Question

What is the surface area of the solid created by revolving `f(x) = (2x-9)^2 , x in [2,3]` around the x axis?

Answer 1

The principle is straight forward but the integral well you have to work a little bit. The key here is to understand the "Surface Area of Revolution". You can always use a computer to crack the integral itself. Good luck!
`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`

Explanation:

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`