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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? 
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2 Answers

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

Of course - you're more than welcome (these are fun for me  :) 
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

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