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Arc area of a sphere? (a piece of r^2*sinθ*ΔrΔθΔφ)

  1. Nov 26, 2012 #1
    Arc area of a sphere??? (a piece of r^2*sinθ*ΔrΔθΔφ)

    Hello, this is not a homework question. I'm trying to self-derive the divergence formula in spherical coordinates, and I'm doing this by taking a small arc-volume about the point (r,θ,φ), where r is the radial distance from the origin, θ is the polar angle from the positive z-axis, and φ is the azimuthal angle from the positive x-axis.

    Consider a small slab of the sphere about (r,θ,φ). The volume of this small slab is: Δr*(r*Δθ)*(r*sinθ*Δφ). Consider the side of the slab that has the normal vector(to the surface) = +r(hat). How would I find the area of this surface? It is curved in two angles; I have no idea how to compute the area. If someone can show me how to compute this area, it would be greatly appreciated.

    I have another area-portion that I can't figure out. Consider the surface with the normal vector = +θ(hat). I initially thought the area to this surface was [itex]\frac{1}{2}[/itex]Δφ*(r+[itex]\frac{Δr}{2}[/itex])2 - [itex]\frac{1}{2}[/itex]Δφ*(r-[itex]\frac{Δr}{2}[/itex])2 = r*Δr*Δφ

    But I realized that i forgot the fact that this area will also be dependent upon θ. So can anyone help me out with these two surface areas? Thanks
     
  2. jcsd
  3. Nov 26, 2012 #2
    Re: Arc area of a sphere??? (a piece of r^2*sinθ*ΔrΔθΔφ)

    I edited the post, and now the LaTeX stuff isn't showing up :/. Here it is again, hopefully unbroken this time

    [itex]\frac{1}{2}[/itex]Δφ*(r+[itex]\frac{Δr}{2}[/itex])2 - [itex]\frac{1}{2}[/itex]Δφ*(r-[itex]\frac{Δr}{2}[/itex])2 = r*Δr*Δφ
     
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