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I am given the following question:

Consider the surface of a sphere (that is the boundary of the sphere) of radius $R>0$ in $\mathbb{R}^3$ and two parallel planes which are $R$ units away from each other and at the same distance of the center of the given sphere. What is the area of the surface of the sphere in-between the two planes?

I'm having some trouble describing the above set analytically.

Let $\mathcal{C}$ be the sphere and $\Pi_1, \Pi_2$ the two planes. Let $S$ be the set I'm looking for.

I know that for some $A, B, C, D, x_0, y_0, z_0\in \mathbb{R}$ $$\mathcal{C}=\{(x,y,z)\in \mathbb{R}^3:(x-x_0)+(y-y_0)+(z-z_0)=R^2\}$$ $$\Pi_1=\{(x,y,z)\in \mathbb{R}^3:A(x-x_0)+B(y-y_0)+C(z-z_0)=D+R \wedge \ldots\}$$ $$\Pi_2=\{(x,y,z)\in \mathbb{R}^3:A(x-x_0)+B(y-y_0)+C(z-z_0)=D \wedge \ldots\}$$ $$S=\{(x,y,z)\in \mathbb{R}^3: (x,y,z)\in \mathcal{C} \wedge D\leq A(x-x_0)+B(y-y_0)+C(z-z_0)\leq D+R \wedge \ldots \}$$

I don't know how to describe the condition 'the planes are at the same distance of the center of the given sphere' nor can I describe $S$.

I think I just need help with finding $S$. I can probably manage the area myself.

This isn't homework, this extra-work on my part and since I can't come back to this computer before Monday I won't be able to interact with the answerers (but I'll be able to check the answers). So I ask of you that if you have to choose between a hint or just giving me $S$, please give me $S$ so I can continue studying. Of course a full answer will be greatly appreciated.

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possible duplicate of Calculating the surface area of sphere above a plane You need $4\pi$ minus twice this. –  Ross Millikan Mar 16 '13 at 13:30
@RossMillikan I'm not sure about how similar questions must be to be called duplicates, but I find this one much harder than the one you suggest as a duplicate. –  Git Gud Mar 16 '13 at 13:36
@GitGud: If you have the area of a spherical cap, you can just subtract twice that from the area of the sphere as I commented. If the approach needs to be as above, I agree it is different and harder. –  Ross Millikan Mar 16 '13 at 13:52

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