What is the surface area of the solid created by revolving #f(x) = x^2-2x+15 , x in [2,3]# around the x axis?

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Answer 1 · 2018-05-14T23:45:45

#S_A=2piint_2^3(x^2-2x+15)*sqrt[1+(2x-2)^2]*dx=52.2146#

To determine the surface area created by revolving y around the
x-axis

#S_A=2piint_a^by*sqrt[1+(y')^2]*dx#

#y=x^2-2x+15#

#y'=2x-2#

#S_A=2piint_2^3(x^2-2x+15)*sqrt[1+(2x-2)^2]*dx#

#=2piint_2^3(x^2-2x+15)*sqrt[1+(4x^2-8x+4)]*dx#

#=2piint_2^3(x^2-2x+15)*sqrt[4x^2-8x+5]*dx#

#=[(223*arsinh((8*x-8)/4))/64+(x*(4*x^2-8*x+5)^(3/2))/16-(4*x^2-8*x+5)^(3/2)/16+(223*x*sqrt(4*x^2-8*x+5))/32-(223*sqrt(4*x^2-8*x+5))/32]_2^3#

#=[(223arsinh((8x-8)/4)+(4x-4)(4x^2-8x+5)^(3/2)+(446x-446)sqrt(4x^2-8x+5))/64]_2^3#

#=[(223*arsinh(4)+1028*sqrt(17))/64-(223*arsinh(2)+466*sqrt(5))/64]#

#[(223*arsinh(4)-223*arsinh(2)+1028*sqrt(17)-466*sqrt(5))/64]#

#=52.2146#