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

Mar 28, 2016

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!
$\pi \left(\textcolor{b r o w n}{{S}_{1}} + \textcolor{p u r p \le}{{S}_{1}}\right) = \pi \left\{\textcolor{b r o w n}{{\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{\frac{3}{2}} / 96} + \textcolor{p u r p \le}{\frac{1}{2} \frac{1}{4} \left[{\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{2} \cdot 4 \frac{2 x - 9}{2} + \frac{1}{2} \ln | \sec {\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{2} + \tan 4 \left(2 x - 9\right) |\right]}\right\} \approx 247.65$

#### Explanation:

Given: $f \left(x\right) = {\left(2 x - 9\right)}^{2} , x \in \left[2 , 3\right]$
Required: the surface area obtained by revolving about the x-axis
Solution Strategy: Use the Definition of Surface Area of Revolution
Definition: $S = \int \text{ } \left(2 \pi \cdot x\right) \mathrm{ds}$ where $\mathrm{ds} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$ thus,
$S = 2 \pi {\int}_{2}^{3} x \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$ ===========> (1)
Now
$y = {\left(2 x - 9\right)}^{2}$ and dy/dx = 4(2x-9); (dy/dx )^2=16(4x^2-36x+81 )
Insert in (1)
$S = 2 \pi {\int}_{2}^{3} x \sqrt{1 + \left(16 \left(4 {x}^{2} - 36 x + 81\right)\right)} \mathrm{dx}$ Integrate
Write $x$ as $\implies 16 \cdot \frac{1}{128} \left(8 x - 36\right) + \frac{9}{2}$ and split the integral to:
$S = \pi \left[\int \left(2 x - 9\right) \sqrt{1 + \left(16 \left(4 {x}^{2} - 36 x + 81\right)\right)} + 9 \int \sqrt{1 + \left(16 \left(4 {x}^{2} - 36 x + 81\right)\right)}\right] = \pi \left[\textcolor{b r o w n}{{S}_{1}} + \textcolor{p u r p \le}{{S}_{2}}\right]$
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:
$= \int \sqrt{16 {\left(2 x - 9\right)}^{2} + 1} \mathrm{dx}$
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]
$= \frac{1}{2} \frac{1}{4} \left[\int \sqrt{{\left(\tan v\right)}^{2} + 1} \left({\sec}^{2} \left(v\right) \left(\mathrm{dv}\right)\right)\right]$
$= \frac{1}{2} \frac{1}{4} \left[\int \left(\sec v\right) \left({\sec}^{2} \left(v\right) \left(\mathrm{dv}\right)\right)\right] = \frac{1}{2} \frac{1}{4} \int {\sec}^{3} \left(v\right) \mathrm{dv}$
Apply Integral reduction:
$= \frac{{\sec}^{n - 2} \left(v\right) \cdot \tan v}{n - 1} + \frac{n - 2}{n - 1} \int {\sec}^{n - 2} \left(v\right) \mathrm{dv}$ for $n = 3$
$= \frac{\sec \left(v\right) \cdot \tan \left(v\right)}{2} + \frac{1}{2} \int \sec \left(v\right) \mathrm{dv} = {S}_{2 a} + {S}_{2 b}$
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 $v = \arctan \left(4 u\right)$:
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 *

$\pi \left(\textcolor{b r o w n}{{S}_{1}} + \textcolor{p u r p \le}{{S}_{1}}\right) = \pi \left\{\textcolor{b r o w n}{{\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{\frac{3}{2}} / 96} + \textcolor{p u r p \le}{\frac{1}{2} \frac{1}{4} \left[{\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{2} \cdot 4 \frac{2 x - 9}{2} + \frac{1}{2} \ln | \sec {\left(1 + 16 {\left(2 x - 9\right)}^{2}\right)}^{2} + \tan 4 \left(2 x - 9\right) |\right]}\right\} \approx 247.65$