Question

What is the area of the region inside the circle:
\[r = \sqrt3 \sin \theta \] and outside the cardioid
\[\ r = 1 + \cos \theta \]

Answer 1

Here's what it looks when graphed.

Answer 2

I'm not sure what the bounds would be, but I think the setup would be along the lines of:
\[\int\limits_{}^{} 1/2* (\sqrt 3 \sin \theta - (1+\cos \theta)^2 d \theta\]

Answer 3

i had a problem kinda like this it took me awhile to sole it... I can show you ow i did that? @shamil98

Answer 4

Sure.

Answer 5

The question was Find the area of the region that lies inside the circle r=3cos theta and outside the cardioid r=1+cos theta.

Answer 6

Let parametric curves be given by
r₂(θ): r = 3cosθ
r₁(θ): r = 1+cosθ

area of bounded region: A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...

the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves.
2cosθ = 1 => θ = ±π/3

A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ
= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ
= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] ..
.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)]
= 3θ/2 +sin(2θ) - sin(θ)

Area = A(π/3) - A(-π/3)
= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3)
= π.

Answer 7

Would the bound be from 0 to pi ?

Answer 8

yup

Answer 9

I suggest finding the angle where they intersect and integrate from 0 to that angle for the circle then add that to the integral from that angle to pi of the cartioid.