Calculating the Area of a Region Bounded by a Cardioid and Circle

[MozAngeles]

Homework Statement

Inside the cardioid r=2(1+sin(theta)) and outside the circle r=2sin(theta)

Homework Equations

A= ∫ (from a..b) 1/2 f(θ) ² dθ

The Attempt at a Solution

A= 2∫(from π/2..3π/2) 1/2 [2(1+sin(θ)]² dθ- 2∫(from0..π/2) 1/2 (2sinθ)²dθ

after working that out i got the answer to be 5pi.

i need a verification for my answer because it is an even problem in the book and I'm studying for my test.

 

[jbunniii]

It looks OK except for the bounds of integration.

Why is the first integral taken over [\pi/2,3\pi/2], and why is the second integral taken over [0,\pi/2]?

 

[Ray Vickson]

The bounds of integration are OK. The range -pi/2 <= theta <= pi/2 gives the right half of the cardioid; equivalently, the range pi/2 <= theta <= 3pi/2 gives the left half. The range 0 <= theta <= pi/2 gives the right half of the circle; equivalently, 0 <= theta <= Pi gives the whole circle. (The whole range 0 <= theta <= 2*Pi goes around the circle twice, so would give twice the desired area.) The answer 5*pi is correct.

RGV

 

[jbunniii]

You're right, I didn't notice the extra factor of 2.