1. ## Polar Double Integration

Consult this picture: http://i43.tinypic.com/f224xz.jpg

I want to find the area of the region enclosed by the green outline (so basically the whole yellow/opaque yellow region)

The bigger curve is r=2cos(2theta+(pi/2)) and the smaller curve is just r=cos(2theta)

I thought it would be easiest to find the area of the blue region or the red region and then multiply by 8 and do some subtraction, etc. to get the area I want.

However, I am not sure what to integrate because the theta values of intersection of the curves are different for each curve.

How do I do it?

2. Hi. Looks like it's:

$\displaystyle 8\left[\int_{\text{red}}^{\text{blue}}\int_0^{2\cos(2t+\p i/2)} r drdt+\int_{\text{blue}}^{\pi/2}\int_{2\cos(2t+\pi/2)}^{\cos(2t)} r drdt\right]$

3. Originally Posted by stones44
Consult this picture: http://i43.tinypic.com/f224xz.jpg

I want to find the area of the region enclosed by the green outline (so basically the whole yellow/opaque yellow region)

The bigger curve is r=2cos(2theta+(pi/2)) and the smaller curve is just r=cos(2theta)

I thought it would be easiest to find the area of the blue region or the red region and then multiply by 8 and do some subtraction, etc. to get the area I want.

However, I am not sure what to integrate because the theta values of intersection of the curves are different for each curve.

How do I do it?
let $\displaystyle r_1 = \cos(2\theta)$

$\displaystyle r_2 = 2\cos\left(2\theta + \frac{\pi}{2}\right) = -2\sin(2\theta)$

to get the two graphs to "sync" in quad I ... let $\displaystyle r_2 = 2\sin(2\theta)$

in quad I, the two polar curves intersect at an angle $\displaystyle \phi = \frac{1}{2}\arctan\left(\frac{1}{2}\right)$ relative to the x-axis.

using symmetry ...

$\displaystyle A = 8 \int_0^\phi \frac{r_1^2}{2} \, d\theta + 8 \int_\phi^\frac{\pi}{4} \frac{r_2^2}{2} \, d\theta$

4. Originally Posted by skeeter
let $\displaystyle r_1 = \cos(2\theta)$

$\displaystyle r_2 = 2\cos\left(2\theta + \frac{\pi}{2}\right) = -2\sin(2\theta)$

to get the two graphs to "sync" in quad I ... let $\displaystyle r_2 = 2\sin(2\theta)$

in quad I, the two polar curves intersect at an angle $\displaystyle \phi = \frac{1}{2}\arctan\left(\frac{1}{2}\right)$ relative to the x-axis.

using symmetry ...

$\displaystyle A = 8 \int_0^\phi \frac{r_1^2}{2} \, d\theta + 8 \int_\phi^\frac{\pi}{4} \frac{r_2^2}{2} \, d\theta$
so in this: $\displaystyle A = 8 \int_0^\phi \frac{r_1^2}{2} \, d\theta + 8 \int_\phi^\frac{\pi}{4} \frac{r_2^2}{2} \, d\theta$

the first integral is the area of the red region * 8. So to get the rest of the area don't you just need to full area of the bigger yellow function? So why integrate from that angle phi to pi/4 (multiplied by 8)? why not just 0 to pi/2 (and multiply by 4)

5. Originally Posted by stones44
so in this: $\displaystyle A = 8 \int_0^\phi \frac{r_1^2}{2} \, d\theta + 8 \int_\phi^\frac{\pi}{4} \frac{r_2^2}{2} \, d\theta$

the first integral is the area of the red region * 8. So to get the rest of the area don't you just need to full area of the bigger yellow function? So why integrate from that angle phi to pi/4 (multiplied by 8)? why not just 0 to pi/2 (and multiply by 4)
because the limits from 0 to pi/2 do not work ...

check out the attached graph of both functions with those limits.

6. as can be seen in that graph, for the bigger 4-petal-rose, those limits would work. If i was just trying to find the area of the bigger rose, I could take 1/2 the integral from 0 to pi/2 and multiply my result by 4. Correct?

and then for my total area I am seeking (both curves), just add on the 8 * [the first integral in your equation from 0 to .2318..]

hmm EDIT: i think I am seeing what your doing now. you arent getting the full areas of curves, but rather the area up to a diagonal line that is shared in both integrals.

7. note the region using the limits I provided ...