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 ->
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

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]

```

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