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Area of a vertically sliced circle

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A circle of radius 7 inches is sliced vertically, parallel to the y-axis, into three pieces. Each piece has an equal area. What is the width, x-axis, of each piece?

    2. Relevant equations

    f(x)= +/- sqrt((r^2)-(x^2)) where "r" is the radius.

    3. The attempt at a solution

    My sign for the integral is $

    Using the equation for a circle above I took advantage of the obvious symmetry about the x-axis to isolate the part of the circle above the x-axis relevant to the given radius.

    g(x)= sqrt((7^2)-(x^2)).

    My attempt was to first define the area as an integral from the origin until x=7. (I chose x=7 because of the symmetry about the y-axis.)

    I(x)= 0->7 $sqrt((7^2)-(x^2))dx

    Here is where I hit a wall. The prof tried explaining to the class about Pythagoras and I thought I understood but apparently I didn't. Any assistance or pushes in the right direction would be greatly appreciated.
     
  2. jcsd
  3. May 30, 2010 #2

    HallsofIvy

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    First the area of the circle is [itex]\pi(7^2)= 49\pi[/itex]. If it is divided into 3 pieces of equal area, each piece must have area [itex]49\pi/3[/itex].

    Taking X to be the x coordinate of the left most slice, its area is given by
    [tex]2\int_{-7}^X \sqrt{49- x^2} dx[/tex]
    and that must be equal to [itex]49\pi/3[/itex] .

    Go ahead and do the integral (the substitution [itex]u= 7sin(t)[/itex] will work), set it equal to [itex]49\pi/3[/itex], and solve for X. The symmetry tells you that the other slice is at 7+ X.
     
  4. May 30, 2010 #3
    Do you even need calculus? If you consider just the part that is above the y-axis, then each slice has to be equal to [itex]49\pi/6[/itex]. and, the two end pieces are quarter ellipses.

    so, if b is the x coordinate of the right slice, then

    49pi/6 = (pi/4)(7-b)(49-b^2)^(1/2)
     
    Last edited: May 30, 2010
  5. May 30, 2010 #4
    I do need to use calculus. I need to use trigonometric substitutions and integration. The method provided by HallsofIvy is the one i must use.

    I followed through with the integral but now I am stuck at trying to solve for X.

    What I have is:

    sin^-1(x/7)+sin(2(sin^-1(x/7))) = pi/6

    How do i break this down?
     
  6. May 30, 2010 #5

    tiny-tim

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    No, a cap of a circle is a sector minus a triangle, and you can easily find the area of each. :wink:
     
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