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WHat's an easier way to set up this integral?

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    I think I overdid this. Find area outside of the polar curve r = 2cosθ and inside r = 1

    3. The attempt at a solution

    http://img651.imageshack.us/img651/2558/44170022.jpg [Broken]

    It is the black region.

    Here is what I did.

    1) Find area of the semi-circle of r = 2cosθ
    2) Find area of r = 2cosθ from 0 to π/3
    3) Find area of r = 1 from π/3 to π/2

    Do some geometry

    Then the integral should be

    Area = [tex]\int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \mathrm{d}\theta - \int_{0}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta + \int_{0}^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta [/tex]

    But this took quit a lot of work. Is there, in general, anyway to get around that curve?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 22, 2011 #2
    By the way, the answer turned out positive. Which was good
  4. Nov 22, 2011 #3
    I'm not sure I entirely understand the question, but shouldn't you consider the area of the r = 1 circle minus the area of the r = 2cos\theta circle?
  5. Nov 23, 2011 #4


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    Note that
    A &= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
    \int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta +
    \int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
    &= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
    \left[\int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta -
    \int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta\right] \\
    &= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
    \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
    &= \frac{1}{2} \int_{\frac{\pi}{3}}^\frac{\pi}{2} [1 - (2\cos\theta)^2] \,d\theta
    By the way, is there a reason you're excluding the area in quadrants II through IV?
  6. Nov 23, 2011 #5
    Nope...but i just realize that was what the question was asking...

    So that integral translates to (I may as well continue with this approach, as going from pi/3 to -pi/3 won't make a difference with what my problem with this question is)

    [tex]\int_{2\cos\theta}^{1}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} r\mathrm{dr}\mathrm{d\theta} [/tex]

    But this doesn't make sense because

    http://img585.imageshack.us/img585/979/60638227.jpg [Broken]

    Basically the green line is the line θ = π/3. The black area is what is the actual area and the red area is what got cut off and that red area is INSIDE the curve r = 2cosθ
    Last edited by a moderator: May 5, 2017
  7. Nov 23, 2011 #6


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    Perhaps it makes more sense to you if you write it as
    A &= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{2\cos\theta}^1 r \,dr\,d\theta \\
    &= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left[\int_0^1 r \,dr - \int_0^{2\cos\theta}r \,dr\right]d\theta \\
    &= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^1 r \,dr\,d\theta - \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^{2\cos \theta}r \,dr\,d\theta
  8. Nov 23, 2011 #7


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    [itex]2cos(\theta)= 1[/itex] at [itex]\theta= \pi/3[/itex] so that integral is just
    [tex]\int_{\pi/3}^{\pi/2} 1- 2cos(\theta) d\theta[/tex]

    That's assuming you want to stop as the positive y axis as your picture shows.
  9. Nov 23, 2011 #8
    Woaw, I feel like such an idiot for not seeing that in the first place.

    I also figured out that if I had drawn a ray (past the line pi/3, and close to pi/2), I could've set up that double integral without all that mess I created

    Thanks y'all
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