Re: Area of an ellipse
- From: no comment <adler.math@xxxxxxxxx>
- Date: Sat, 26 Jan 2008 21:38:29 -0800 (PST)
On Jan 26, 9:05 pm, no comment <adler.m...@xxxxxxxxx> wrote:
On Jan 26, 8:01 pm, ms <silvert...@xxxxxxxxx> wrote:
Is there an equation for a certain shaded area of an ellipse, in which
the shaded area is not necessarily a part of the major or minor axes?
I am trying to calculate the angle at which the area of a quadrant of
the ellipse is equal for both halves of the quadrant. I know that the
area is pi*a*b (in which my a = 7.5 and b = 6.25; entire long axis is
15 and shorter axis is 12.5 in length). So, the area for the entire
ellipse would be pi*7.5*6.25 = 147.26. So the area of a quadrant
would be 147.26/4 = 36.82. So I would like to know what the angle
would be to cut this quadrant of the ellipse at a diagonal (through
the center or intersection of a and b) so that the area of the two
portions are equal (18.41 for each portion of the quadrant).
This is not a homework problem, and would be more confusing to
describe what this is for, but any help would be greatful.
Hello,
I will assume that you know some calculus, and can look up any
details.
The polar equation of your ellipse is r = ab/sqrt{a^2 sin^2(t) + b^2
cos^2(t)}.
(I am using the letter t where most calculs books would use the Greek
letter theta.)
Let the desired angle be a. By the usual method for finding area in
polar coordinates, we want
(Integral from 0 to a of) (1/2) r^2 dt = (1/8) pi ab.
Now we want to find an antiderivative of (1/2)(a^2b^2)/(a^2 sin^2(t) +
b^2 cos^2(t).
This integral is not as bad as it looks. One can look it up in
standard tables of integrals.
If I were to do it by hand, I would probably express sin^2(t) and
cos^2(t) in terms of cos(2t). After a little effort you end up with a
simple expression involving tan(2t), so the required angle can be
expressed in terms of the arctan function.
In the above post I was having a major attack of stupidity. The
method was correct, but the prroblem can be solved much more simply.
Think of an ellipse in standard orientation, with a and b having the
usual meanings.
IMAGINE making the ellipse into a circle by stretching everything in
the y-direction by a factor a/b. Now find the point P on the circle
that gives you the desired "ratio of areas," 1/2 in your case but
anything will do.
TRANSFORM BACK, by scaling everything in the y-direction by a factor
of b/a. This scaling will produce the desired point on the ellipse.
.
- References:
- Area of an ellipse
- From: ms
- Re: Area of an ellipse
- From: no comment
- Area of an ellipse
- Prev by Date: Re: JSH: A solution to the factoring problem
- Next by Date: Re: JSH: Generalizing to factors mod N
- Previous by thread: Re: Area of an ellipse
- Next by thread: Re: Area of an ellipse
- Index(es):
Relevant Pages
|
