# Area Inside Lemniscate Vector Calculus

1. Sep 17, 2007

Find the area inside the lemniscate ,which is described by the equation (x^2 + y^2) = 2a^2 (x^2 - y^2)

I have no idea where to start so I found out what a lemniscate is.. but this didn't help me much. I have rearranged the equation for a:

a = sqrt ( (x^2 + y^2)^2 / 2(x^2 - y^2) )
a = x^2 + y^2 / sqrt (2(x^2 - y^2))

How do I proceed from here?

2. Sep 17, 2007

### EnumaElish

First, the equation should have been (x^2 + y^2)^2 = 2a^2 (x^2 - y^2).

3. Sep 17, 2007

oh yes, when i isolated for a i did remember the ^2
but i have no idea how to proceed from there

4. Sep 18, 2007

### a_Vatar

Hint: convert to Lemniscate equation into polar form and calculate area from there

5. Sep 19, 2007

what does the a even mean..
my prof did not say anything about a lemniscate, I am confused out of my mind

6. Sep 19, 2007

okay so i can only put it in polar coordinates, but where are the limits of integration in which i am suppose to integrate??

(r^2)^2 = 2a^2 (rcos^2theta - rsin^2theta) r
r^4 = 2a^2 (rcos^2theta - rsin^2theta) r
r^3 = 2a^2 (r^2cos^2theta - r^2sin^2theta)
r = 2a^2 (cos^2theta - sin^2theta)
r = 2a^2 (1-sin^2theta - sin^2theta)
r = 2a^2 (1-2sin^2theta)
Using half angle rule:
r = 2a^2 (1-2(1-cos^2theta / 2))
r = 2a^2 (1-(1-cos^2theta))
r = 2a^2 (cos^2theta)

i still do not know the meaning of a.. and the boundaries

7. Sep 19, 2007

### a_Vatar

should be r^2 = 2a^2 (cos^2theta);

sketch the curve to find the limits of intergration, I'd say for the right loop you integrate
from -pi/4 to pi/4.
Constant a controls the form of the curve.

Try sketching it here http://graph.seriesmathstudy.com/

8. Sep 19, 2007

so how do i get rid of the constant?

9. Sep 19, 2007

### a_Vatar

Well, you don't.
What is the integral of y = ax ? ax^2 / 2, right? Can you integrate r^2 = 2a^2 (cos^2theta) then, integrate wrt to theta not a?

10. Sep 19, 2007

### hanhan

i was doing the same problem and i got 2a^2 as my answer

r^2 = 2a^2cos(2theta)

the integral of r^2 should be the area hence integrating the right side gives u an asnwer wrt a

11. Sep 19, 2007

### hanhan

anyone know how to compute Integral 0 to 1 ; integral y to 1 (1+x^2)^(1/2) dxdy

its a double integral problem, im stuck....

12. Sep 19, 2007

### hanhan

wait its a^2 not 2a^2

13. Sep 19, 2007

i found examples in class where it is integrating the double integral w.r.t dr dtheta
so do i rearrange for a to get rid of a... or do as u say and integrate w.r.t dtheta
i still have not found the limits of integration yet, i dont see how i am suppose to figure all this stuff out with no information and no knowledge of a lemniscate

14. Sep 19, 2007

### hanhan

yeh hahaha ethans class

15. Sep 23, 2007

i still need help on this,
anyone have any ideas?
is hanhan correct that r^2 is the area so u can integrate ??

16. Sep 23, 2007

### a_Vatar

17. Sep 23, 2007