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

##### 1 Answer

#### Explanation:

The surface area of the solid generated by rotating

#S=2piint_a^bf(x)sqrt(1+(f'(x))^2)dx#

Here,

#S=2piint_2^3(x-9)^2sqrt(1+4(x-9)^2)dx#

Let

#S=2piint_(-7)^(-6)u^2sqrt(1+4u^2)du#

Now we should use the trigonometric substitution

#I=2piintu^2sqrt(1+4u^2)du#

#I=2piint1/4tan^2thetasqrt(1+tan^2theta)(1/2sec^2thetad theta)#

#I=pi/4inttan^2thetasec^3thetad theta#

Rewrite

#I=pi/4int(sec^5theta-sec^3theta)d theta#

These are two known integrals. They can be found using iterative integration by parts and are quite cumbersome. I recommend looking them up or committing them to memory, since they're fairly common in trigonometric substitution questions:

- [Derivation]
#intsec^5thetad theta=1/4sec^3thetatantheta+3/8secthetatantheta+3/8lnabs(sectheta+tantheta)# - [Derivation]
#intsec^3thetad theta=1/2secthetatantheta+1/2lnabs(sectheta+tantheta)#

Then:

#I=pi/4(1/4sec^3thetatantheta-1/8secthetatantheta-1/8lnabs(sectheta+tantheta))#

Let's revert to the variable

#I=pi/16(1+4u^2)^(3/2)(2u)-pi/32sqrt(1+4u^2)(2u)-pi/32lnabs(sqrt(1+4u^2)+2u)#

#I=pi/8u(1+4u^2)^(3/2)-pi/16usqrt(1+4u^2)-pi/32lnabs(sqrt(1+4u^2)+2u)#

Thus:

#S=[pi/8u(1+4u^2)^(3/2)-pi/16usqrt(1+4u^2)-pi/32lnabs(sqrt(1+4u^2)+2u)]_(-7)^(-6)#

Moving through this quickly:

#S=(-3pi)/4(145)^(3/2)+(3pi)/8sqrt145-pi/32lnabs(sqrt145-12)-((-7pi)/8(197)^(3/2)+(7pi)/16sqrt197-pi/32lnabs(sqrt197-14))#

#S=pisqrt145(3/8-3/4(145))+pisqrt197(7/8(197)-7/16)+pi/32ln((sqrt197-14)/(sqrt145-12))#

#S=(-867pisqrt145)/8+(2751pisqrt197)/16+pi/32ln((sqrt197-14)/(sqrt145-12))#

#Sapprox3481.65496968#