How do you find the region inside cardioid #r=1+cos(theta)# and outside the circle #r=3cos(theta)#?

Question

Elijah Abram · Accepted Answer

It is #pi/4#

Find the intersection points of the curves hence we have that

#3cosθ=1+cosθ=>cosθ=1/2=>θ=+-pi/3#

The saded area is

(cardiod area from pi/3 to pi)-(cricle area from pi/3 to pi/2)

The cardiod area is

#int_(pi/3)^(pi) 1/2*(1+cosθ)^2dθ=pi/2-9/6*sqrt3#

and the circle area is

#int_(pi/3)^(pi/2) 1/2*(3*cosθ)^2dθ=(3pi/8)-9/16*sqrt3#

Hence the shaded area is #pi/8#

The total amount is #2pi/8=pi/4#

A graph for the curves is

Isaiah Allen · Answer

undefined To find the region inside the cardioid $ r = 1 + \cos(\theta) $ and outside the circle $ r = 3\cos(\theta) $, you need to set up the integral for the area enclosed by these curves. First, find the points of intersection by setting the equations equal to each other and solving for $ \theta $. Then, integrate the difference of the equations from the smaller angle to the larger angle to find the area enclosed. The integral setup would be as follows:

$$ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left[(1+\cos(\theta))^2 - (3\cos(\theta))^2\right] d\theta $$

Where $ \theta_1 $ and $ \theta_2 $ are the angles of intersection obtained from solving the equations $ 1+\cos(\theta) = 3\cos(\theta) $.