# Thread: find area of polar graph

1. ## find area of polar graph

i'll use x for theta because i don't know how to make the theta symbol

find the area inside the oval limacon r=4+2cos(x)

so i've squared r to get 16 +16cos(x) +4cos(x)^2

factor out a 4 to get 4(4+4cos(x)+cos(x)^2)

then change to 4(4+4cos(x)+(1+cos2x)/2
factor out a 1/2 and integrate, which comes to
2(3x + 2sin(x)+(sin2x)/2

so i think the limits of integration are from 0 to pi. but i'm not getting the correct answer using those. can anyone help me out?

2. 4[4+4cos(x)+(1+cos2x)/2]

= 18 + 16cos(x) + 2cos(2x)

1/2 integral {18 + 16cos(x) + 2cos(2x)dx} = 9x + 8sin(x) +1/2 sin(2x)

also the graph is generated by letting theta vary from 0 to 2pi

so you can integrate from 0 to pi and double it

3. Originally Posted by isuckatcalc
i'll use x for theta because i don't know how to make the theta symbol

find the area inside the oval limacon r=4+2cos(x)

so i've squared r to get 16 +16cos(x) +4cos(x)^2

factor out a 4 to get 4(4+4cos(x)+cos(x)^2)

then change to 4(4+4cos(x)+(1+cos2x)/2
factor out a 1/2 and integrate, which comes to
2(3x + 2sin(x)+(sin2x)/2

so i think the limits of integration are from 0 to pi. but i'm not getting the correct answer using those. can anyone help me out?
$A = \int_0^{2\pi} \frac{(4+2\cos{\theta})^2}{2} \, d\theta
$

$A = \int_0^{2\pi} \frac{16+16\cos{\theta}+4\cos^2{\theta}}{2} \, d\theta$

$A = \int_0^{2\pi} 8+8\cos{\theta}+2\cos^2{\theta} \, d\theta$

$A = \int_0^{2\pi} 9+8\cos{\theta}+\cos(2\theta) \, d\theta$

$A = \left[9\theta + 8\sin{\theta} + \frac{\sin(2\theta)}{2}\right]_0^{2\pi} = 18\pi$

4. how do you get that 18? i don't see where that's coming from. thanks for the help. replying to calculus 26.

5. or to skeeter, how do you get the 9 there? i'm guessing it's some kind of identity that i'm not familiar with. or i'm just dumb today.

6. distribute the 4 through 4[4+4cos(x)+(1+cos2x)/2]

you get 16 + 4(1/2) = 18

7. ok thank you very much to both of you, much appreciated

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