# Find the Area between Polar Curves

johnhuntsman
r = sin 2θ, r = cos 2θ.

I'm having some trouble setting this up.

$$1/2 \int_{\ -pi/8}^{\pi/8} cos^2 2θ~d\theta - 1/2 \int_{\ -pi/8}^{\pi/8} sin^2 2θ~d\theta$$

Which can be:

$$\int_0^{\pi/8} cos^2 2θ~d\theta - \int_0^{\pi/8} sin^2 2θ~d\theta$$

Since there are 8 petals.

$$8 \int_0^{\pi/8} cos^2 2θ~d\theta - 8 \int_0^{\pi/8} sin^2 2θ~d\theta$$

Which becomes:

____(π / 8)
[2sin 4θ]
______0

Which gives me the wrong answer. Somehow the answer is (π / 2) - 1. Can someone please tell me what's wrong with the way I set it up?

Last edited:

Mentor
r = sin 2θ, r = cos 2θ.

I'm having some trouble setting this up.

____(π / 8)________________________(π / 8)
(1 / 2) ∫ cos2 2θ dθ - (1 / 2) ∫ sin2 2θ dθ
___-(π / 8)_______________________-(π / 8)

Which can be:

(π / 8)_______________(π / 8)
∫ cos2 2θ dθ - ∫ sin2 2θ dθ
0______________________0

Since there are 8 petals.

(π / 8)___________________(π / 8)
8 ∫ cos2 2θ dθ - 8 ∫ sin2 2θ dθ
__0_______________________0

Which becomes:

____(π / 8)
[2sin 4θ]
______0

Which gives me the wrong answer. Somehow the answer is (π / 2) - 1. Can someone please tell me what's wrong with the way I set it up?

Let's focus on the parts of the graphs in the first quadrant. The petals of the two graphs intersect where θ = ##\pi/8## and at the origin (which corresponds to different values of θ for each function.

As θ runs from 0 to ##\pi/8##, the graph of r = sin(2θ) traces out the lower boundary of the intersecting region. As θ runs from ##\pi/8## to ##\pi/4##, the graph of r = cos(2θ) traces out the upper boundary of the intersecting region.

The area of this region (again, just the part in the first quadrant) would be:
$$1/2 \int_0^{\pi/8} sin^2(2\theta)~d\theta + 1/2 \int_{\pi/8}^{\pi/4} cos^2(2\theta)~d\theta$$

Click the integral above to see how I did it in LaTeX.

johnhuntsman
Let's focus on the parts of the graphs in the first quadrant. The petals of the two graphs intersect where θ = ##\pi/8## and at the origin (which corresponds to different values of θ for each function.

As θ runs from 0 to ##\pi/8##, the graph of r = sin(2θ) traces out the lower boundary of the intersecting region. As θ runs from ##\pi/8## to ##\pi/4##, the graph of r = cos(2θ) traces out the upper boundary of the intersecting region.

The area of this region (again, just the part in the first quadrant) would be:
$$1/2 \int_0^{\pi/8} sin^2(2\theta)~d\theta + 1/2 \int_{\pi/8}^{\pi/4} cos^2(2\theta)~d\theta$$

Click the integral above to see how I did it in LaTeX.

I gotcha.

Taking that integral and building on it:
$$4 \int_0^{\pi/8} sin^2(2\theta)~d\theta + 4 \int_{\pi/8}^{\pi/4} cos^2(2\theta)~d\theta$$
Since there are 8 petals. That should give me the area in all 8 petals formed by the two curves?

 Changed integral to reflect $8$ being multiplied by $1/2$. 

[Edit 2] When I evaluate this integral I get $π/2 + 3/2$. [Edit 2]

Last edited:
Staff Emeritus
Homework Helper
r = sin 2θ, r = cos 2θ.

I'm having some trouble setting this up.

____(π / 8)________________________(π / 8)
(1 / 2) ∫ cos2 2θ dθ - (1 / 2) ∫ sin2 2θ dθ
___-(π / 8)_______________________-(π / 8)

Which can be:

(π / 8)_______________(π / 8)
∫ cos2 2θ dθ - ∫ sin2 2θ dθ
0______________________0

Since there are 8 petals.

(π / 8)___________________(π / 8)
8 ∫ cos2 2θ dθ - 8 ∫ sin2 2θ dθ
__0_______________________0

Which becomes:

____(π / 8)
[2sin 4θ]
______0

Which gives me the wrong answer. Somehow the answer is (π / 2) - 1. Can someone please tell me what's wrong with the way I set it up?

Typing in LaTeX makes it much easier for us.

johnhuntsman
Typing in LaTeX makes it much easier for us.

I'm sorry. My humblest of apologies. I've always been to hasty in wanting to get help for my question I've never bothered learning how to do that. My apologies once again.

$$4 \int_0^{\pi/8} sin^2(2\theta)~d\theta + 4 \int_{\pi/8}^{\pi/4} cos^2(2\theta)~d\theta$$
[Edit 2] When I evaluate this integral I get $π/2 + 3/2$. [Edit 2]
I get $π/2 - 1$. The two integrals yield the same value, agreed?
It seems I integrated incorrectly : ( I do get $π/2 - 1$. Thanks. I appreciate it.