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Find the area of the region that lies inside both curves.  `r=3+cos theta` , `r=3-cos...

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dawks | Student, Undergraduate | Honors

Posted October 6, 2012 at 1:37 AM via web

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Find the area of the region that lies inside both curves.

 `r=3+cos theta` , `r=3-cos theta`

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 6, 2012 at 4:47 PM (Answer #1)

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You need to find the limits of integration setting up the following equation such that:

`3 + cos theta = 3 - cos theta => 2cos theta = 0 => cos theta = 0 => {(theta = pi/2) , (theta = 3pi/2):}`

You should evaluate the area based on symmetry such that:

`A = 2 int_0^(pi/2) ((3 + cos theta)^2 - (3 - cos theta)^2) d theta`

`A = 2 int_0^(pi/2) (9 + 6cos theta + cos^2 theta - 9 + 6cos theta + cos^2 theta) d theta`

`A = 2 int_0^(pi/2) (12 cos theta + 2 cos^2 theta) d theta`

`A =24 int_0^(pi/2) cos theta d theta+ 4 int_0^(pi/2) cos^2 theta d theta`

Using the half angle trigonometric identity yields:

`A = 24 sin theta|_0^(pi/2) + 4int_0^(pi/2) (1 + cos 2 theta)/2 d theta`

`A = 24 (sin (pi/2) - sin 0) + 2(pi/2 - 0) + sin 2(pi/2) - sin 0`

Since `sin (pi/2)= 1`  and `sin 0 = sin pi = 0`  yields:

`A = 24 + pi`

Hence, evaluating the area found inside both given curves yields `A = 24 + pi.`

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