1, and you run t from tlow to thigh, then {x[r, t], y[r, t]} describes (Subscript[R, xy]).  This tells you that on rt-paper, (Subscript[R, rt]) is the rectangle 
       -1 ≤ r ≤ 1 and tlow ≤ t ≤ thigh,
so
       (∫_^)(∫_( (Subscript[R, xy]))^) 
            dx dy = 
        (∫_^)(∫_( 
              (Subscript[R,  r t]))^) (Subscript[A, xy])[r, t] dr 
              dt
                      = 
        (∫_tlow^thigh)(∫_(-1)^1) 
            (Subscript[A, xy])[r, t] dr dt.

\(Clear[gradx, grady, Axy]; 

    gradx[r_, t_] = {D[x[r, t], r], D[x[r, t], t]}; 

    grady[r_, t_] = {D[y[r, t], r], D[y[r, t], t]}; 

    Axy[r_, t_] = Det[{gradx[r, t], grady[r, t]}]\)

Analyzing this mess see whether it can ever go negative is a scary prospect. Take the easy way out and integrate its absolute value,using NIntegrate to calculate
       (∫_^)(∫_( (Subscript[R, xy]))^) 
            dx dy = 
        (∫_^)(∫_( 
              (Subscript[R,  r t]))^) (Subscript[A, xy])[r, t] dr 
              dt
                      = 
        (∫_tlow^thigh)(∫_(-1)^1) 
            (Subscript[A, xy])[r, t] dr dt.

NIntegrate[Evaluate[Abs[Axy[r, t]]], {t, tlow, thigh}, {r, -1, 1}, 
      AccuracyGoal -> 2]

About 17.3 square units.
Pity those poor souls in the traditional calculus course.  Most of them couldn't even dream of making this measurement. 
And it's so easy because of rt -paper.

T.2)  Transforming  (∫_^)(∫_( (Subscript[R, xy]))^) 
          f[x, y] dx dy when the boundary of (Subscript[R, xy]) is not given with parametric formulas

When the boundary of a region (Subscript[R, xy]) is plotted with nonparametric formulas, things are not always as simple as they were in T.1).
But even in this case, there are situations that allow you to inspect the boundary curves to help come up with favorable uv-paper.
Here's one such:

T.2.a.i)

(Subscript[R, xy]) is the region plotted below which is bounded by the curves
       y = 0.5  (x^2) +  1, y = 0.5  (x^2) -  1, 
       y = 3  x +  2, and y = 3  x -  2:

\(Clear[x]; 

    Rxyplot = 
      Plot[{0.5\ x\^2 + 1, 0.5\ x\^2 - 1, 3\ x + 2, 3\ x - 2}, {x, \(-1\), 
          1.4}, PlotStyle -> {{Red, Thickness[0.01]}}, 
        AxesLabel -> {"x", "y"}, 
        PlotRange -> {\(-1\), 1.8}, 
        Epilog -> 
          Text[StyleForm["\!\(R\_xy\)", FontSize -> 16], {0.5, 
              0.7}]]; \)

The region (Subscript[R, xy]) under scrutiny is the four sided figure you see above and everything inside it.
Transform (Subscript[R, xy]) into a rectangle on uv-paper to help come up with a quick, easy calculation of
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          ((x^2) +  (y^2)) dx dy.

Enter f[x, y] = (x^2) +  (y^2):

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

Look at the formulas for the functions whose plots make up the boundary of (Subscript[R, xy]).  They are:
       y = 0.5  (x^2) +  1, y = 0.5  (x^2) -  1, 
       y = 3  x +  2, and y = 3  x -  2:
Put 
       u[x, y] = y -  0.5  (x^2) and v[x, y] = y -  3  x:

\(Clear[u, v, x, y]; 
    
{u[x_, y_], v[x_, y_]} = {y - 0.5\ x\^2, y - 3\ x}\)

