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

[JJHK]

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 \frac{1}{2}Δφ*(r+\frac{Δr}{2})² - \frac{1}{2}Δφ*(r-\frac{Δr}{2})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

 

[JJHK]

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

\frac{1}{2}Δφ*(r+\frac{Δr}{2})2 - \frac{1}{2}Δφ*(r-\frac{Δr}{2})2 = r*Δr*Δφ