What is the surface area of the solid created by revolving #f(x) = 3x, x in [2,5]# around the x axis?

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Answer 1 · 2016-05-24T10:17:17

#A=63sqrt10 pi#

#A=2pi int_2^5 f(x)*sqrt(1+(d f(x))/(d x))^2*d x#

#f(x)=3x#

#d/(d x) f(x)=3#

#A=2 pi int_2^5 3x*sqrt(1+3^2)* d x#

#A=2 pi int_2^5 3xsqrt 10 *d x#

#A=6sqrt10 pi int_2^5 x *d x#

#A=6 sqrt10[1/2 x^2]_2^5#

#A=6sqrt10 pi[(1/2*5^2-1/2*2^2)]#

#A=6sqrt 10 pi[25/2-4/2]#

#A=6sqrt 10 pi[21/2]#

#A=63sqrt10 pi#