The original boundary curves are level curves of u[x, y] and v[x, y].  In fact,
-> y = 0.5  (x^2) +  1 is the level curve u[x, y] = 1,
-> y = 0.5  (x^2) -  1 is the level curve u[x, y] = -1,
-> y = 3  x -  2 is the level curve v[x, y] = 2, and
-> y = 3  x +  2 is the level curve v[x, y] = -2.
This is very good news because this tells you that (Subscript[R, uv]) is the rectangle 
       -1 ≤ u ≤ 1 and -2 ≤ v ≤ 2.  
The upshot:
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          ((x^2) +  (y^2)) dx dy
       (∫_^)(∫_( (Subscript[R, uv]))^) 
          (x[u, v]^2 +  y[u, v]^2) (Subscript[A, xy])[u, v] du 
            dv
       (∫_(-2)^2)(∫_(-1)^1) 
          (x[u, v]^2 +  y[u, v]^2) (Subscript[A, xy])[u, v] du 
            dv.
First, you gotta to come up with formulas for x[u, v] and y[u, v]:

solutions = Solve[{u == u[x, y], v == v[x, y]}, {x, y}]

This gives two choices:

\(Clear[x1, y1, x2, y2]; 
    
{x1[u_, v_], y1[u_, v_]} = {
        0.5\ \((6 - \@\(36 - 4\ \((2\ u - 2\ v)\)\))\), 
        9 + v - 1.5\ \@\(36 - 8\ u + 8\ v\)}\)

{x2[u_, v_], y2[u_, v_]} = {
        0.5 (6 + (Sqrt[36 - 4 (2 u - 2 v)])), 
        9 + v + 1.5 (Sqrt[36 - 8 u + 8 v])}

You know that (Subscript[R, uv]) is the rectangle 
       -1 ≤ u ≤ 1 and -2 ≤ v ≤ 2.
The uv-point {0.5, 1} is in this rectangle. 
See which solution makes the uv-point {0.5, 1} plot out inside (Subscript[R, xy]) on xy-paper by seeing where {x1[0.5, 1], y1[0.5, 1]} and {x2[0.5, 1], y2[0.5, 1]} land on xy-paper:

Show[Rxyplot, 
      Graphics[{Blue, PointSize[0.1], Point[{x1[0.5, 1], y1[0.5, 1]}]}], 
      PlotRange -> All];

Good; this lands inside (Subscript[R, xy]). Just for the heck of it, try out the other pair of solutions:

Show[Rxyplot, 
      Graphics[{Blue, PointSize[0.1], Point[{x2[0.5, 1], y2[0.5, 1]}]}], 
      PlotRange -> All];

Way outside (Subscript[R, xy]).  This means you definitely want to go with:

{x[u_, v_], y[u_, v_]} = {x1[u, v], y1[u, v]}

Calculate the area conversion factor (Subscript[A, xy])[u, v]:

\(Clear[Axy, gradx, grady]; 

    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]}]\)

Calculate
       (∫_^)(∫_( (Subscript[R, xy]))^) 
          ((x^2) +  (y^2)) dx dy
       (∫_^)(∫_( (Subscript[R, uv]))^) 
          (x[u, v]^2 +  y[u, v]^2) (Subscript[A, xy])[u, v] du 
            dv
       (∫_(-2)^2)(∫_(-1)^1) 
          (x[u, v]^2 +  y[u, v]^2) (Subscript[A, xy])[u, v] du 
            dv.

NIntegrate[
      (x[u, v]^2 + y[u, v]^2) Axy[u, v], {v, -2, 2}, {u, -1, 
        1}, AccuracyGoal -> 2]

Finished.

T.2.a.ii) The Achilles heel

When you have a set-up like the problem in part i), what can go wrong?

In theory, nothing much can go wrong. 
In practice, this technique can grind to a quick halt. The hitch is that you specify u[x, y] and v[x, y], and then you have to solve the simultaneous equations
       u = u[x, y] and v = v[x, y] 
for x and y to get the formulas for 
       x[u, v] and y[u, v]. 
Solving u = u[x, y] and v = v[x, y] for x and y is possible only in simple special situations.
Samples:

\(Clear[x, y, u, v]; 
{u[x_, y_], v[x_, y_]} = {y - Sin[x], y - x}; 

    Solve[{u == u[x, y], v == v[x, y]}, {x, y}]\)

No dice.  That transcendental function Sin[x] screws up the algebra.

