Area of surface of revolution

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  • #1
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When I learned about volumes of solids of revolution, I never really memorized any formulas for specific cases per se. I used two expressions for area, either ##A = \pi (R^2 - r^2)## and ##A = 2\pi r h##.
Those expressions worked for rotations about any horizontal/vertical axis (not necessarily ##x## or ##y##) and for all functions.
Now I'm learning about areas of surfaces of revolution, but all the "formulas" or integrals online seem to be solely for rotations about one of the axes (##x## or ##y##). Is it possible to use a formula that is less specific, for instance, ##2\pi r## for the circumference? This would work for all rotations and it does not obscure the essence of deriving formulas for areas of surfaces of revolution. Also, in the "washer" (for lack of a better word) case, aren't we supposed to add the areas, unlike the volume case where we had to "subtract the volumes"? It seems very obvious to be, but I wanted to make sure since there is no reference to this particular case online.
Thank you in advance.
 

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  • #2
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You can always replace x by y.
Usually you have a function y(x), and rotating this function around different axes makes the problem different. Sometimes you can find a function x(y).
Also, in the "washer" (for lack of a better word) case, aren't we supposed to add the areas, unlike the volume case where we had to "subtract the volumes"?
I don't understand that question.
 
  • #3
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Also, in the "washer" (for lack of a better word) case, aren't we supposed to add the areas, unlike the volume case where we had to "subtract the volumes"?
mfb said:
I don't understand that question.
I don't, either. If you are using washers to calculate the volume of a solid of revolution, with a curve that is revolved around the x-axis, the volume of a typical volume element is ##\Delta V = \pi (R^2 - r^2)\Delta x##. For a given volume element you're subtracting the volume in the hole from the total volume. In the integral, you're essentially adding all of the volume increments, so perhaps this is what you meant.
 

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