Which of the following answers actually calculates the surface area of the Steinmetz solid?

Question

To calculate the value of the surface area of the Steinmetz solid generated by two cylinders of radius 1, in Surface Area of Two Cylinders Calculus 3 the answer is proposed to be 8, while in Intersection of two cylinders the answer is said to be 16. Since both answers are already accepted, naturally one of the two must be wrong, but I can't figure out which one and why. After a little searching it seems to me that the correct answer really is 16, but I can't figure out why the answer that gives 8 is wrong. Could you please help me find the error in his solution?

Answer 1

Let me repeat whole task here: so, we have two cylinders $$y^2 + z^2 = 1 \\ y^2 + x^2 = 1$$

and we would like to calculate surface area of created by them bounded figure, known as Steinmetz solid (bicylinder).

One way is consider projection on plane $Oxy$ and count integral as it is partly done on Surface Area of Two Cylinders Calculus 3, for $z=\pm\sqrt{1-y^2}$ i.e. $$2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dxdy = 8$$ But, this is not enough, because by this integral we calculate only surface area for floor and ceiling with respect to $Oxy$, but forget walls for our figure. To obtain surface area for walls we can consider them, for example, as floor and ceiling, given by $x=\pm\sqrt{1-y^2}$, with respect to $Oyz$ plane. Now we obtain exactly same as above integral, but with respect to $y,z$ variables

$$2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dzdy = 8$$ So, at last, we have $16$ for whole surface.

This, of course, can be done by considering projection with respect to $Oxz$ plane, which way is fulfilled on Intersection of two cylinders and have same answer $16$.