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Integral of polar curve

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    The graph of the polar curve r=2-cosΘ for 0≤θ≤2pi is shown in the figure. (attached)

    a) write an integral expression for the area of the region inside the curve

    b) write expressions for dx/dΘ and dy/dΘ in terms of Θ

    c) find dy/dx as a function of Θ

    d) write an equation in terms of x and y for the line tangent to the graph of the polar curve at the point where Θ=pi/2. show the work that lead to your answer.

    2. Relevant equations

    to find are under a curve you find integral under the curve

    3. The attempt at a solution
    the solutions are:
    a) 2*(1/2) ∫(0 to pi) (2-cosΘ)2

    b) dx/dΘ = -2sinΘ + 2(cosΘ)(sinΘ)

    dy/dΘ = 2cosΘ-cos2Θ + sin2Θ

    c) dy/dx = (2cosΘ-cos2Θ +sin2θ) / (-2sinθ + 2(cosθ)(sinθ))


    So, i tried part A but could not figure out the area of the region because if you did a regular integral of ∫(2-cosθ), according to the graph the region above and below the x-axis would cancel out, resulting in an answer of 0.
    i understand that you might have to change the equation into a semi-circle by multiplying 1/2 then later have to undo that by multiplying by 2.
    i dont understand why they took (2-cosθ) and squared it...? also why the limits of integration are 0 to pi, not 0 to 2pi???
     

    Attached Files:

  2. jcsd
  3. Apr 16, 2014 #2

    SammyS

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    You have answered your own final question.

    Do you really understand as the above statement claims? It almost answers the following.

    Except that taking the limits of integration from 0 to π is what gives a semi-circle rather than multiplying by 1/2 .
     
  4. Apr 16, 2014 #3

    BiGyElLoWhAt

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    it looks like to me they (for a)they just simplified.

    What you want is the area here: r=2-cosΘ for 0≤θ≤2pi

    What you start out with is: [itex]∫_{E}dA[/itex] Then you begin to make substitutions. and you're right, you can't do a standard integral, so you have to break your region (E) up into to parts, from 0 to pi, and from pi to 2 pi, so now we have [itex]2∫_{0}^{\pi}dA[/itex], but what is dA? In polar it's rdrdΘ.

    Now what we have is [itex]2∫_{0}^{\pi}∫_{0}^{2-cos(\theta
    )}rdrd\theta[/itex]
    This gives you [itex]2∫_{0}^{pi}\frac{r^2}{2}d\theta|_{r=0}^{r=2-cos(\theta)}[/itex]

    evaluate it and theres a)
     
  5. Apr 16, 2014 #4

    haruspex

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    I don't think that's what they did. Consider a triangle subtended at the origin by the segment of the curve starting at (r, θ) and length rdθ. What is its area?
     
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