RE: Area of ellipse between major axis and ray through focus, given angle

*To*: mathgroup at smc.vnet.net*Subject*: [mg71676] RE: [mg71601] Area of ellipse between major axis and ray through focus, given angle*From*: "David Park" <djmp at earthlink.net>*Date*: Sun, 26 Nov 2006 03:49:04 -0500 (EST)

The following is the parametrization of an ellipse in terms of the parameter Theta. ellipse[a_, b_][\[Theta]_] := {a*Cos[\[Theta]], b*Sin[\[Theta]]} We must remember that the parameter Theta is not the same as the angle Phi from the x axis. They are related by \[Theta] == ArcTan[(a/b)*Tan[\[Phi]]]; \[Phi] == ArcTan[(b/a)*Tan[\[Theta]]]; The following makes a diagram if you have DrawGraphics. Otherwise skip the diagram. Your diagram was fairly good. Needs["DrawGraphics`DrawingMaster`"] Module[{\[Phi]0 = Pi/4, a = 2, b = 1, segment, \[Phi]}, segment = FineGrainLines[0.05, 4][ FineGrainPolygons[0.05, 4][ FilledDraw[Norm[ellipse[2, 1][ ArcTan[2*Tan[\[Phi]]]]], {\[Phi], 0, \[Phi]0}, Fills -> DodgerBlue]]]; Draw2D[{ParametricDraw[ellipse[a, b][ \[Theta]], {\[Theta], 0, 2*Pi}], segment /. DrawingTransform[ #2*Cos[#1] & , #2*Sin[#1] & ], Line[{{0, 0}, ellipse[a, b][ ArcTan[2*Tan[\[Phi]0]]]}], AngleArc[{0, 0}, {0, \[Phi]0}, 0.65, "\[Phi]", AAArrowOptions -> {HeadCenter -> 1/2, HeadLength -> 0.2}, AATextOptions -> {Background -> DodgerBlue}, AAStyleOptions -> {FontFamily -> "Courier", FontSize -> 12, FontWeight -> "Bold"}], Text["a", {a/2, 0}, {0, 1}], Text["b", {0, b/2}, {1, 0}]}, AspectRatio -> Automatic, Frame -> False, Axes -> True, AxesStyle -> Gray, AxesOrigin -> {0, 0}, Ticks -> None, PlotLabel -> "Ellipse Segment Area", TextStyle -> {FontFamily -> "Courier", FontSize -> 12, FontWeight -> "Bold"}, Background -> Linen, ImageSize -> 450]]; We will first calculate the area of the segment in terms of a and b, the major and minor semiaxes. The following calculates the radius of the ellipse as a function of the angle from the x axis, Phi. Norm[ellipse[a, b][ArcTan[(a/b)*Tan[\[Phi]]]]]; r[a, b][\[Phi]_] = Simplify[%, 0 < \[Phi] < Pi/2 && 0 < b < a] (a*b*Sec[\[Phi]])/Sqrt[b^2 + a^2*Tan[\[Phi]]^2] We can then use the standard integration formula for area in polar coordinates to obtain the segment area. areasolution[a_, b_][\[Phi]_] = Assuming[0 < \[Phi] < Pi/2 && 0 < b < a, Integrate[(1/2)*r[a, b][t]^2, {t, 0, \[Phi]}]] (1/2)*a*b*ArcTan[(a*Tan[\[Phi]])/b] We can check this out by taking 4 times the area of the first quadrant. We have to use Limit to obtain an expression there. We obtain the standard handbook solution for the area of an ellipse. 4 Limit[areasolution[a, b][\[Phi]], \[Phi] -> \[Pi]/2, Direction -> 1]; Simplify[%, 0 < b < a] a*b*Pi We can also calculate in terms of a and e, the eccentricity of the ellipse. We substitute b in terms of e and a and simplify to obtain a radius expression r2. r[a, b][\[Phi]] /. b -> a*Sqrt[1 - e^2]; r2[a_, e_][\[Phi]_] = Simplify[%, a > 0 && 0 < e < 1] (a*Sec[\[Phi]])/Sqrt[(-1 + e^2 - Tan[\[Phi]]^2)/ (-1 + e^2)] Evaluating the integral and simplifying takes a little more work. step1 = Assuming[0 < \[Phi] < Pi/2 && a > 0 && 0 < e < 1, Integrate[ (1/2)*r2[a, e][t]^2, {t, 0, \[Phi]}]]; step2 = ComplexExpand[Re[step1], TargetFunctions -> {Re, Im}]; areasolution2[a_, e_][\[Phi]_] = Simplify[step2, 0 < \[Phi] < Pi/2 && a > 0 && 0 < e < 1] (1/2)*a^2*Sqrt[1 - e^2]* ArcTan[Tan[\[Phi]]/Sqrt[1 - e^2]] Calculating the area of the entire ellipse, we again obtain the handbook solution. 4*Limit[areasolution2[a, e][\[Phi]], \[Phi] -> Pi/2, Direction -> 1]; Simplify[%, 0 < \[Phi] < Pi/2 && a > 0 && 0 < e < 1] a^2*Sqrt[1 - e^2]*Pi David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Kelly Jones [mailto:kelly.terry.jones at gmail.com] Given: 1) an ellipse with eccentricity "ec", one focus on the origin, and the major axis along the x-axis 2) a ray through the origin at angle theta to the x-axis Question: What Mathematica function gives the relation/inverse relation between the angle theta and the area of the ellipse between the x-axis and the ray? I'm guessing one of EllipticE/EllipticF/EllipticK gives the area as a function of theta, but I can't figure out which one. I also can't figure out what function gives theta as a function of the area? Finally, for a fixed value of ec (eccentricity), what are the power series expansions for the functions taking theta to area and vica versa? Ugly drawing: (* numbers chosen "randomly" for drawing purposes only *) lower = ParametricPlot[{x,-(Sqrt[3]*Sqrt[3 - 8*x - 16*x^2])/8}, {x,-3/4,1/4}] upper = ParametricPlot[{x, (Sqrt[3]*Sqrt[3 - 8*x - 16*x^2])/8}, {x,-3/4,1/4}] line = Line[{{0,0},{.14,.27097}}] arc = Circle[{0,0},.05,{0,1.09391}] Show[lower,upper,Graphics[line],Graphics[arc], AspectRatio->Automatic,Ticks->None]