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A degenerate cuboid is a rectangle. What about its surface area? Is it twice the rectangle's area or only once the rectangle's area? What is the convention here?

The latter is more natural but the first makes the function, which assigns a surface area to side lengths $a,b,c$ continuous.

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    $\begingroup$ What's natural about the latter? To me, it is not natural at all. You have two surfaces here, only they are really close, so there is almost no space between them. Well, not even "almost". $\endgroup$ Feb 6 '19 at 7:43
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Twice the rectangle's area

Let's say you want to paint a cuboid of size $100 \times 80 \times 0.0001$ and you want to calculate how much paint you need. To simplify the calculation you use a cuboid of size $100 \times 80 \times 0$ instead. For this to work the 'twice the rectangle's area' convention needs to be used. You and Ivan also gave good arguments.

To solve the paradox you've come across, we say that a degenerate cuboid is technically not a rectangle.

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