Given: #f(x) = (2x-9)^2 , x in [2,3]#
Required: the surface area obtained by revolving about the x-axis
Solution Strategy: Use the Definition of Surface Area of Revolution
Definition: #S= int" "(2pi*x) ds# where #ds=sqrt(1+(dy/dx )^2) dx# thus,
#S = 2piint_2^3 x sqrt(1+(dy/dx )^2) dx # ===========> (1)
Now
#y=(2x-9)^2# and #dy/dx = 4(2x-9); (dy/dx )^2=16(4x^2-36x+81 )#
Insert in (1)
#S = 2piint_2^3 x sqrt(1+(16(4x^2-36x+81))) dx # Integrate
Write #x# as #=> 16*1/128(8x-36)+9/2# and split the integral to:
#S= pi[int(2x-9)sqrt(1+(16(4x^2-36x+81)))+ 9intsqrt(1+(16(4x^2-36x+81)))]= pi[color(brown)(S_1)+color(purple)(S_2)]#
Solving for #color(brown)(S_1= int(2x-9)sqrt(1+(16(4x^2-36x+81)))dx#
#=color(brown)(int(8x-36)/4sqrt(1+(16(4x^2-36x+81)))dx#
let #=>u=1+16(4x^2-36x+81); du=16(8x-36)dx #
#:. color(brown)(S_1 = 1/64 intsqrt(u) du = 1/64 (2/3u^(3/2))=u^(3/2)/96# undo substitution:
#color(brown)(S_1=(1+16(2x-9)^2)^(3/2)/96 #
Solving for #color(purple)(S_2=intsqrt(1+(16(4x^2-36x+81)))#
Complete the square and factor:
#= intsqrt(16(2x-9)^2+1) dx#
let #=> u= 2x-9; du = 2dx#
#color(purple)(S_2=1/2intsqrt(16u^2+1) du#
Further let #u=tan(v)/4 -> v=arctan(4u); du = (sec^2(v) (dv))/4#
#color(purple)(S_2=1/2[intsqrt(16(1/4tanv)^2+1)*(sec^2(v) (dv))/4]#
#=1/2 1/4[intsqrt((tanv)^2+1) (sec^2(v) (dv))]#
#=1/2 1/4[int (secv) (sec^2(v) (dv))]=1/2 1/4 int sec^3(v) dv#
Apply Integral reduction:
#=(sec^(n-2) (v)*tanv)/(n-1) +(n-2)/(n-1) int sec^(n-2)(v) dv# for #n=3#
#=(sec (v)*tan(v))/(2) +(1)/(2) int sec(v) dv = S_(2a) + S_(2b)#
#color(red)(S_(2b) =(1)/(2) int sec(v) dv # from table of integrals
#color(red)(S_(2b) = 1/2ln|secv + tanv|#
#color(purple)(S_2=1/2 1/4 [(sec(v)*tan(v))/2+1/2ln|secv + tanv |]#
Now undoing the substitutions #v=arctan(4u)#:
#color(purple)(S_2=1/2 1/4[(1+16u^2)^2*4u/2+1/2ln|sec(1+16u^2)^2 + tan4u|]#
#color(purple)(S_2=1/2 1/4[(1+16(2x-9)^2)^2*4(2x-9)/2+1/2ln|sec(1+16(2x-9)^2)^2 + tan4(2x-9)|]#
* Putting it all together *
#pi(color(brown)(S_1)+color(purple)(S_1))= pi{color(brown)((1+16(2x-9)^2)^(3/2)/96)+ color(purple)(1/2 1/4[(1+16(2x-9)^2)^2*4(2x-9)/2+1/2ln|sec(1+16(2x-9)^2)^2 + tan4(2x-9)|]}}~~247.65#
