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Suface Area of a Section of a Sphere

  1. Feb 23, 2013 #1
    If a V-shaped cut were to be made through an entire sphere, radius DR (Dome Radius) with a Cutting Angle (CA- angle of the cut), what would be the surface area of the sphere section that would be cut? This is assuming that CA is less than 180 degrees. Also the Cut Depth (CD- depth of the V-shaped cut from the top of the sphere to the bottom of the "V") is to be less than the Dome Radius (DR). So, the Area of the cut-out section (A) would be in terms of DR, CA, and CD. Any help in figuring out this problem would be greatly appreciated!

    If CD were to be equal to DR, this problem would not be too difficult, because the shape of the cut-out section would be a "loon," but since CD must be less than DR, it is quite tougher. I don't think that spherical triangles could be used either because since CD is not equal to DR, the circles that would cut out the section would not be Great Circles (not straight lines, but curved lines I believe) so this adds a great deal of difficulty to the problem.
     
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  3. Feb 23, 2013 #2

    SteamKing

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    A loon is a bird. The cut made in the surface of a sphere is a lune.
     
  4. Feb 24, 2013 #3

    HallsofIvy

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    The total surface area of a hemisphere, of radius [itex]\rho[/itex] is [itex]2\pi\rho^2[/itex]. If the lune has angle [itex]\theta[/itex], at the center of the sphere, it cuts off a section of the surface [itex](\theta)/\pi)(2\pi\rho^2= 2\theta\rho^2[/itex].

    The only way "cut depth" comes into this is that you have the angle measured at depth r rather than [itex]\rho[/itex]. We can calculate the difference by imagining two isosceles triangles, one with side length [itex]\rho[/itex], the other with side length r, both having the same base. Dropping a perpendicular from the vertex to the base in each triangle, we have, for one, [itex]sin(\theta/2)= (b/2)/\rho[/itex] and for the other, [itex]sin(\theta'/2)= (b/2)/r[/itex]. So [itex]sin(\theta/2)/sin(\theta/2)= (b/2\rho)/(b/2r)= r/\rho[/itex]. From that [itex]sin(\theta/2)= (\rho/r)sin(\theta'/2)[/itex] so that [itex]\theta= arcsin((2\rho/ r)sin(\theta'/2))[/itex].

    Putting that into the original formula, the area cut off the surface of a sphere of radius [itex]\rho[/itex], at angle [itex]\theta[/itex] going to r "cut depth" is
    [tex]2arcsin((2\rho/r)sin(\theta/2)\rho^2[/tex].
     
  5. Feb 25, 2013 #4
    Thank you so much for your reply HallsofIvy. I had a feeling that a formula could be derived using the formula for a loon, I just could not quite find a way to relate them, but your method using isosceles triangles definitely makes a lot of sense to me.

    What if there were another dimension, RR (Root Radius), that was the radius of the bottom of the "V-shaped" cut that was slicing through the sphere? I believe that this would make the area of the desired surface smaller, but how would you go about accounting for the change in area in the form of an equation? Is there a way that this formula, 2arcsin((2ρ/r)sin(θ/2)ρ2, could be changed to account for RR? The area would now be in terms of RR, CA, CD, and DR.
     
  6. Feb 26, 2013 #5
    I'm sure that there is a way, but I have no idea how to do it.
     
  7. Feb 27, 2013 #6
    Can someone get me started on the right path?
     
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