Question

What is the surface area of the solid created by revolving `f(x) = x^3 , x in [1,2]` around the x axis?

Answer 1

`S=2piint_1^2x^3sqrt(1+9x^4)dx=pi/27(145^(3/2)-10^(3/2))approx199.4805`

Explanation:

The surface area of a solid created by a function revolved the `x`-axis on `x in[a,b]` is:

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

Here we see that `f(x)=x^3` and `f'(x)=3x^2` so:

`S=2piint_1^2x^3sqrt(1+9x^4)dx`

We can integrate this. Let `u=1+9x^4` so `du=36x^3dx`. Due to this, we need to multiply the interior of the integral by `36` and the exterior by `1//36`.

`S=(2pi)/36int_1^2 36x^3sqrt(1+9x^4)dx`

Before going from `x` to `u`, we will have to transform the bounds. `x=1` becomes `u=1+9(1^4)=10`. `x=2` becomes `u=1+9(2^4)=145`.

`S=pi/18int_10^145sqrtudu`

`S=pi/18int_10^145u^(1/2)du`

`S=pi/18[2/3u^(3/2)]_10^145`

`S=pi/18(2/3(145)^(3/2)-2/3(10)^(3/2))`

`S=pi/27(145^(3/2)-10^(3/2))approx199.4805`