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

Question

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

Cameron Alexander · Accepted Answer

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

The surface area of the solid generated by rotating #f# around the #x#-axis on #x in[a,b]# is given by:

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

Here, #f(x)=(x-9)^2#, so #f'(x)=2(x-9)#. We're working on the interval #x in[2,3]#. Thus, our surface area is given by:

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

Let #u=x-9#. This will make what we need to do more obvious by making the integrand not so ugly. Note that #du=dx#, so this is an easy substitution. Don't forget to change the bounds by plugging #x=2# and #x=3# into #u=x-9#.

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

Now we should use the trigonometric substitution #u=1/2tantheta#. This was chosen because #1+4u^2=1+tan^2theta=sec^2theta#. This substitution also implies that #du=1/2sec^2thetad theta#. Let's ignore the bounds for now and come back later:

#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 #tan^2theta# using #tan^2theta=sec^2theta-1#:

#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:

Then:

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

Let's revert to the variable #u#, which our definite integral is in. Recall that #tantheta=2u# so #sectheta=sqrt(1+tan^2theta)=sqrt(1+4u^2)#.

#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#

William Abdalla · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = (x-9)^2 $ over the interval $[2,3]$ around the x-axis, you can use the formula for the surface area of a solid of revolution:

$$ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{{dy}}{{dx}}\right)^2} \, dx $$

where $ y = f(x) $, and $ a $ and $ b $ are the limits of integration.

Evaluate $ \frac{{dy}}{{dx}} $, which is the derivative of $ f(x) $. Then, compute $ 2\pi y \sqrt{1 + \left(\frac{{dy}}{{dx}}\right)^2} $ for each $ x $ in the interval $[2,3]$. Finally, integrate this expression from 2 to 3 with respect to $ x $ to find the surface area of the solid.