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Area Inside Lemniscate Vector Calculus 
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#1
Sep1707, 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? 


#2
Sep1707, 01:45 PM

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P: 2,481

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



#3
Sep1707, 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 


#4
Sep1807, 07:43 AM

P: 17

Area Inside Lemniscate Vector Calculus
Hint: convert to Lemniscate equation into polar form and calculate area from there



#5
Sep1907, 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 


#6
Sep1907, 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 (1sin^2theta  sin^2theta) r = 2a^2 (12sin^2theta) Using half angle rule: r = 2a^2 (12(1cos^2theta / 2)) r = 2a^2 (1(1cos^2theta)) r = 2a^2 (cos^2theta) i still do not know the meaning of a.. and the boundaries 


#7
Sep1907, 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/ 


#8
Sep1907, 08:04 PM

P: 162

so how do i get rid of the constant?



#9
Sep1907, 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? 


#10
Sep1907, 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 


#11
Sep1907, 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.... 


#12
Sep1907, 10:59 PM

P: 4

wait its a^2 not 2a^2



#13
Sep1907, 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 


#14
Sep1907, 11:23 PM

P: 4

yeh hahaha ethans class



#15
Sep2307, 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 ?? 


#16
Sep2307, 09:25 AM

P: 17

http://mathworld.wolfram.com/Lemniscate.html
eq. 13  15 


#17
Sep2307, 03:21 PM

P: 162

but i am learning about double integrals and it is only a single integration on the website link



#18
Sep2407, 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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