\(Clear[x, y, u, v]; 
{u[x_, y_], v[x_, y_]} = {y - E\^\(-x\), y - x}; 

    Solve[{u == u[x, y], v == v[x, y]}, {x, y}]\)

No dice.  That transcendental function (e^(-x)) screws up the algebra.

\(Clear[x, y, u, v]; 
{u[x_, y_], v[x_, y_]} = {y - 2\ x, y + x}; 

    ExpandAll[Solve[{u == u[x, y], v == v[x, y]}, {x, y}]]\)

No sweat.  This gives you 
       x[u, v]=((-u  +  v)/3) and y[u, v]=((u  +  2 v)/3).

T.2.a.iii)

What is a transcendental function?

Transcendental functions are those that transcend the ordinary processes of algebra.
The basic calculus functions Sin[x], Cos[x], and (e^x) are all transcendental.
That's why you are guaranteed to fail when you try simple things like:

\(Clear[x]; 
Solve[x == Sin[x], x]\)

\(Clear[x]; 
Solve[x == Cos[x], x]\)

\(Clear[x]; 
Solve[x == E\^x, x]\)

Line functions like f[x] = 3  x +  2 are not transcendental:

\(Clear[x]; 
Solve[x == 3\ x + 2, x]\)

Determining whether a given function is transcendental is part of the stuff of advanced mathematics.

T.3)  The area conversion factor (Subscript[A, xy])[u, v]

T.3.a) Expansion and compression

Polar coordinates {u, v} are related to xy-coordinates via
       x = u Cos[v],
       y = u Sin[v].

Here is a plot of a random bunch of points {u, v} on uv-paper with 
       0 ≤ u ≤ 10 and 0 ≤ v ≤ 2  π:

\(Clear[k]; 

    uvpoints = 
      Table[{Random[Real, {0, 10}], Random[Real, {0, N[2\ π]}]}, {k, 1, 
          150}]; 
\(uvpaperplot = 
      ListPlot[uvpoints, PlotStyle -> {Blue, PointSize[0.02]}, 
        AspectRatio -> Automatic, AxesLabel -> {"u", "v"}]; 
    \)\)

Should be fairly well scattered. If not, then rerun.
Now look at the xy-paper plot of these uv-paper points:

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

    xypoints = uvtoxy/@uvpoints; 


    xypaperplot = 
      ListPlot[xypoints, PlotStyle -> {Blue, PointSize[0.03]}, 
        AspectRatio -> Automatic, AxesLabel -> {"x", "y"}]; 
    \)

On the xy-paper, why are the points so bunched up near the origin and sparsely scattered far from the origin?

Look at the area conversion factor:

\(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]}]]\)

At a point with uv-paper coordinates {u, v}, 
       \(\(xy - paper\ area\ measurements\)
\ \ \ \ \ \ \ \ \ \ \ \ 
        \( = u\ times\ uv - paper\ area\ measurements\)\).

Look at:

{x[u, v], y[u, v]}

The uv-paper coordinate, u, of the xy-paper point, {x[u, v],  y[u, v]}, measures the distance from {0, 0} to {x[u, v],  y[u, v]} on xy-paper.
-> When (Subscript[A, xy])[u, v] = u is big, then xy-paper area measurements are a lot bigger than the corresponding uv-area measurements, so uv-paper points with u big are flung apart when they are plotted on  xy-paper.
-> When (Subscript[A, xy])[u, v] = u  is wee little, uv-paper points {u, v} with u small are compressed together when they are plotted on xy-paper.
Since u is small for points near the origin, you see a pile-up near the origin on the xy-paper plot.
Take another look:

Show[uvpaperplot, PlotLabel -> "uv-paper plot"];

Show[xypaperplot, PlotLabel -> "xy-paper plot"];

Uh-huh.

