Given: f(x) = (2x-9)^2 , x in [2,3]f(x)=(2x−9)2,x∈[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) dsS=∫ (2π⋅x)ds where ds=sqrt(1+(dy/dx )^2) dxds=√1+(dydx)2dx thus,
S = 2piint_2^3 x sqrt(1+(dy/dx )^2) dx S=2π∫32x√1+(dydx)2dx ===========> (1)
Now
y=(2x-9)^2y=(2x−9)2 and dy/dx = 4(2x-9); (dy/dx )^2=16(4x^2-36x+81 )dydx=4(2x−9);(dydx)2=16(4x2−36x+81)
Insert in (1)
S = 2piint_2^3 x sqrt(1+(16(4x^2-36x+81))) dx S=2π∫32x√1+(16(4x2−36x+81))dx Integrate
Write xx as => 16*1/128(8x-36)+9/2⇒16⋅1128(8x−36)+92 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)]S=π[∫(2x−9)√1+(16(4x2−36x+81))+9∫√1+(16(4x2−36x+81))]=π[S1+S2]
Solving for color(brown)(S_1= int(2x-9)sqrt(1+(16(4x^2-36x+81)))dxS1=∫(2x−9)√1+(16(4x2−36x+81))dx
=color(brown)(int(8x-36)/4sqrt(1+(16(4x^2-36x+81)))dx=∫8x−364√1+(16(4x2−36x+81))dx
let =>u=1+16(4x^2-36x+81); du=16(8x-36)dx ⇒u=1+16(4x2−36x+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
