A torus with major radius, ##R##, and minor radius, ##r##, has a total surface area given by ##4\pi^2 Rr##. If one slices the torus on its midline (i.e. at a line on a poloidal angle of ##-\pi/2## and ##\pi/2##), I was told the inner half of the torus has a smaller surface area than the outer half of the torus.(adsbygoogle = window.adsbygoogle || []).push({});

Although I am having some trouble visualizing this. If we were to simply cut the torus at one point and stretch it out, it would simply look like a cylinder, where the inner and outer halves have equal surface areas. But in this case, the curvature supposedly changes that. Any insight on why exactly the surface area changes for the inner and outer halves, and how one can quantify the change in surface area due to curvature in this case, would be greatly appreciated.

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# I Surface area of inner and outer halves of torus

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