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Double integrals over general regions.

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

    The domain D is the intersection of two disks x^2 +y^2 = 1 and x^2 + (y-1)^2 =1

    use polar coordinates to find the double integral ∫∫(x)dA

    2. Relevant equations

    x = rcosθ y = rsinθ r^2 = x^2 + y^2

    3. The attempt at a solution

    I have drawn the circles. also i set the two circle equations to equal each other.

    x^2 +y^2 = x^2 + (y-1)^2

    which yielded points of intersection.. [-√(3/4), (1/2)] and [√(3/4),(1/2)]

    however i am unsure how to find my intervals of θ and I'm not sure of my intervals of r. I think it may be 0≤r≤1.


    any help would be appreciated, even just to put me on the right track.. I have been looking over this problem for 45 minutes now
     
  2. jcsd
  3. Nov 1, 2011 #2
    im looking at it now. could i use simple geometry to find the intervals of theta?

    Pythagorean theorem to get the angle on each side with respect to the x axis?
     
  4. Nov 1, 2011 #3

    Deveno

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    it should be obvious that from symmetry, we can just let θ run from 0 to π/2, and double the result.

    also, after θ = π/6, r is just a constant function r(θ) = 1.

    so the real challenge is capturing that thin sliver under the line y = (1/√3)x. the lower limit of r will be 0, of course, but the upper limit will be the circle centered at (0,1).

    use x = rcos(θ) and y = rsin(θ) to derive a formula for r in terms of θ for the upper circle (it should be a fairly nice function of θ), which will give you your upper limit for r.

    then just integrate the polar form (times r dr dθ, as the polar form of dA) of your function from θ = 0 to θ = π/6, and r = 0 to (upper limit from above). after that, you just have a pie slice (nifty pun, eh?) of π/3 radians, to integrate your function over (and then double everything).
     
    Last edited: Nov 1, 2011
  5. Nov 1, 2011 #4

    HallsofIvy

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    I agree with Deveno. Do this integral in three parts. For [itex]\theta[/itex] from 0 to [itex]\pi/6[/itex], your boundary is the upper circle, [itex]x^2+ (y-1)^2= 1[/itex] or [itex]x^2+ y^2- 2y+ 1= r^2- 2r sin(\theta)+ 1= 1[/itex] which reduces to [itex]r^2= 2rsin(\theta)[/itex]. Since r is not 0 anywhere except (0, 0), we must have [itex]r= 2sin(\theta)[/itex]. That is, for [itex]\theta[/itex] from 0 to [itex]\pi/2[/itex], integrate with r from 0 to [itex]2sin(\theta)[/itex].

    For [itex]\theta[/itex] from 0 to [itex]\pi- \pi/6= 5\pi/6[/itex], the boundary is the circle [itex]x^2+ y^2= r^2= 1[/itex] so the "r" integral is from 0 to 1. Finally, for [itex]\theta[/itex] from [itex]5\pi/6[/itex] to [itex]\pi[/itex], use r from 0 to [itex]2sin(\theta)[/itex] again.
     
  6. Nov 1, 2011 #5
    I'm adding all 3 integrals up to compute my final area?
     
  7. Nov 1, 2011 #6
    i see it more of as, computing 0 to pi/2 and having my upperbound be my bottom circle.
    and then computing 0 to pi/6 and having my upperbound be my top circle.

    subtracting the second from the first and then doubling it.
     
  8. Nov 1, 2011 #7

    SammyS

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    What do you mean by the "upper circle"? ... lower circle?
     
  9. Nov 1, 2011 #8

    Deveno

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    by "upper circle" i believe he means the one centered at (0,1).

    however, if he wants to compute an integral over an area to be subtracted from the integral over the (lower) semi-circle, the "lower bound" for r, needs to be
    r = 2 sin(θ), because the area to be subtracted is the part "outside" the upper circle.

    see here (and ignore the funky scaling these are circles, not ellipses): http://www.wolframalpha.com/input/?i=plot{(x^2+y^2=1),(x^2+(y-1)^2=1),(y=(1/√3)x)}
     
  10. Nov 1, 2011 #9
    correct, i want to compute the area of the oval type shape in the middle.

    i took the integral with 0 ≤ r ≤ 2sin(θ) 0≤ θ ≤ ∏/2

    ∫∫ (r2cosθ) dr dθ

    and got 2/3


    then i took the integral
    (5/6)∏ ≤θ ≤ ∏ 0 ≤ r ≤ 2sinθ

    ∫∫(r2cosθ) dr dθ

    and got -1/24



    i took the third integral.
    0 ≤ θ ≤ (5/6)∏ 0≤r≤1
    ∫∫ (r2cosθ) dr dθ

    and got -1/6.



    im stumped after that, i feel as though i would add them all up.

    doing so i got my total area to be 11/24
     
  11. Nov 1, 2011 #10

    Deveno

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    i think you are taking the integral over the wrong regions.

    above the line y = x/√3, the region in question is entirely within the unit circle. this line corresponds to θ= π/6 in the upper two quadrants.

    so we first split our integral:

    [tex]\int \int_D x dA = \int \int_D r^2cos(\theta) dr d\theta[/tex]

    [tex]=2\left(\int_0^{\pi/6}\int_0^{2sin(\theta)} r^2cos(\theta) drd\theta + \int_{\pi/6}^{\pi/2}\int_0^1 r^2cos(\theta) drd\theta\right)[/tex]

    that is, we are splitting half of D into:

    [tex]D_1 = \{(x,y) : x^2+y^2 \leq 1,\ y > \frac{x}{\sqrt{3}},\ x \geq 0\}[/tex] and

    [tex]D_2 = \{(x,y) : x^2 + (y-1)^2 \leq 1,\ y \leq \frac{x}{\sqrt{3}}\}[/tex]

    these are disjoint, so the integral over half of D is the integral over D1 + the integral over D2.

    i don't believe your answer is correct.
     
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