Calculating area within two ellipses?

Question

two ellipses are drawn on the same graph, x^2/3 + y^2 = 1, and x^2 + y^2/3 = 1. How would I find the area contained within the two ellipses? ie. the area in the middle that both ellipses overlap on. I'm trying to use integrals and the area between curves, but i'm getting these ridiculously complicated integrals. Can anyone help? Thanks very much!

Answer 1

hi

intersection
x=y= √3/2

polar coordinates
r^2 = 1 / ( cos^2 a + 1/3 sin^2 a)
= 3/2 * [1 / ( 3/2 - cos^2 a)]

A = 8 int (1/2 r^2 da ) (from π/4 to π/2)
= 6 int [1 / ( 3/2 - cos^2 a)] (from π/4 to π/2)
= (6 * 2 / √3) * arctg ( √3 tg a) (from π/4 to π/2)
= 4 * √3 * [ arc tg (inf) - arc tg (√3) ]
= 4 * √3 * ( π/2 - π/3)
= 2/3 π * √3
= 3.6276 (e<0.0001)


bye

Answer 2

Ellipses Area

Answer 3

First, let's find out where those two ellipses intersect one another.

(1/3)x^2 + y^2 = x^2 + (1/3)y^2 --> (2/3)x^2 = (2/3)y^2 --> x^2 = y^2 --> y = +/-x.

(1/3)x^2 + y^2 = 1 --> (4/3)x^2 = 1 --> x^2 = 3/4 --> x = +/- sqrt(3)/2.

So, the ellipses intersect at (sqrt(3)/2, sqrt(3)/2), (-sqrt(3)/2, sqrt(3)/2), (-sqrt(3)/2, -sqrt(3)/2), and (sqrt(3)/2, -sqrt(3)/2). *whew*

Note that both ellipses have the same center at (0,0). If you draw the ellipses and the lines y = x and y = -x then you can see that if we find the area of one little section inside the intersection then we have 1/8 of the total area.

Thus, integrate dx*dy from y = [x to sqrt[1-(1/3)x^2] and then from x = [0, sqrt(3)/2] and multiply by 8.

I'll have to work out the details a little later.

Answer 4

You will be getting complicated integrals, where you probably going to have to use trig. substitution in order to solve the integral. You should first solve the two ellipses in terms of y=. Wish i had more time, but this might take too long for me to do.

Answer 5

Okay, iff you can look at it graphically it helps.
First, see where they intersect in 1sT quadrant.
They intersect at ((sqr3)/2,(sqr3)/2).
Now, the one with major axis on x-axis is
x^2/3+y^2=1, or y=sqr(1-x^2/3).
Now the area for the eclipse part, or
from the top of graph to where it intersect with other
graph, is INT from 0 to (sqr3)/2 of pi (sqr(1-x^2/3))^2,
or pi (1-x^2/3) minus ((sqr3)//2)^2.
Or from 0 to (sqr3)/2 of x-(x^3/9) minus ((sqr3)/2)^2..
SO, I get ((2pi*sqr3)/6)-(3/4) = 1.0638 approximately.

(SOrry, i made a mistake, it would be 8 times instead of 4 times of 1.0638, I didnt take into account both sides of what i integrated. SO its 8*1.0638=8.51. So, then add (2sqr3/2)^2= 3 to that, getting 11.51)