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Area Inside Lemniscate Vector Calculus

by braindead101
Tags: calculus, inside, lemniscate, vector
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braindead101
#1
Sep17-07, 12:49 AM
P: 162
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?
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EnumaElish
#2
Sep17-07, 01:45 PM
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P: 2,482
First, the equation should have been (x^2 + y^2)^2 = 2a^2 (x^2 - y^2).
braindead101
#3
Sep17-07, 09:47 PM
P: 162
oh yes, when i isolated for a i did remember the ^2
but i have no idea how to proceed from there

a_Vatar
#4
Sep18-07, 07:43 AM
P: 17
Area Inside Lemniscate Vector Calculus

Hint: convert to Lemniscate equation into polar form and calculate area from there
braindead101
#5
Sep19-07, 06:19 PM
P: 162
what does the a even mean..
my prof did not say anything about a lemniscate, I am confused out of my mind
braindead101
#6
Sep19-07, 06:47 PM
P: 162
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
a_Vatar
#7
Sep19-07, 07:14 PM
P: 17
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/
braindead101
#8
Sep19-07, 08:04 PM
P: 162
so how do i get rid of the constant?
a_Vatar
#9
Sep19-07, 10:42 PM
P: 17
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?
hanhan
#10
Sep19-07, 10:53 PM
P: 4
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
hanhan
#11
Sep19-07, 10:55 PM
P: 4
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....
hanhan
#12
Sep19-07, 10:59 PM
P: 4
wait its a^2 not 2a^2
braindead101
#13
Sep19-07, 11:20 PM
P: 162
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
hanhan
#14
Sep19-07, 11:23 PM
P: 4
yeh hahaha ethans class
braindead101
#15
Sep23-07, 06:12 AM
P: 162
i still need help on this,
anyone have any ideas?
is hanhan correct that r^2 is the area so u can integrate ??
a_Vatar
#16
Sep23-07, 09:25 AM
P: 17
http://mathworld.wolfram.com/Lemniscate.html
eq. 13 - 15
braindead101
#17
Sep23-07, 03:21 PM
P: 162
but i am learning about double integrals and it is only a single integration on the website link
braindead101
#18
Sep24-07, 01:05 AM
P: 162
anyone know how i can change it into a double integral to solve for the area inside the lemniscate?


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