Finding the area bounded by a logspiral curve and two straight lines

Question

I'm facing some trouble with finding the area of a region which is described by x-y coordinates (or line equations) and a curved line represented by logspiral. I tried my best in coming up with the following illustration below to show what I mean:

• B = Angle between Horizontal (at x2,y2) and the line described by r
• r = Equation of the curve region, described by: r = a*exp(B)
• a = Value of r when B = 0
• (x,y) = Coordinates of the vertices

I understand that I could find the intersection of the 2 half-spaces using 2 equations describing the 2 straight line portions (say using RegionIntersection), but I'm not too sure how I could incorporate the logspiral equation to get the bounding area using RegionMeasure.

I'm inclined to convert the logspiral equation to parametric form for this purpose, but I have no idea how to move on from there.

I would really appreciate any help or advice regarding this problem, I have been losing some sleep thinking about this question.

Answer 1

I don't really understand what is known and what is not, but assuming that you know x1, y1, x2, y2 and a you can do (with v8, I assume v10 makes it much easier):

{x1, y1} = {0, 0}; {x2, y2} = {2, 3};
angle[v1_, v2_] := N@ArcCos[(First@v1*First@v2 + Last@v1*Last@v2)/(Norm@v1*Norm@v2)]
eqn[k_, l_, a_, x_, α_] := k + l*a*Exp[x*α]

kk[l_, a_] := k /. Solve[eqn[k, l, a, x1, angle[{x2 - x1, y2 - y1}, {a, 0}]] == y1, {k}]
ll[a_] := l /. Solve[eqn[kk[l, a], l, a, x2 + a, angle[{x2 - x1, y2 - y1}, {a, 0}]] == y2, {l}];
pts[a_] := Join[First@
   Cases[Plot[kk[ll[a], a] + ll[a]*a*Exp[x*angle[{x2 - x1, y2 - y1}, {a, 0}]], {x, x1, x2 + a}], Line[x_] :> x, ∞], 
   {{x1, y1}, {x2, y2}, {x2 + a, y2}}]

Here I'm using this and this to make sure the Polygon will have the right shape:

SignedArea[p1_, p2_, p3_] := 
  0.5 (#1[[2]] #2[[1]] - #1[[1]] #2[[2]]) &[p2 - p1, p3 - p1];
IntersectionQ[p1_, p2_, p3_, p4_] := 
  SignedArea[p1, p2, p3] SignedArea[p1, p2, p4] < 0 && 
   SignedArea[p3, p4, p1] SignedArea[p3, p4, p2] < 0;
Deintersect[p_] := Append[p, p[[1]]] //. {s1___, p1_, p2_, s2___, p3_, p4_, s3___} /; 
      IntersectionQ[p1, p2, p3, p4] :> ({s1, p1, p3, 
       Sequence @@ Reverse@{s2}, p2, p4, s3}) // Most;
fpts[a_] := Deintersect[pts[a]];
Graphics`Mesh`MeshInit[];
Abs@PolygonArea[fpts[2]]

6.18614

Block[{a = 2}, 
 Graphics[Polygon@fpts[a], Axes -> True, 
  Epilog -> {Red, Point[{{x1, y1}, {x2, y2}, {x2 + a, y2}}], 
    Line@pts@2, Green, Dashed, Line@fpts@2}]]

Answer 2

In this I have made $(x_2,y_2)$ the origin:

f[a_, t_] := a Exp[t]
Manipulate[
 Column[{
   ParametricPlot[{s f[a, t] Cos[t], s f[a, t] Sin[t]}, {t, angle, 
     0}, {s, 0, 1}, PlotRange -> {{-1, 2}, {-1, 0.1}}, 
    ImageSize -> 400, BoundaryStyle -> {Red, Thick}],
   Row[{"Area: ", Integrate[0.5 f[a, u]^2, {u, angle, 0}]}]
   }, Alignment -> Center]
 , {angle, -Pi/4, -Pi}, {a, 1, 2}]

UPDATE

Manipulate[Column[{ParametricPlot[s {f[a, t] Cos[t],
      -f[a, t] Sin[t]}, {t, 0, angle}, {s, 0, 1}, 
    PlotRange -> {{-10, 4}, {-10, 1}}, ImageSize -> 600, 
    BoundaryStyle -> {Red, Thick}, 
    Epilog -> {Circle[{0, 0}, 1/2, {0, -angle}], 
      Text[Style[
        "(\!\(\*SubscriptBox[\(x\), \
\(2\)]\),\!\(\*SubscriptBox[\(y\), \(2\)]\))", 16], {0, 
        0}, {0, -1.5}], 
      Text[Style[
        "(\!\(\*SubscriptBox[\(x\), \(2\)]\)+a,\!\(\*SubscriptBox[\(y\
\), \(2\)]\))", 16], {1.4 a, 0}, {0, -1.5}], 
      Text[Style[
        "(\!\(\*SubscriptBox[\(x\), \
\(1\)]\),\!\(\*SubscriptBox[\(y\), \(1\)]\))", 
        16], {a Exp[angle] Cos[angle], -a Exp[angle] Sin[angle]}, {0, 
        1.5}], Text[Style["B", 16], 
       0.8 {Cos[angle/2], -Sin[angle/2]}, {0, 0}]}], 
   Row[{"Area: ", Integrate[0.5 f[a, u]^2, {u, 0, angle}]}]}, 
  Alignment -> Center], {angle, Pi/4, Pi}, {a, 1, 2}]