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?

Answer 1

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

Explanation:

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:

• [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 `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`