Can you please verify the correct answer? (Just check my work)

Question

Which of the following is equal to the area of the region inside the polar curve r=2cos(θ) and outside the polar curve r=cos(θ)?

 

a) 3 ∫ cos^2 (θ) dθ from 0 to pi/2

 

b) 3 ∫ cos^2 (θ) dθ from 0 to pi

 

c) (3/2) ∫ cos^2 (θ) dθ from 0 to pi/2

 

d) 3 ∫ cos(θ) dθ from 0 to pi/2

 

e) 3 ∫ cos(θ) dθ from 0 to pi

 

I worked through the problem and got ∫ 3*cos^2 (θ) dθ from 0 to pi. The answer in my textbook says is A. But I don't know why the answer isn't B. Can anyone tell me how to simplify to get answer A? 

Answer 1

Because we trace each circle out twice as θ varies from 0 to 2pi:

 

Consider: r=2cos(θ)

 

i) At  θ=0, we have r=2

 

ii) At θ=(pi/2) we have r=0

 

iii) At θ=pi we have r=-2 (at which point we've gone full circle, as the radius points in the opposite direction; i.e. back towards the line θ=0)

 

iv) The same analysis holds true for r=cos(θ)

 

 

Hence the integration is as follows:

 

∫[outer circle] - ∫[inner circle]

 

=∫(1/2)[2cos(θ)]^2 dθ -∫(1/2)[cos(θ)]^2 dθ (each evaluated from 0 to pi)

 

= ∫2cos^2(θ) dθ -∫(1/2)cos^2(θ) dθ (each evaluated from 0 to pi)

 

= ∫(3/2)cos^2(θ) dθ (evaluated from 0 to pi)

 

= 2∫(3/2)cos^2(θ) dθ (evaluate from 0 to pi/2)

 

=3∫cos^2(θ) dθ (evaluated from 0 to pi/2)

 

= a)

 

Answer 2

Sun,

        The area is the area between two concentric circles larger one AL=π(2)²=4π and the smaller one AS=π

The area between is 4π-π=3π

 

The indefinite integral of 3cos²(θ)dθ=3/2(θ+sin(θ)cos(θ)) if this is evaluated from 0 to π/2 you get 3π/4 but when evaluated between 0 and π you get 3π/2 neither one of which is right. If you integrate from 0 to 2π you get 3π which is the correct answer. 

 

This is what I got for the answer: 3∫cos²(θ)dθ from 0 to 2π which is not any of the choices a-e ?? What am I missing?

Regards

Jim