If we rotate the region below ( the picture shown below ) about the $x$-axis, we obtain a coin, and I want to calculate the surface area of this coin, using
$$SA=2\pi \int_{a}^{b} f(x)\sqrt{1+f'(x)^2}$$
you can see that there are two semicircles, the upper semicircle is given by $y=2+\sqrt{1-x^{2}}$, the one at the bottom is given by $y=-2-\sqrt{1-x^{2}}$, what I don't understand is, in the solution manual, only the upper circle was taken into consideration.
I am confused, because in a different problem, I had a torus obtained by rotating a circle around the the $x$-axis, and both the upper half and the bottom half of circle was taken into consideration when doing the calculations… I don't understand however why in the case of a coin this is different? Why are we only taking the upper semicircle? What am I missing here exactly? I am not asking for a solution, but rather an explanation as to why.
Best Answer
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2\pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.