Question

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

Answer 1

`pi/512(1032sqrt65-ln(sqrt65+8))`

Explanation:

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))`