Find the Area between Polar Curves

  • #1
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:

Answers and Replies

  • #2
35,028
6,774
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.
 
  • #3
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?

[Edit] Changed integral to reflect [itex]8[/itex] being multiplied by [itex]1/2[/itex]. [Edit]

[Edit 2] When I evaluate this integral I get [itex]π/2 + 3/2[/itex]. [Edit 2]
 
Last edited:
  • #4
22,089
3,297
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?

Please read our FAQ on typing mathematical equations: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
Typing in LaTeX makes it much easier for us.
 
  • #6
haruspex
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$$ 4 \int_0^{\pi/8} sin^2(2\theta)~d\theta + 4 \int_{\pi/8}^{\pi/4} cos^2(2\theta)~d\theta $$

[Edit 2] When I evaluate this integral I get [itex]π/2 + 3/2[/itex]. [Edit 2]
I get [itex]π/2 - 1[/itex]. The two integrals yield the same value, agreed?
 
  • #7
It seems I integrated incorrectly : ( I do get [itex]π/2 - 1[/itex]. Thanks. I appreciate it.
 

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