What is the area of a limaçon left after being carved out by a circle?
I got this far: (these are polar functions)
r1= 4 + 4cos(θ) and r2=6
4 + 4cos(θ)= 6 when θ = ± π/3
So:
-π/3 ∫ π/3 (1/2)(4+4cos(θ))^2 – (1/2)(6)^2 dθ
I cut the limits in half (0 to π/3 instead of -π/3 to π/3) and then doubled the integrand since both equations are symmetric about the polar axis, and then I somehow came up with the negative answer 6(√3) - 4π. I checked my calculator and it came up with a completely different answer.
Please help, and if you can, please show work
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pi/3 is correct for the point of intersection, and your area formula looks correct. My guess is that your integration algebra was not correct.
A = integral (from t=-pi/3 to t=+pi/3)
{ (1/2) [(outer radius)^2- (inner radius)^2] } dt
Here
outer radius = 4+4cos(t)
inner radius = 6
so the quantity in braces is
(1/2) {(4+4cos(t))^2 - 6^2} = 8+16cos(t)+8cos^2(t)-18
= 16cos(t) + 8cos^2(t) - 10
As you did, we can use symmetry and integrate from 0 to pi/3:
A = 2 * integral (from t=0 to t=+pi/3)
{ 16cos(t) + 8cos^2(t) - 10 } dt
= 4 * integral (from t=0 to t=+pi/3)
{ 8cos(t) + 4cos^2(t) - 5 } dt
= 4 * [8sin(t) + 2*(t+sin(2t)/2) - 10t] {from t=0 to t=+pi/3}
= 4* [8 sqrt(3)/2 + 2*pi/3 + sqrt(3)/2 - 10*pi/3]
= 4*[ (9/2) sqrt(3) - (8/3) pi ]
= 28.844....