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

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?

### 2 Answers by Expert Tutors

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Kahroline D. | I CAN teach you Mathematics!!I CAN teach you Mathematics!!
5.0 5.0 (3 lesson ratings) (3)
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)

### Comments

Thank you so much, Kahroline D.
Of course - you're more than welcome (these are fun for me  :)
Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (186 lesson ratings) (186)
-1
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

### Comments

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