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?

1 Answer
May 14, 2018

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

Explanation:

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