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Homework Help: Integration of part of a circle

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Draw a circle centered on the origin with radius 2. Draw the vertical line x = 1. Find the area bounded on the left by x=1 and on the right by the circle.


    2. Relevant equations
    Area of a circle: x[tex]^{2}[/tex] + y[tex]^{2}[/tex] = r[tex]^{2}[/tex]. In this race, r = 2

    Polar coordinates:

    x = p cos([tex]\theta[/tex])
    y = p sin([tex]\theta[/tex])


    3. The attempt at a solution

    At first I tried to solve this in rectangular coordinates, using the equation y = [tex]\sqrt{4 - x^{2}}[/tex] and integrating it from 1 to 2. However the antiderivative for that is messy as heck, and this is supposed to be a simple problem. I'm trying to do it in polar coords, but I'm at a loss as how to bound the integral in that case.
     
  2. jcsd
  3. Aug 4, 2010 #2

    Mark44

    Staff: Mentor

    In the region of concern, r ranges from r = 1/cos(theta) = sec(theta) to r = 2. Theta ranges from theta = -arctan(sqrt(3)) to theta = +arctan(sqrt(3)), but you can let theta run from 0 to arctan(sqrt(3)) and double the value, using the symmetry of the circle.

    The integral in rectangular coordinates isn't that bad if you know trig substitution.
     
  4. Aug 7, 2010 #3
    I was hoping to avoid having to use arctan if at all possible...I'm having more luck with the small triangles method in polar coordinates.

    You take a triangle with height along the x axis, "r". The angle from the x-axis is assumed to be very small, d[tex]\theta[/tex]. Because this is very small, the base of the triangle is approximated as the arc length, given in this case by r * d[tex]\theta[/tex]. The area of a triangle is 1/2 * base * height, or in this case 1/2*r[tex]^{2}[/tex]d[tex]\theta[/tex]. Integrating this from 0 to 2pi gives you your answer.
     
  5. Aug 8, 2010 #4

    Mark44

    Staff: Mentor

    Arctan doesn't enter into the picture at all. If you stay with rectangular coordinates, the substitution is sin(theta) = x/4
    You lost me here. How can the height be along the x axis? And what does this have to do with r?
    Why would you integrate from 0 to 2pi? In polar coordinates, the line x = 1 intersects the upper part of the circle at (2, pi/3).
     
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