What is the surface area of the solid created by revolving #f(x) = (2x-2)^2 , x in [1,2]# around the x axis?

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

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

We have #f(x)=(2x-2)^2=4(x-1)^2#, with a derivative of #f'(x)=8(x-1)#. So, the surface area in question is:

#A=2piint_1^2 4(x-1)^2sqrt(1+(8(x-1))^2)dx#

#color(white)A=8piint_1^2(x-1)^2sqrt(64(x-1)^2+1)dx#

Let #x-1=1/8tantheta#. This implies that #64(x-1)^2+1=tan^2theta+1=sec^2theta#. Differentiating, it also implies that #dx=1/8sec^2thetad theta#.

Let's momentarily leave out the bounds and return to the bounds once we complete integration.

#A=8piint1/64tan^2thetasqrt(tan^2theta+1)(1/8sec^2thetad theta)#

#color(white)A=pi/64inttan^2thetasec^3thetad theta#

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

#A=pi/64intsec^5thetad theta-pi/64intsec^3thetad theta#

Unfortunately, these integrals are both quite messy, so attached are links to finding both the integrals:

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

Thus, #intsec^5thetad theta-intsec^3thetad theta=1/4sec^3thetatantheta-1/8secthetatantheta-1/8lnabs(sectheta+tantheta)# and

#A=pi/64(1/4sec^3thetatantheta-1/8secthetatantheta-1/8lnabs(sectheta+tantheta))#

#color(white)A=pi/512(2sec^3thetatantheta-secthetatantheta-lnabs(sectheta+tantheta))#

From #x-1=1/8tantheta# we see that #tantheta=8(x-1)#, which implies that #sectheta=sqrt(tan^2theta+1)=sqrt(64(x-1)^2+1)#.

#A=pi/512(2(8(x-1))(64(x-1)^2+1)^(3/2)-8(x-1)sqrt(64(x-1)^2+1)-lnabs(sqrt(64(x-1)^2+1)+8(x-1)))#

Wow. Now apply the bounds from #x=1# to #x=2#. I'm not gonna write out the biggest part since it just takes up too much space:

#A=pi/512(16(65^(3/2))-8sqrt65-lnabs(sqrt65+8))-pi/512(0-0-ln1)#

#color(white)A=pi/512(1032sqrt65-ln(sqrt65+8))#