# Find the Area between Polar Curves

1. Nov 10, 2012

### 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: Nov 10, 2012
2. Nov 10, 2012

### Staff: Mentor

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. Nov 10, 2012

### johnhuntsman

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: Nov 10, 2012
4. Nov 10, 2012

### micromass

Staff Emeritus
Typing in LaTeX makes it much easier for us.

5. Nov 10, 2012

### johnhuntsman

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.

6. Nov 10, 2012

### haruspex

I get $π/2 - 1$. The two integrals yield the same value, agreed?

7. Nov 10, 2012

### johnhuntsman

It seems I integrated incorrectly : ( I do get $π/2 - 1$. Thanks. I appreciate it.