T.3.b.i) The sizer

Take x[u, v] = (u^2) -  (v^2) and y[u, v] = 2  u v and calculate the area conversion factor (Subscript[A, xy])[u, v]:

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

    \(y[u_, v_] = 2\ u\ v; \)\)\(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]}]\)

Here's a table of random points inside the uv-paper square with -2 ≤ u ≤ 2 and -2 ≤ v ≤ 2 plotted as dots on uv-paper:

\(Clear[k]; 
pointcount = 150; 

    \(uvpoints = 
      Table[{Random[Real, {\(-2\), 2}], Random[Real, {\(-2\), 2}]}, {k, 1, 
          pointcount}]; 


    uvpointplot = 
      Show[Table[
          Graphics[{Blue, PointSize[0.02], 
              Point[uvpoints[[k]]]}], {
            k, 1, pointcount}], AspectRatio -> Automatic, 
        Axes -> Automatic, AxesLabel -> {"u", "v"}]; \)\)

Should be fairly well scattered. If not, then rerun.
Here is the same plot with the size of each plotted point {u, v} adjusted by a factor proportional to (Sqrt[(Subscript[A, xy])])[u, v].

\(Clear[sizer]; 
sizer[u_, v_] = 0.015\ \@Axy[u, v]; 

    \(sizeduvpointplot = 
      Show[Graphics[
          Table[{Blue, 
              PointSize[
                sizer@@uvpoints[[k]]], 
              Point[uvpoints[[k]]]}, {k, 
              1, pointcount}]], AspectRatio -> Automatic, 
        Axes -> Automatic, AxesLabel -> {"u", "v"}]; \)\)

A humdinger.
See the plots side-by-side:

Show[GraphicsArray[{uvpointplot, sizeduvpointplot}]];

What information is conveyed by these plots?

The area conversion factor in going from {u, v} to {x[u, v], y[u, v]} is 
       (Subscript[A, xy])[u, v]. 
The points are sized in proportion to (Sqrt[(Subscript[A, xy])])[u, v].
Consequently, this second plot shows what the relative sizes of the plotted uv-points will be after they have been plotted on xy-paper.
Evidently as {u, v} gets farther and farther from {0, 0},
       (Subscript[A, xy])[u, v]
gets bigger and bigger.  This fact is suggested by the plot, and is confirmed by the formula for (Subscript[A, xy])[u, v]:

Axy[u, v]

If you wonder why the points are sized proportionally to 
       (Sqrt[(Subscript[A, xy])])[u, v]
instead of proportionally to 
       (Subscript[A, xy])[u, v], 
then click the little box.

In Mathematica, the PointSize instruction governs the radius of the plotted point.  To make the area of the plotted point proportional to (Subscript[A, xy])[u, v], you have to make the radius of the plotted point proportional to 
       (Sqrt[(Subscript[A, xy])])[u, v],
because the area of a circle of radius r is proportional to (r^2).

T.3.b.ii)

Here is a plot of the same uv points as above in part i) but plotted on xy-paper with 
       x[u, v] = (u^2) -  (v^2) and y[u, v] = 2  u v:

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

    \(xypoints = uvtoxy/@uvpoints; 


    xypointplot = 
      Show[Graphics[{Blue, PointSize[0.02], 
            Table[Point[xypoints[[k]]], {
                k, 1, pointcount}]}], AspectRatio -> Automatic, 
        Axes -> Automatic, AxesLabel -> {"x", "y"}]; \)\)

And look at:

sizedxypointplot = 
      Show[Table[
          Graphics[{Blue, 
              PointSize[
                sizer@@xypoints[[k]]], 
              Point[xypoints[[k]]]}], {
            k, 1, pointcount}], AspectRatio -> Automatic, 
        Axes -> Automatic, AxesLabel -> {"x", "y"}];

What information do these plots convey?

Take a look at the original uv-points and the xy-points side-by-side:

Show[GraphicsArray[{uvpointplot, xypointplot}]];

On the left you see the original points on uv-paper; on the right you see the same points plotted on xy-paper. 
Now take a look at the original uv-points and the sized xy-points side-by-side:

Show[GraphicsArray[{uvpointplot, sizedxypointplot}]];

This is the final product.
On the left you see the original points on uv-paper; on the right you see the same points plotted on xy-paper sized according to their relative sizes on uv-paper.

T.4)  Measurements of volume, mass, and density

T.4.a)

You make an object by distributing a substance over a certain region R on xy-paper. What does it mean to say that the density of the resulting object measures out at p[x, y] (grams/(unit^2)) at location {x, y}?

