r = 1 + cos θ on the same axis. a) Find the area inside both the circle and cardioid. b) Find the arc length of that part of the cardioid outside the circle.

Question

Sketch the circle r = 3 cos θ and cardioids  r = 1 + cos θ on the same axis.

a)         Find the area inside both the circle and cardioid.

b)         Find the arc length of that part of the cardioid outside the circle.

Answer 1

OK, so let me continue part a)!

Part A) Don't forget to multiply by 2 to take advantage of the symmetry of the areas.

2 * ( 0.5 * integral(0, pi/3, (1 + cosx)^2, dx) + 0.5 * integral(pi/3, pi/2, (3cosx)^2, dx) ) =

∫(0, pi/3, 1 + 2cosx + cos^2(x) , dx) + ∫(pi/3, pi/2, 9cos^2(x) , dx) =

integral(0, pi/3, 1.5 + 2cosx + 0.5cos(2x) , dx) + ∫(pi/3, pi/2, 4.5 + 4.5cos(2x) , dx) =

(1.5x + 2sinx + 0.25sin(2x)) | (from 0 to pi/3) + (4.5x + 2.25sin(2x)) | (from pi/3, pi/2) =

(after a little algebra and calculation): 5pi/4 (Ans)

Part b) As (sort of) shown in the video, the arc length mentioned in the problem refers to the cardioid (shown in red in the graph in the video) from the intersection points, starting from pi/3 to 5pi/3.

Remember the arc length formula with respect to Θ is:

Arc length = ∫(Θ1, Θ2, sqrt((dx/dΘ)^2 + (dy/dΘ)^2), dΘ)

We can define x = rcosΘ, and y = rsinΘ.

Therefore, x = (1 + cosΘ)(cosΘ) and y = (1 + cosΘ)(sinΘ). (FYI: 2sinΘcosΘ = sin(2Θ))

Taking each derivative with respect to Θ is: dx/dΘ = -sinΘ - sin(2Θ), and dy/dΘ = cosΘ + cos(2Θ).

Plugging that back into the integral yields:

Arc length = ∫(pi/3, 5pi/3, sqrt((dx/dΘ)^2 + (dy/dΘ)^2), dΘ). At this point, I recommend you use a calculator. Just make sure your calculator is in raidans before you plug in the numbers! I hope this helped! Any feedback is highly appreciated! :)