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Homework Help: Circle to sphere integration

  1. Aug 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Derive the formula for surface area of a sphere using integration of circles

    2. Relevant equations

    Need to get : S = 4πr2

    3. The attempt at a solution

    Consider a sphere of radius r centred on the origin of a 3D space. Let y be an axis thru the origin. The sphere can be sliced into a row of circles with the y axis at their centres. Consider one of these circles. The circumference c of the circle depends on it centre's distance from the origin such that
    c = 2πr cos θ , where θ is the angle from the origin between the (x,z) plane to any point on the circle's circumference. r cosθ is the radius of the circle.

    For -r to r, y= r sinθ
    The Integration the circles circumferences along y from -r to r is
    2πr cos θ ∫dy
    Since y = r sinθ, then dy / dθ = r cosθ , so dy = r cos θ dθ
    so the integration formula is
    2πr2 ∫ (cos θ)2

    I don't get the right formula from that integral.
    There's something wrong with my setting up the integral. I can't see what though. Please give me a hint or two.
    Last edited: Aug 30, 2013
  2. jcsd
  3. Aug 31, 2013 #2


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    The problem is in this last step. You're using ##dA = 2\pi r \cos \theta\,dy##, but that's only good when ##\theta \approx 0##, where a strip of the sphere looks like a piece of a cylinder. As you near the top and bottom of the sphere, the strip becomes flatter and flatter. Its area is no longer well approximated by that of cylinder of radius r cos θ and height dy.

  4. Aug 31, 2013 #3
    thks. Looks like I was hoping the circles would sort themselves out. Now I think I need to start by approximating a slice of the sphere as a frustum. The lower limit ( no min element ) of frustum of height h is a circle, and an array of frustums can approximate a sphere.
  5. Aug 31, 2013 #4


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    If I understand your attempt correctly, you're basically summing the surface area of open cylindrical elements with radius ##r\cos\theta## and an infinitesimal height.

    Unfortunately, the element you used for the height is not correct. I suggest you use the arc length along the sphere given by ##\mathrm{d s} = r\mathrm{d \theta}## as a better approximation for the height of the cylindrical element.
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