It means that the mass of the object is given by 
       (∫_^)(∫_( R)^) 
          p[x, y] dx dy
where R is the same region that you distributed the substance over to begin with.  
It also means that if (Subscript[R, 1]) is a region inside R, then the total mass of the substance that was spread over (Subscript[R, 1]) is  
       (∫_^)(∫_( (Subscript[R, 1]))^) 
          p[x, y] dx dy.
As a result p[(Subscript[x, 0]), (Subscript[y, 0])] is the conversion factor that converts area at {(Subscript[x, 0]), (Subscript[y, 0])} on  xy-paper to mass of the object at {(Subscript[x, 0]), (Subscript[y, 0])}.

T.4.b.i)

An object is made by forming a uniform substance that weighs 2(grams/(unit^3)) in the shape of the paraboloid 
       f[x, y] = 9 -  (x^2) -  (y^2) 
over the region inside and on the circle (x^2) +  (y^2) = 9 on  xy-paper.
Here's a look at it:

\(Clear[x, y]; 

    Plot3D[9 - x\^2 - y\^2, {x, \(-3\), 3}, {y, \(-3\), 3}, 
      PlotRange -> {0, 9}, ViewPoint -> CMView, 
      AxesLabel -> {"x", "y", "z"}]; \)

Measure the total mass of this object.
Measure the total volume of this object.

The density of the object, p[x, y], at {x, y} is 2  f[x, y].
The total mass is given by
       (∫_^)(∫_( R)^) 
          2  f[x, y] dx dy
where R is the region inside and on the circle (x^2) +  (y^2) = 9 on xy-paper.
No one likes to integrate over circular regions, so calculate this by moving to uv-paper with 
       x[u, v] = u Cos[v], and  
       y[u, v] = u Sin[v] (polar coordinates).
On uv-paper, R plots out as the rectangle with corners at 
       {0, 0}, {3, 0}, {3, 2  π} and {0, 2  π}
       (0 ≤ u ≤ 3 and 0 ≤ v ≤ 2  π). 
The total weight is
       (∫_^)(∫_( R)^) 
          2  f[x, y] dx dy
        = 
        (∫_0^(2  π))(∫_0^3) 
            2  f[x[u, v], y[u, v]] (Subscript[A, xy])[u, v] du 
              dv:

\(Clear[f, u, v, x, y, gradx, grady, Axy]; 
f[x_, y_] = 9 - x\^2 - y\^2; 
    
\(x[u_, v_] = u\ Cos[v]; \)\)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]}]]\)

The total mass is:

mass = 
      (∫_0
          ^(2 π))(∫_0^3)
             2 f[x[u, v], y[u, v]] Axy[u, v] \[DifferentialD]u 
            \[DifferentialD]v

The total volume of this object is:

volume = 
      (∫_0
          ^(2 π))(∫_0^3)
             f[x[u, v], y[u, v]] Axy[u, v] \[DifferentialD]u 
            \[DifferentialD]v

Routine stuff.

T.4.b.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).  The deformation affected measurements on the base, but had no effect on the height measurements.
What does the resulting deformed object look like when it is plotted on uv-paper for the functions 
       x = x[u, v] and y = y[u, v] 
you 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 ≤ 3 and 0 ≤ v ≤ 2  π?

Here is how it looks on uv-paper:

Plot3D[f[x[u, v], y[u, v]], {v, 0, 2 π}, {u, 0, 3}, 
      PlotRange -> {0, 9}, ViewPoint -> CMView, 
      AxesLabel -> {"x", "y", "z"}];

Quite a change of shape.
You can figure out what the deformed object's density is at a point {u, v} within the uv-paper rectangle 0 ≤ u ≤ 3 and 0 ≤ v ≤ 2  π.  It is just what you integrate to calculate its mass:

TrigExpand[2 f[x[u, v], y[u, v]] Axy[u, v]]

You've got to multiply by the area conversion factor because 
       2  f[x[u, v], y[u, v]] (Subscript[A, xy])[u, v] 
is what you integrate to calculate mass.