Question

Find the surface area of the solid of revolution obtained by rotating the curve y=4x^3 from x=1 to x=5 about the x-axis?

Answer 1

`36597.46` units^2

Explanation:

Surface area of revolution is given by, `A` = `2piintf[x]sqrt[1+[f'[x]]^2dx`.

So from the question, `A` = `2piint_1^5[4x^3]sqrt[1+[12x^2]^2dx`
Since `d/dx[4x^3]=12x^2`,[ by the power rule, i.e, `d/dx ax^n=nax^[n-1]`.where `a` is a constant].

We have `A = 2piint_1^5[4x^3]sqrt[1+144x^4 dx].........`[1]
let `u = 1+144x^4.....`[2]`,` then `[du]/[dx] = 4[144]x^3`, ie, `[du]/[4[144x^3]]=dx`

substituting for `dx` and `u` in.........`[1]`,

`A = 2piint_1^5[4x^3]sqrt[udu]/[4[144x^3]`, cancelling the the terms in `x^3` and simplifying will result in,

`A = pi/[72]int_1^5 sqrt[udu]`, and after integrating , `A=pi/108sqrt[u^3]`

substituting back for `x` from .....`[2]` `A=pi/108sqrt[[1+144x^4]^3` and when evaluated for the bounds of `x=1` to `x = 5` will give the result in the answer. [ hopefully this was helpful.]