What are the Areas and Volumes of Inner-Half and Outer-Half of a Donut Shape (Toroid)?
Andrew Mclaughlin 
A tube of radius r is rolled (its ends joined) so as to form a Donut. When looked from top, you will see the Donut having inner radius R and outer radius R+2r
Still looking from the top, if now you pass a thin knife into the Donut along R+r completely, all around, then you will have sliced the Donut into 2 parts -- the inner half and the outer half.
Question is: what are the (a) surface areas and (b) the Volumes of the Inner-half and the Outer-half of the Donut?
(Note: If your first instinct is to reply saying that they are both equal then you haven't understood the description fully; please read it again. The inner half obviously has smaller Surface Area and Volume.)
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$\begingroup$Pappus' first centroid theorem says that the surface area is the arc length of the generating curve times the distance traveled by the centroid. The second centroid theorem says that the volume is the cross-sectional area of the generating slice times the distance traveled by the centroid. The centroid of a half-circle lies $\frac{2r}{\pi}$ past the cut and of a half-disk lies $\frac{4r}{3\pi}$ past the cut.
By the first theorem, the surface area of the outer-half would be $2\pi(R + r + \frac{2r}{\pi})(\pi r)$ and the inner-half would be $2\pi (R + r - \frac{2r}{\pi}) (\pi r)$. This doesn't count the surface area of the cut, but that's just a cylinder if you want to add it.
By the second theorem, the volume of the outer-half would be $2\pi(R+r + \frac{4r}{3\pi})(\frac{\pi r^2}{2})$ and the inner-half would be $2\pi(R + r - \frac{4r}{3\pi})(\frac{\pi r^2}{2})$
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