What is the surface area of the solid created by revolving #f(x) = 2x^2-x+5 , x in [1,2]# around the x axis?
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"What is the equation of a neutralization reaction between sodium hydroxide and hydrochloric acid?"
The surface area of the solid created by revolving a positive function, #f(x)#, for #x in [a,b]# around the x-axis is always
#\int_a^b 2pif(x)sqrt(1+f'(x)^2)dx#
The derivation of this is fairly simple. It involves the surface area of a frustrum which is #2pirl#. I would highly recommend looking up the derivation, as it make this formula make a lot more sense. This link is pretty good, http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
Anyways, in the case of this problem, the first thing to notice is that #f(x)# is negative over the interval indicated.
#SA = \int_1^2 2pif(x)sqrt(1+f'(x)^2)dx#
#= \int_1^2 2pi(2x^2-2x+5)sqrt(1+(4x-2)^2)dx#
at this point I would use a calculator to find the value of this integral which would end up being the answer.
#SA \approx 181.265#
