TITLE: Surface area of a solid of revolution: Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ work?
QUESTION [5 upvotes]: Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ yield the correct answer when calculating the surface area of a solid of revolution?

REPLY [2 votes]: For a thin slice surface area of a cone = $ 2 \pi \,f(x) \,ds $ which you should integrate. 
If it is a flat disc of height tending to zero there would be a big error without multiplying by $ \sec \phi $ due to slope. 
$ \int_{b}^{a} 2\pi \,f(x) \,dx $ yields correct answer only when calculating the surface area of a cylinder, zero slope, constant radius $ a = f(x) $.
No area is represented by $ \int_{b}^{a} 2\pi \,f(x) \,dx,$ for variable  $ f(x). $