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Math Help - Polar Double Integration

  1. #1
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    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?
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  2. #2
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    Hi. Looks like it's:

    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]
    Attached Thumbnails Attached Thumbnails Polar Double Integration-polar-int.jpg  
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  3. #3
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    Quote Originally Posted by stones44 View Post
    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 r_1 = \cos(2\theta)<br />

    r_2 = 2\cos\left(2\theta + \frac{\pi}{2}\right) = -2\sin(2\theta)<br />

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

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

    using symmetry ...

    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
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  4. #4
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    Quote Originally Posted by skeeter View Post
    let r_1 = \cos(2\theta)<br />

    r_2 = 2\cos\left(2\theta + \frac{\pi}{2}\right) = -2\sin(2\theta)<br />

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

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

    using symmetry ...

    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: 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)
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  5. #5
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    Quote Originally Posted by stones44 View Post
    so in this: 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.
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  6. #6
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    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.
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  7. #7
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    note the region using the limits I provided ...
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