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

Question

1243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T13:22:10

#A=-sqrt2 pi#

#A=2piint_2^3 f(x)sqrt(1+(d/(d x) f(x))^2 )d x#

#d/(d x)f(x)=-1#

#(d/(d x)f(x))^2=1#

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

#A=2 pi int _2^3 (2-x)sqrt2* d x#

#A=2sqrt2 pi int_2^3 (2-x)d x#

#A=2 sqrt2 pi|2x-x^2/2|_2^3#

#A=2 sqrt2 pi[(2*3-3^2/2)-(2*2-2^2/2)]#

#A=2 sqrt2 pi[(6-9/2)-(4-2)]#

#A=2sqrt2 pi[3/2-2]#

#A=2sqrt2 pi(-1/2)#

#A=-sqrt2 pi#