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Find the area that lies inside both curves r = a and r2 = 2a2 cos(2θ). Here a is a...

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haqc | eNoter

Posted October 2, 2013 at 4:36 AM via web

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Find the area that lies inside both curves r = a and r2 = 2a2 cos(2θ). Here a is a positive constant. 

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aruv | High School Teacher | Valedictorian

Posted October 2, 2013 at 5:38 AM (Answer #1)

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Curves are  `r=a`  and `r^2=2a^2cos(2theta).`

Thus

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

`cos(2theta)=1/2`

`=> theta=pi/6,(5pi)/6`

Thus required area A,

`A=int_(pi/6)^((5pi)/6)(1/2)(2a^2cos(2theta)-a^2)d theta`

`` `=(a^2/2)(sin(2theta)-theta)_(pi/6)^((5pi)/6)`

`=(a^2/2){-sqrt(3)-(2pi)/3}`

Since area is not considered as negative, ignore negative sign.

Thus

`A=(sqrt(3)+(2pi)/3)(a^2/2)`

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