area within two ellipses

[ACLerok]

i am given the equations for two ellipses:
(x^2)/3 + y^2 =1 and x^2 + (y^2)/3 =1
and am told to find the area within both.

i can just find the volume of both ellipses by taking the integral of the equations and then add them right? That's what first comes to mind when i saw this problem. Thanks!!!

[faust9]

Your told to find the area of both ellipses? Why are you even thinking about volume? Why do you want to add the results together? By inspection, shouldn't the are of both ellipses be the same. The only difference is one ellipse has the major axis along the x-axis and the othe major axis falls along the y-axis. The magnitude of the major and minor axis for both ellipses is the same though.

Good luck.

[Tide]

Even if it is volume you can easily use Pappus' (2nd) Theorem.

In either case it won't be a simple difference.

[diegojco]

well, first of all, in this kind of problems try to sketch the situation. for example in this case you have two ellipses that intersects, you have to find the area of the intersection (since I don't have a scanner I can't put a draw of it, but you can do it since you must have knowledge of analytic geometry and conics). One of the ellipses is horizontal and the other vertical, in fact both are the same but one rotated 90º.

Since you have the situation in mind calculate the points where ellipses intersect in order to have limits of the integration. you will find the points are along the lines y=x and y=-x and are four. they x coordinate is +-sqrt(3) (where sqrt is square root).

Now you can integrate, I have done an integration respect to x (you can do it respect y but generally the intersection points will be other, in this case because of symmetry they are the same, sketch it and you will see it), then i have solved two ellipses' equations for y to obtain a function of x, this causes the area you obtain is only the one above x-axis. but you have symmetry, then the area above is equal to the area down of x-axis. in resume the total area is two times the area you obtain in integration, hence:

A=2*(int(sqrt(3-(x^2))/sqrt(3))+2*int(sqrt(3-3*(x^2))))

where int(f(x)) is the integral of f(x). The first integral is evaluated from -sqrt(3)/2 to sqrt(3)/2. And the second integral is evaluated from -1 to -sqrt(3)/2 (Why this last evaluation interval? if you sketch the functions you will have the answer).

[ACLerok]

how do i integrate sqrt(3-(x^2))/sqrt(3)? i put the sqrt(3) outside the integral and tried using the substitution method to integrate it but there is still an x in there?

same thing with integrating sqrt(3-3*(x^2))

[maverick280857]

how do i integrate sqrt(3-(x^2))/sqrt(3)? i put the sqrt(3) outside the integral and tried using the substitution method to integrate it but there is still an x in there?

same thing with integrating sqrt(3-3*(x^2))

If you mean \int \sqrt{3-3x^2}dx then this is a standard form. You need to do this by parts if you do not know it already. Take sqrt(1-x^2) as the first function and x as the second.

Cheers
Vivek

[poolwin2001]

Easier method to integrate, take 3 out and Sub,x=Siny/Cosy
regards

[maverick280857]

Easier method to integrate, take 3 out and Sub,x=Siny/Cosy
regards

Yes, put

x = \sin\theta

so that

\int \sqrt{1-x^2}dx = \int \cos^2\theta d\theta

You should be able to do this using substitution now.

Note: your integration limits must change accordingly.

Cheers
Vivek