**2**∫(3/2)cos^2(θ) dθ (evaluate from

**0 to pi/2**)

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?

Tutors, sign in to answer this question.

Kahroline D. | I CAN teach you Mathematics!!I CAN teach you Mathematics!!

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)

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 3cos^{2}(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∫cos^{2}(θ)dθ from 0 to 2π which is not any of the choices a-e ?? What am I missing?

Regards

Jim

Jim - these are roses with a single petal of length a:

i) r=2cos(θ) is a rose with one petal, of length 2 (hence a circle of radius 1) - the area contained within this circle is pi

ii) r=cos(θ) is a rose with one petal, of length 1 (hence of radius 1/2) - the area contained within this circle is pi/4

The difference in area between the two circles (or roses with one petal) is 3 pi/4

If we integrate answer a) the result is as follows:

a) 3∫cos^2(θ) dθ = 3/2∫[1+cos(2θ)] dθ (evaluated from 0 to pi/2)

=3/2[θ+1/2 sin(2θ)] (evaluated from 0 to pi/2)

=3/2[pi/2]

=3 pi/4

Hi Kahroline

thanks. I missed that point totally. I assumed the curves were concentric circles duh.

thanks again

jim

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments