Re: Area of sector in ellipse

From: Oscar Lanzi III (ol3_at_webtv.net)
Date: 01/09/05 

Date: Sat, 8 Jan 2005 19:39:35 -0600

You don't really need calculus to get the area.

Let P be a point on the ellipse coresponding to your angle theta,
henceforth written @, in your diagram. Then the length of the "radius"
vector is b^2/a(1+e cos @) where the eccentricity e is sqrt(1-b^2/a^2).
Multiply this by sin @ to get y, the altitude of P above the major axis.
Very important: Keep track of the sign of y! Define phi = arcsin(y/b)
for @ in the first quadrant and make appropriate symmetry adjustments
for the other quadrants. Phi is always in the same quadrant as @. Then
the area of a CENTRAL sector is (1/2)*a*b*phi with phi in radians.
Subtract the signed triangular area (1/2)*sqrt(a^2-b^2)*y where the
triangle has a base running from the center to the focus and signed
altitude y, and the result is your focal sector area.

--OL